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the twentieth century. And within the context of these questions, Max Planck discovered something of tremendous importance. Planck’s Revolution The prevailing theory at the time of Max Planck’s discovery was that intensity and frequency were related by the equation This equation, derived from classical physics and using wave phenomena, infers that as wavelength increases, the intensity of energy provided will decrease with an inverse-squared relationship. This relationship is graphed in Figure 21.3 and shows a troubling trend. For starters, it should be apparent that the graph from this equation does not match the blackbody graphs found experimentally. Additionally, it shows that for an object of any temperature, there should be an infinite amount of energy quickly emitted in the shortest wavelengths. When theory and experimental results clash, it is important to re-evaluate both models. The disconnect between theory and reality was termed the ultraviolet catastrophe. 694 Chapter 21 • The Quantum Nature of Light Figure 21.3 The graph above shows the true spectral measurements by a blackbody against those predicted by the classical theory at the time. The discord between the predicted classical theory line and the actual results is known as the ultraviolet catastrophe. Due to concerns over the ultraviolet catastrophe, Max Planck began to question whether another factor impacted the relationship between intensity and wavelength. This factor, he posited, should affect the probability that short wavelength light would be emitted. Should this factor reduce the probability of short wavelength light, it would cause the radiance curve to not progress infinitely as in the classical theory, but would instead cause the curve to precipitate back downward as is shown in the 5,000 K, 4,000 K, and 3,000 K temperature lines of the graph in Figure 21.3. Planck noted that this factor, whatever it may be, must also be dependent on temperature, as the intensity decreases at lower and lower wavelengths as the temperature increases. The determination of this probability factorwas a groundbreaking discovery in physics, yielding insight not just into light but also into energy and matter itself. It would be the basis for Planck’s 1918 Nobel Prize in Physics and would result in the transition of physics from classical to modern understanding. In an attempt to determine the cause of the probability factor,Max Planck constructed a new theory. This theory, which created the branch of physics called quantum mechanics, speculated that the energy radiated by the blackbody could exist only in specific numerical, or quantum, states. This theory is described by the where nis any nonnegative integer (0
, 1, 2, 3, …) and his Planck’s constant, given by equation and fis frequency. Through this equation, Planck’s probability factor can be more clearly understood. Each frequency of light provides a specific quantized amount of energy. Low frequency light, associated with longer wavelengths would provide a smaller amount of energy, while high frequency light, associated with shorter wavelengths, would provide a larger amount of energy. For specified temperatures with specific total energies, it makes sense that more low frequency light would be radiated than high frequency light. To a degree, the relationship is like pouring coins through a funnel. More of the smaller pennies would be able to pass through the funnel than the larger quarters. In other words, because the value of the coin is somewhat related to the size of the coin, the probability of a quarter passing through the funnel is reduced! Furthermore, an increase in temperature would signify the presence of higher energy. As a result, the greater amount of total blackbody energy would allow for more of the high frequency, short wavelength, energies to be radiated. This permits the peak of the blackbody curve to drift leftward as the temperature increases, as it does from the 3,000 K to 4,000 K to 5,000 K values. Furthering our coin analogy, consider a wider funnel. This funnel would permit more quarters to pass through and allow for a reduction in concern about the probability factor. In summary, it is the interplay between the predicted classical model and the quantum probability that creates the curve depicted in Figure 21.3. Just as quarters have a higher currency denomination than pennies, higher frequencies come with larger Access for free at openstax.org. 21.1 • Planck and Quantum Nature of Light 695 amounts of energy. However, just as the probability of a quarter passing through a fixed diameter funnel is reduced, so is the probability of a high frequency light existing in a fixed temperature object. As is often the case in physics, it is the balancing of multiple incredible ideas that finally allows for better understanding. Quantization It may be helpful at this point to further consider the idea of quantum states. Atoms, molecules, and fundamental electron and proton charges are all examples of physical entities that are quantized—that is, they appear only in certain discrete values and do not have every conceivable value. On the macroscopic scale, this is not a revolutionary concept. A standing wave on a string allows only particular harmonics described by integers
. Going up and down a hill using discrete stair steps causes your potential energy to take on discrete values as you move from step to step. Furthermore, we cannot have a fraction of an atom, or part of an electron’s charge, or 14.33 cents. Rather, everything is built of integral multiples of these substructures. That said, to discover quantum states within a phenomenon that science had always considered continuous would certainly be surprising. When Max Planck was able to use quantization to correctly describe the experimentally known shape of the blackbody spectrum, it was the first indication that energy was quantized on a small scale as well. This discovery earned Planck the Nobel Prize in Physics in 1918 and was such a revolutionary departure from classical physics that Planck himself was reluctant to accept his own idea. The general acceptance of Planck’s energy quantization was greatly enhanced by Einstein’s explanation of the photoelectric effect (discussed in the next section), which took energy quantization a step further. Figure 21.4 The German physicist Max Planck had a major influence on the early development of quantum mechanics, being the first to recognize that energy is sometimes quantized. Planck also made important contributions to special relativity and classical physics. (credit: Library of Congress, Prints and Photographs Division, Wikimedia Commons) WORKED EXAMPLE How Many Photons per Second Does a Typical Light Bulb Produce? Assuming that 10 percent of a 100-W light bulb’s energy output is in the visible range (typical for incandescent bulbs) with an average wavelength of 580 nm, calculate the number of visible photons emitted per second. Strategy The number of visible photons per second is directly related to the amount of energy emitted each second, also known as the bulb’s power. By determining the bulb’s power, the energy emitted each second can be found. Since the power is given in watts, which is joules per second, the energy will be in joules. By comparing this to the amount of energy associated with each photon, the number of photons emitted each second can be determined. Solution The power in visible light production is 10.0 percent of 100 W, or 10.0 J/s. The energy of the average visible photon is found by substituting the given average wavelength into the formula By rearranging the above formula to determine energy per photon, this produces 21.1 The number of visible photons per second is thus 696 Chapter 21 •
The Quantum Nature of Light Discussion This incredible number of photons per second is verification that individual photons are insignificant in ordinary human experience. However, it is also a verification of our everyday experience—on the macroscopic scale, photons are so small that quantization becomes essentially continuous. WORKED EXAMPLE How does Photon Energy Change with Various Portions of the EM Spectrum? Refer to the Graphs of Blackbody Radiation shown in the first figure in this section. Compare the energy necessary to radiate one photon of infrared light and one photon of visible light. Strategy To determine the energy radiated, it is necessary to use the equation frequency for infrared light and visible light. It is also necessary to find a representative Solution According to the first figure in this section, one representative wavelength for infrared light is 2000 nm (2.000 × 10-6 m). The associated frequency of an infrared light is Using the equation, the energy associated with one photon of representative infrared light is The same process above can be used to determine the energy associated with one photon of representative visible light. According to the first figure in this section, one representative wavelength for visible light is 500 nm. 21.2 21.3 21.4 21.5 Discussion This example verifies that as the wavelength of light decreases, the quantum energy increases. This explains why a fire burning with a blue flame is considered more dangerous than a fire with a red flame. Each photon of short-wavelength blue light emitted carries a greater amount of energy than a long-wavelength red light. This example also helps explain the differences in the 3,000 K, 4,000 K, and 6,000 K lines shown in the first figure in this section. As the temperature is increased, more energy is available for a greater number of short-wavelength photons to be emitted. Practice Problems 1. An AM radio station broadcasts at a frequency of 1,530 kHz. What is the energy in Joules of a photon emitted from this station? a. b. c. d. 10.1 × 10-26 J 1.01 × 10-28 J 1.01 × 10-29 J 1.01 × 10-27 J 2. A photon travels with energy of 1.0 eV. What type of EM radiation is this photon? a. visible radiation Access for free at openstax.org. 21.1 • Planck and Quantum Nature of Light 697 b. microwave radiation infrared radiation c. d. ultraviolet radiation Check Your Understanding 3. Do reflective
or absorptive surfaces more closely model a perfect blackbody? reflective surfaces a. b. absorptive surfaces 4. A black T-shirt is a good model of a blackbody. However, it is not perfect. What prevents a black T-shirt from being considered a perfect blackbody? a. The T-shirt reflects some light. b. The T-shirt absorbs all incident light. c. The T-shirt re-emits all the incident light. d. The T-shirt does not reflect light. 5. What is the mathematical relationship linking the energy of a photon to its frequency? a. b. c. d. 6. Why do we not notice quantization of photons in everyday experience? a. because the size of each photon is very large b. because the mass of each photon is so small c. because the energy provided by photons is very large d. because the energy provided by photons is very small 7. Two flames are observed on a stove. One is red while the other is blue. Which flame is hotter? a. The red flame is hotter because red light has lower frequency. b. The red flame is hotter because red light has higher frequency. c. The blue flame is hotter because blue light has lower frequency. d. The blue flame is hotter because blue light has higher frequency. 8. Your pupils dilate when visible light intensity is reduced. Does wearing sunglasses that lack UV blockers increase or decrease the UV hazard to your eyes? Explain. Increase, because more high-energy UV photons can enter the eye. a. b. Increase, because less high-energy UV photons can enter the eye. c. Decrease, because more high-energy UV photons can enter the eye. d. Decrease, because less high-energy UV photons can enter the eye. 9. The temperature of a blackbody radiator is increased. What will happen to the most intense wavelength of light emitted as this increase occurs? a. The wavelength of the most intense radiation will vary randomly. b. The wavelength of the most intense radiation will increase. c. The wavelength of the most intense radiation will remain unchanged. d. The wavelength of the most intense radiation will decrease. 698 Chapter 21 • The Quantum Nature of Light 21.2 Einstein and the Photoelectric Effect Section Learning Objectives By the end of this section, you will be able to do the following: • Describe Einstein’s explanation of the photoelectric effect • Describe how the photoelectric effect could not be explained by classical
physics • Calculate the energy of a photoelectron under given conditions • Describe use of the photoelectric effect in biological applications, photoelectric devices and movie soundtracks Section Key Terms electric eye photoelectric effect photoelectron photon The Photoelectric Effect Teacher Support [EL]Ask the students what they think the term photoelectricmeans. How does the term relate to its definition? When light strikes certain materials, it can eject electrons from them. This is called the photoelectric effect, meaning that light (photo) produces electricity. One common use of the photoelectric effect is in light meters, such as those that adjust the automatic iris in various types of cameras. Another use is in solar cells, as you probably have in your calculator or have seen on a rooftop or a roadside sign. These make use of the photoelectric effect to convert light into electricity for running different devices. Figure 21.5 The photoelectric effect can be observed by allowing light to fall on the metal plate in this evacuated tube. Electrons ejected by the light are collected on the collector wire and measured as a current. A retarding voltage between the collector wire and plate can then be adjusted so as to determine the energy of the ejected electrons. (credit: P. P. Urone) Revolutionary Properties of the Photoelectric Effect When Max Planck theorized that energy was quantized in a blackbody radiator, it is unlikely that he would have recognized just how revolutionary his idea was. Using tools similar to the light meter in Figure 21.5, it would take a scientist of Albert Einstein’s stature to fully discover the implications of Max Planck’s radical concept. Through careful observations of the photoelectric effect, Albert Einstein realized that there were several characteristics that could be explained only if EM radiation is itself quantized. While these characteristics will be explained a bit later in this section, you can already begin to appreciate why Einstein’s idea is very important. It means that the apparently continuous stream of energy in an EM wave is actually not a continuous stream at all. In fact, the EM wave itself is actually composed of tiny quantum packets of energy called photons. In equation form, Einstein found the energy of a photon or photoelectron to be where Eis the energy of a photon of frequency fand his Planck’s constant. A beam from a flashlight, which to this point had been considered a wave, instead could now be viewed as a series of photons, each providing a specific amount of energy see Figure 21.6
. Furthermore, the amount of energy within each individual photon is based upon its individual frequency, as Access for free at openstax.org. dictated by frequency-dependent photon energies added together. As a result, the total amount of energy provided by the beam could now be viewed as the sum of all 21.2 • Einstein and the Photoelectric Effect 699 Figure 21.6 An EM wave of frequency fis composed of photons, or individual quanta of EM radiation. The energy of each photon is, where his Planck’s constant and fis the frequency of the EM radiation. Higher intensity means more photons per unit area per second. The flashlight emits large numbers of photons of many different frequencies, hence others have energy, and so on. Just as with Planck’s blackbody radiation, Einstein’s concept of the photon could take hold in the scientific community only if it could succeed where classical physics failed. The photoelectric effect would be a key to demonstrating Einstein’s brilliance. Consider the following five properties of the photoelectric effect. All of these properties are consistent with the idea that individual photons of EM radiation are absorbed by individual electrons in a material, with the electron gaining the photon’s energy. Some of these properties are inconsistent with the idea that EM radiation is a simple wave. For simplicity, let us consider what happens with monochromatic EM radiation in which all photons have the same energy hf. Figure 21.7 Incident radiation strikes a clean metal surface, ejecting multiple electrons from it. The manner in which the frequency and intensity of the incoming radiation affect the ejected electrons strongly suggests that electromagnetic radiation is quantized. This event, called the photoelectric effect, is strong evidence for the existence of photons. 1. If we vary the frequency of the EM radiation falling on a clean metal surface, we find the following: For a given material, there is a threshold frequency f0 for the EM radiation below which no electrons are ejected, regardless of intensity. Using the photon model, the explanation for this is clear. Individual photons interact with individual electrons. Thus if the energy of an individual photon is too low to break an electron away, no electrons will be ejected. However, if EM radiation were a simple wave, sufficient energy could be obtained simply by increasing the intensity. 2. Once EM radiation falls on a material, electrons are ejected without delay. As soon as an individual photon of sufficiently high frequency is absorbed by an individual electron, the electron is ejected. If
the EM radiation were a simple wave, several minutes would be required for sufficient energy to be deposited at the metal surface in order to eject an electron. 3. The number of electrons ejected per unit time is proportional to the intensity of the EM radiation and to no other 4. characteristic. High-intensity EM radiation consists of large numbers of photons per unit area, with all photons having the same characteristic energy, hf. The increased number of photons per unit area results in an increased number of electrons per unit area ejected. If we vary the intensity of the EM radiation and measure the energy of ejected electrons, we find the following: The maximum kinetic energy of ejected electrons is independent of the intensity of the EM radiation. Instead, as noted in point 3 above, increased intensity results in more electrons of the same energy being ejected. If EM radiation were a simple wave, a higher intensity could transfer more energy, and higher-energy electrons would be ejected. 5. The kinetic energy KE of an ejected electron equals the photon energy minus the binding energy BE of the electron in the 700 Chapter 21 • The Quantum Nature of Light specific material. An individual photon can give all of its energy to an electron. The photon’s energy is partly used to break the electron away from the material. The remainder goes into the ejected electron’s kinetic energy. In equation form, this is given by 21.6 is the maximum kinetic energy of the ejected electron, where electron to the particular material. This equation explains the properties of the photoelectric effect quantitatively and demonstrates that BE is the minimum amount of energy necessary to eject an electron. If the energy supplied is less than BE, the electron cannot be ejected. The binding energy can also be written as particular material. Figure 21.8 shows a graph of maximum particular material. versus the frequency of incident EM radiation falling on a is the photon’s energy, and BE is the binding energy of the is the threshold frequency for the where Figure 21.8 A graph of the kinetic energy of an ejected electron, KEe, versus the frequency of EM radiation impinging on a certain material. There is a threshold frequency below which no electrons are ejected, because the individual photon interacting with an individual electron has insufficient energy to break it away. Above the threshold energy, KEe increases linearly with f, consistent with KEe = hf− BE. The slope of this line is h, so the data can be used to determine Planck’s constant experimentally. TIPS FOR SUCCESS
The following five pieces of information can be difficult to follow without some organization. It may be useful to create a table of expected results of each of the five properties, with one column showing the classical wave model result and one column showing the modern photon model result. The table may look something like Table 21.1 Classical Wave Model Modern Photon Model Threshold Frequency Electron Ejection Delay Intensity of EM Radiation Speed of Ejected Electrons Relationship between Kinetic Energy and Binding Energy Table 21.1 Table of Expected Results Virtual Physics Photoelectric Effect Click to view content (http://www.openstax.org/l/28photoelectric) Access for free at openstax.org. In this demonstration, see how light knocks electrons off a metal target, and recreate the experiment that spawned the field of quantum mechanics. 21.2 • Einstein and the Photoelectric Effect 701 GRASP CHECK In the circuit provided, what are the three ways to increase the current? a. decrease the intensity, decrease the frequency, alter the target b. decrease the intensity, decrease the frequency, don’t alter the target increase the intensity, increase the frequency, alter the target c. increase the intensity, increase the frequency, alter the target d. WORKED EXAMPLE Photon Energy and the Photoelectric Effect: A Violet Light (a) What is the energy in joules and electron volts of a photon of 420-nm violet light? (b) What is the maximum kinetic energy of electrons ejected from calcium by 420 nm violet light, given that the binding energy of electrons for calcium metal is 2.71 eV? Strategy To solve part (a), note that the energy of a photon is given by is a straightforward application of to find the ejected electron’s maximum kinetic energy, since BE is given.. For part (b), once the energy of the photon is calculated, it Solution for (a) Photon energy is given by Since we are given the wavelength rather than the frequency, we solve the familiar relationship yielding for the frequency, Combining these two equations gives the useful relationship Now substituting known values yields Converting to eV, the energy of the photon is Solution for (b) Finding the kinetic energy of the ejected electron is now a simple application of the equation the photon energy and binding energy yields 21.7 21.8 21.9 21.10. Substituting 21.11 Discussion The energy of this 420 nm photon of violet light is a tiny fraction of a j
oule, and so it is no wonder that a single photon would be difficult for us to sense directly—humans are more attuned to energies on the order of joules. But looking at the energy in electron volts, we can see that this photon has enough energy to affect atoms and molecules. A DNA molecule can be broken with about 1 eV of energy, for example, and typical atomic and molecular energies are on the order of eV, so that the photon in this example could have biological effects, such as sunburn. The ejected electron has rather low energy, and it would not travel far, 702 Chapter 21 • The Quantum Nature of Light except in a vacuum. The electron would be stopped by a retarding potential of only 0.26 eV, a slightly larger KE than calculated above. In fact, if the photon wavelength were longer and its energy less than 2.71 eV, then the formula would give a negative kinetic energy, an impossibility. This simply means that the 420 nm photons with their 2.96 eV energy are not much above the frequency threshold. You can see for yourself that the threshold wavelength is 458 nm (blue light). This means that if calcium metal were used in a light meter, the meter would be insensitive to wavelengths longer than those of blue light. Such a light meter would be completely insensitive to red light, for example. Practice Problems 10. What is the longest-wavelength EM radiation that can eject a photoelectron from silver, given that the bonding energy is 4.73 eV? Is this radiation in the visible range? a. 2.63 × 10−7 m; No, the radiation is in microwave region. b. 2.63 × 10−7 m; No, the radiation is in visible region. c. 2.63 × 10−7 m; No, the radiation is in infrared region. d. 2.63 × 10-7 m; No, the radiation is in ultraviolet region. 11. What is the maximum kinetic energy in eV of electrons ejected from sodium metal by 450-nm EM radiation, given that the binding energy is 2.28 eV? a. 0.48 V b. 0.82 eV c. 1.21 eV d. 0.48 eV Technological Applications of the Photoelectric Effect While Einstein’s understanding of the photoelectric effect was a transformative discovery in the early 1900s, its presence is ubiquitous today. If you have watched streetlights turn on automatically in response
to the setting sun, stopped elevator doors from closing simply by putting your hands between them, or turned on a water faucet by sliding your hands near it, you are familiar with the electric eye, a name given to a group of devices that use the photoelectric effect for detection. All these devices rely on photoconductive cells. These cells are activated when light is absorbed by a semi-conductive material, knocking off a free electron. When this happens, an electron void is left behind, which attracts a nearby electron. The movement of this electron, and the resultant chain of electron movements, produces a current. If electron ejection continues, further holes are created, thereby increasing the electrical conductivity of the cell. This current can turn switches on and off and activate various familiar mechanisms. One such mechanism takes place where you may not expect it. Next time you are at the movie theater, pay close attention to the sound coming out of the speakers. This sound is actually created using the photoelectric effect! The audiotape in the projector booth is a transparent piece of film of varying width. This film is fed between a photocell and a bright light produced by an exciter lamp. As the transparent portion of the film varies in width, the amount of light that strikes the photocell varies as well. As a result, the current in the photoconductive circuit changes with the width of the filmstrip. This changing current is converted to a changing frequency, which creates the soundtrack commonly heard in the theater. WORK IN PHYSICS Solar Energy Physicist According to the U.S. Department of Energy, Earth receives enough sunlight each hour to power the entire globe for a year. While converting all of this energy is impossible, the job of the solar energy physicist is to explore and improve upon solar energy conversion technologies so that we may harness more of this abundant resource. The field of solar energy is not a new one. For over half a century, satellites and spacecraft have utilized photovoltaic cells to create current and power their operations. As time has gone on, scientists have worked to adapt this process so that it may be used in homes, businesses, and full-scale power stations using solar cells like the one shown in Figure 21.9. Access for free at openstax.org. 21.2 • Einstein and the Photoelectric Effect 703 Figure 21.9 A solar cell is an example of a photovoltaic cell. As light strikes the cell, the cell absorbs the energy of the photons
. If this energy exceeds the binding energy of the electrons, then electrons will be forced to move in the cell, thereby producing a current. This current may be used for a variety of purposes. (credit: U.S. Department of Energy) Solar energy is converted to electrical energy in one of two manners: direct transfer through photovoltaic cells or thermal conversion through the use of a CSP, concentrating solar power, system. Unlike electric eyes, which trip a mechanism when current is lost, photovoltaic cells utilize semiconductors to directly transfer the electrons released through the photoelectric effect into a directed current. The energy from this current can then be converted for storage, or immediately used in an electric process. A CSP system is an indirect method of energy conversion. In this process, light from the Sun is channeled using parabolic mirrors. The light from these mirrors strikes a thermally conductive material, which then heats a pool of water. This water, in turn, is converted to steam, which turns a turbine and creates electricity. While indirect, this method has long been the traditional means of large-scale power generation. There are, of course, limitations to the efficacy of solar power. Cloud cover, nightfall, and incident angle strike at high altitudes are all factors that directly influence the amount of light energy available. Additionally, the creation of photovoltaic cells requires rare-earth minerals that can be difficult to obtain. However, the major role of a solar energy physicist is to find ways to improve the efficiency of the solar energy conversion process. Currently, this is done by experimenting with new semi conductive materials, by refining current energy transfer methods, and by determining new ways of incorporating solar structures into the current power grid. Additionally, many solar physicists are looking into ways to allow for increased solar use in impoverished, more remote locations. Because solar energy conversion does not require a connection to a large-scale power grid, research into thinner, more mobile materials will permit remote cultures to use solar cells to convert sunlight collected during the day into stored energy that can then be used at night. Regardless of the application, solar energy physicists are an important part of the future in responsible energy growth. While a doctoral degree is often necessary for advanced research applications, a bachelor's or master's degree in a related science or engineering field is typically enough to gain access into the industry. Computer skills are very important for energy modeling, including knowledge of CAD software for design purposes. In addition, the ability to collaborate and communicate
with others is critical to becoming a solar energy physicist. GRASP CHECK What role does the photoelectric effect play in the research of a solar energy physicist? a. The understanding of photoelectric effect allows the physicist to understand the generation of light energy when using photovoltaic cells. b. The understanding of photoelectric effect allows the physicist to understand the generation of electrical energy when using photovoltaic cells. c. The understanding of photoelectric effect allows the physicist to understand the generation of electromagnetic energy when using photovoltaic cells. d. The understanding of photoelectric effect allows the physicist to understand the generation of magnetic energy when using photovoltaic cells. 704 Chapter 21 • The Quantum Nature of Light Check Your Understanding 12. How did Einstein’s model of photons change the view of a beam of energy leaving a flashlight? a. A beam of light energy is now considered a continual stream of wave energy, not photons. b. A beam of light energy is now considered a collection of photons, each carrying its own individual energy. 13. True or false—Visible light is the only type of electromagnetic radiation that can cause the photoelectric effect. a. b. false true 14. Is the photoelectric effect a direct consequence of the wave character of EM radiation or the particle character of EM radiation? a. The photoelectric effect is a direct consequence of the particle nature of EM radiation. b. The photoelectric effect is a direct consequence of the wave nature of EM radiation. c. The photoelectric effect is a direct consequence of both the wave and particle nature of EM radiation. d. The photoelectric effect is a direct consequence of neither the wave nor the particle nature of EM radiation. 15. Which aspects of the photoelectric effect can only be explained using photons? a. aspects 1, 2, and 3 b. aspects 1, 2, and 4 c. aspects 1, 2, 4 and 5 d. aspects 1, 2, 3, 4 and 5 16. In a photovoltaic cell, what energy transformation takes place? a. Solar energy transforms into electric energy. b. Solar energy transforms into mechanical energy. c. Solar energy transforms into thermal energy. d. In a photovoltaic cell, thermal energy transforms into electric energy. 17. True or false—A current is created in a photoconductive cell, even if only one electron is expelled from a photon strike. a. b. false true 18. What is a photon and how is it different
from other fundamental particles? a. A photon is a quantum packet of energy; it has infinite mass. b. A photon is a quantum packet of energy; it is massless. c. A photon is a fundamental particle of an atom; it has infinite mass. d. A photon is a fundamental particle of an atom; it is massless. 21.3 The Dual Nature of Light Section Learning Objectives By the end of this section, you will be able to do the following: • Describe the Compton effect • Calculate the momentum of a photon • Explain how photon momentum is used in solar sails • Explain the particle-wave duality of light Section Key Terms Compton effect particle-wave duality photon momentum Photon Momentum Do photons abide by the fundamental properties of physics? Can packets of electromagnetic energy possibly follow the same rules as a ping-pong ball or an electron? Although strange to consider, the answer to both questions is yes. Despite the odd nature of photons, scientists prior to Einstein had long suspected that the fundamental particle of Access for free at openstax.org. electromagnetic radiation shared properties with our more macroscopic particles. This is no clearer than when considering the photoelectric effect, where photons knock electrons out of a substance. While it is strange to think of a massless particle exhibiting momentum, it is now a well-established fact within the scientific community. Figure 21.10 shows macroscopic evidence of photon momentum. 21.3 • The Dual Nature of Light 705 Figure 21.10 The tails of the Hale-Bopp comet point away from the Sun, evidence that light has momentum. Dust emanating from the body of the comet forms this tail. Particles of dust are pushed away from the Sun by light reflecting from them. The blue, ionized gas tail is also produced by photons interacting with atoms in the comet material. (credit: Geoff Chester, U.S. Navy, via Wikimedia Commons) Figure 21.10 shows a comet with two prominent tails. Comet tails are composed of gases and dust evaporated from the body of the comet and ionized gas. What most people do not know about the tails is that they always point awayfrom the Sun rather than trailing behind the comet. This can be seen in the diagram. Why would this be the case? The evidence indicates that the dust particles of the comet are forced away from the Sun when photons strike them. Evidently, photons carry momentum in the direction of their motion away from the Sun, and some of this momentum
is transferred to dust particles in collisions. The blue tail is caused by the solar wind, a stream of plasma consisting primarily of protons and electrons evaporating from the corona of the Sun. Momentum, The Compton Effect, and Solar Sails Momentum is conserved in quantum mechanics, just as it is in relativity and classical physics. Some of the earliest direct experimental evidence of this came from the scattering of X-ray photons by electrons in substances, a phenomenon discovered by American physicist Arthur H. Compton (1892–1962). Around 1923, Compton observed that X-rays reflecting from materials had decreased energy and correctly interpreted this as being due to the scattering of the X-ray photons by electrons. This phenomenon could be handled as a collision between two particles—a photon and an electron at rest in the material. After careful observation, it was found that both energy and momentum were conserved in the collision. See Figure 21.11. For the discovery of this conserved scattering, now known as the Compton effect, Arthur Compton was awarded the Nobel Prize in 1929. Shortly after the discovery of Compton scattering, the value of the photon momentum, was determined by Louis de Broglie. In this equation, called the de Broglie relation, hrepresents Planck’s constant and λis the photon wavelength. Figure 21.11 The Compton effect is the name given to the scattering of a photon by an electron. Energy and momentum are conserved, resulting in a reduction of both for the scattered photon. 706 Chapter 21 • The Quantum Nature of Light We can see that photon momentum is small, since observe photon momentum. Our mirrors do not recoil when light reflects from them, except perhaps in cartoons. Compton saw the effects of photon momentum because he was observing X-rays, which have a small wavelength and a relatively large momentum, interacting with the lightest of particles, the electron. and his very small. It is for this reason that we do not ordinarily WORKED EXAMPLE Electron and Photon Momentum Compared (a) Calculate the momentum of a visible photon that has a wavelength of 500 nm. (b) Find the velocity of an electron having the same momentum. (c) What is the energy of the electron, and how does it compare with the energy of the photon? Strategy Finding the photon momentum is a straightforward application of its definition: small, we can assume that an electron with the same momentum will be nonrelativistic, making it easy to find
its velocity and kinetic energy from the classical formulas. If we find the photon momentum is Solution for (a) Photon momentum is given by the de Broglie relation. Entering the given photon wavelength yields 21.12 21.13 Solution for (b) Since this momentum is indeed small, we will use the classical expression momentum. Solving for vand using the known value for the mass of an electron gives to find the velocity of an electron with this Solution for (c) The electron has kinetic energy, which is classically given by Thus, Converting this to eV by multiplying by yields The photon energy Eis 21.14 21.15 21.16 21.17 21.18 which is about five orders of magnitude greater. Discussion Even in huge numbers, the total momentum that photons carry is small. An electron that carries the same momentum as a 500-nm photon will have a 1,460 m/s velocity, which is clearly nonrelativistic. This is borne out by the experimental observation that it takes far less energy to give an electron the same momentum as a photon. That said, for high-energy photons interacting with small masses, photon momentum may be significant. Even on a large scale, photon momentum can have an effect if there Access for free at openstax.org. 21.3 • The Dual Nature of Light 707 are enough of them and if there is nothing to prevent the slow recoil of matter. Comet tails are one example, but there are also proposals to build space sails that use huge low-mass mirrors (made of aluminized Mylar) to reflect sunlight. In the vacuum of space, the mirrors would gradually recoil and could actually accelerate spacecraft within the solar system. See the following figure. TIPS FOR SUCCESS When determining energies in particle physics, it is more sensible to use the unit eV instead of Joules. Using eV will help you to recognize differences in magnitude more easily and will make calculations simpler. Also, eV is used by scientists to describe the binding energy of particles and their rest mass, so using eV will eliminate the need to convert energy quantities. Finally, eV is a convenient unit when linking electromagnetic forces to particle physics, as one eV is the amount energy given to an electron when placed in a field of 1-V potential difference. Practice Problems 19. Find the momentum of a 4.00-cm wavelength microwave photon. a. 0.83 × 10−32 kg ⋅ m/s 1
.66 × 10−34 kg ⋅ m/s b. c. 0.83 × 10−34 kg ⋅ m/s 1.66 × 10-32 kg ⋅ m/s d. 20. Calculate the wavelength of a photon that has the same momentum of a proton moving at 1.00 percent of the speed of light. a. 2.43 × 10−10 m b. 2.43 × 10−12 m 1.32 × 10−15 m c. 1.32 × 10−13 m d. Figure 21.12 (a) Space sails have been proposed that use the momentum of sunlight reflecting from gigantic low-mass sails to propel spacecraft about the solar system. A Russian test model of this (the Cosmos 1) was launched in 2005, but did not make it into orbit due to a rocket failure. (b) A U.S. version of this, labeled LightSail-1, is scheduled for trial launches in 2016. It will have a 40 m2 sail. (credit: Kim Newton/NASA) LINKS TO PHYSICS LightSail-1 Project “Provide ships or sails adapted to the heavenly breezes, and there will be some who will brave even that void.” — Johannes Kepler (in a letter to Galileo Galilei in 1608) 708 Chapter 21 • The Quantum Nature of Light Figure 21.13 NASA’s NanoSail-D, a precursor to LightSail-1, with its sails deployed. The Planetary Society will be launching LightSail-1 in early 2016. (credit: NASA/MSFC/D, Wikimedia Commons) Traversing the Solar System using nothing but the Sun’s power has long been a fantasy of scientists and science fiction writers alike. Though physicists like Compton, Einstein, and Planck all provided evidence of light’s propulsive capacity, it is only recently that the technology has become available to truly put these visions into motion. In 2016, by sending a lightweight satellite into space, the LightSail-1 project is designed to do just that. A citizen-funded project headed by the Planetary Society, the 5.45-million-dollar LightSail-1 project is set to launch two crafts into orbit around the Earth. Each craft is equipped with a 32-square-meter solar sail prepared to unfurl once a rocket has launched it to an appropriate altitude. The sails are made of large mirrors, each
a quarter of the thickness of a trash bag, which will receive an impulse from the Sun’s reflecting photons. Each time the Sun’s photon strikes the craft’s reflective surface and bounces off, it will provide a momentum to the sail much greater than if the photon were simply absorbed. Attached to three tiny satellites called CubeSats, whose combined volume is no larger than a loaf of bread, the received momentum from the Sun’s photons should be enough to record a substantial increase in orbital speed. The intent of the LightSail-1 mission is to prove that the technology behind photon momentum travel is sound and can be done cheaply. A test flight in May 2015 showed that the craft’s Mylar sails could unfurl on command. With another successful result in 2016, the Planetary Society will be planning future versions of the craft with the hopes of eventually achieving interplanetary satellite travel. Though a few centuries premature, Kepler’s fantastic vision may not be that far away. If eventually set into interplanetary launch, what will be the effect of continual photon bombardment on the motion of a craft similar to LightSail-1? a. b. c. d. It will result in continual acceleration of the craft. It will first accelerate and then decelerate the craft. It will first decelerate and then accelerate the craft. It will result in the craft moving at constant velocity. Particle-Wave Duality We have long known that EM radiation is like a wave, capable of interference and diffraction. We now see that light can also be modeled as particles—massless photons of discrete energy and momentum. We call this twofold nature the particle-wave duality, meaning that EM radiation has properties of both particles and waves. This may seem contradictory, since we ordinarily deal with large objects that never act like both waves and particles. An ocean wave, for example, looks nothing like a grain of sand. However, this so-called duality is simply a term for properties of the photon analogous to phenomena we can observe directly, on a macroscopic scale. See Figure 21.14. If this term seems strange, it is because we do not ordinarily observe details on the quantum level directly, and our observations yield either particle-like orwave-like properties, but never both simultaneously. Access for free at openstax.org. 21.3 • The Dual Nature of Light 709 Figure 21.14 (a) The interference pattern for light through a
double slit is a wave property understood by analogy to water waves. (b) The properties of photons having quantized energy and momentum and acting as a concentrated unit are understood by analogy to macroscopic particles. Since we have a particle-wave duality for photons, and since we have seen connections between photons and matter in that both have momentum, it is reasonable to ask whether there is a particle-wave duality for matter as well. If the EM radiation we once thought to be a pure wave has particle properties, is it possible that matter has wave properties? The answer, strangely, is yes. The consequences of this are tremendous, as particle-wave duality has been a constant source of scientific wonder during the twentieth and twenty-first centuries. Check Your Understanding 21. What fundamental physics properties were found to be conserved in Compton scattering? a. energy and wavelength b. energy and momentum c. mass and energy d. energy and angle 22. Why do classical or relativistic momentum equations not work in explaining the conservation of momentum that occurs in Compton scattering? a. because neither classical nor relativistic momentum equations utilize mass as a variable in their equations b. because relativistic momentum equations utilize mass as a variable in their formulas but classical momentum equations do not c. because classical momentum equations utilize mass as a variable in their formulas but relativistic momentum equations do not d. because both classical and relativistic momentum equations utilize mass as a variable in their formulas 23. If solar sails were constructed with more massive materials, how would this influence their effectiveness? a. The effect of the momentum would increase due to the decreased inertia of the sails. b. The effect of the momentum would reduce due to the decreased inertia of the sails. c. The effect of the momentum would increase due to the increased inertia of the sails. d. The effect of the momentum would be reduced due to the increased inertia of the sails. 24. True or false—It is possible to propel a solar sail craft using just particles within the solar wind. a. b. true false 25. True or false—Photon momentum more directly supports the wave model of light. a. b. false true 710 Chapter 21 • The Quantum Nature of Light 26. True or false—wave-particle duality exists for objects on the macroscopic scale. a. b. false true 27. What type of electromagnetic radiation was used in Compton scattering? a. visible light b. ultraviolet radiation c. d. X-rays
radio waves Access for free at openstax.org. Chapter 21 • Key Terms 711 KEY TERMS blackbody object that absorbs all radiated energy that strikes it and also emits energy across all wavelengths of the electromagnetic spectrum material by a photon of light photon a quantum, or particle, of electromagnetic radiation Compton effect phenomenon whereby X-rays scattered photon momentum amount of momentum of a photon, from materials have decreased energy calculated by electric eye group of devices that use the photoelectric effect for detection particle-wave duality property of behaving like either a particle or a wave; the term for the phenomenon that all particles have wave-like characteristics and waves have particle-like characteristics photoelectric effect phenomenon whereby some materials eject electrons when exposed to light photoelectron electron that has been ejected from a SECTION SUMMARY 21.1 Planck and Quantum Nature of Light • A blackbody will radiate energy across all wavelengths of the electromagnetic spectrum. • Radiation of a blackbody will peak at a particular wavelength, dependent on the temperature of the blackbody. • Analysis of blackbody radiation led to the field of quantum mechanics, which states that radiated energy can only exist in discrete quantum states. 21.2 Einstein and the Photoelectric Effect • The photoelectric effect is the process in which EM radiation ejects electrons from a material. • Einstein proposed photons to be quanta of EM where fis the radiation having energy frequency of the radiation. • All EM radiation is composed of photons. As Einstein KEY EQUATIONS 21.1 Planck and Quantum Nature of Light quantum energy quantized the fact that certain physical entities exist only with particular discrete values and not every conceivable value quantum discrete packet or bundle of a physical entity such as energy ultraviolet catastrophe misconception that blackbodies would radiate high frequency energy at a much higher rate than energy radiated at lower frequencies explained, all characteristics of the photoelectric effect are due to the interaction of individual photons with individual electrons. • The maximum kinetic energy KEeof ejected electrons (photoelectrons) is given by is the photon energy and BE is the binding energy (or work function) of the electron in the particular material. where hf 21.3 The Dual Nature of Light • Compton scattering provided evidence that photon- electron interactions abide by the principles of conservation of momentum and conservation of energy. • The momentum of individual photons, quantified by, can be used to explain observations of comets and may lead to future space technologies. • Electromagnetic waves and matter have both wave-like and particle-like properties.
This phenomenon is defined as particle-wave duality. maximum kinetic energy of a photoelectron binding energy of an electron 21.2 Einstein and the Photoelectric Effect 21.3 The Dual Nature of Light energy of a photon momentum of a photon (deBroglie relation) 712 Chapter 21 • Chapter Review CHAPTER REVIEW Concept Items 21.1 Planck and Quantum Nature of Light 1. What aspect of the blackbody spectrum forced Planck to propose quantization of energy levels in atoms and molecules? a. Radiation occurs at a particular frequency that does not change with the energy supplied. b. Certain radiation occurs at a particular frequency that changes with the energy supplied. c. Maximum radiation would occur at a particular frequency that does not change with the energy supplied. d. Maximum radiation would occur at a particular frequency that changes with the energy supplied. 2. Two lasers shine red light at 650 nm. One laser is twice as bright as the other. Explain this difference using photons and photon energy. a. The brighter laser emits twice the number of photons and more energy per photon. b. The brighter laser emits twice the number of photons and less energy per photon. c. Both lasers emit equal numbers of photons and equivalent amounts of energy per photon. d. The brighter laser emits twice the number of photons but both lasers emit equivalent amounts of energy per photon. 3. Consider four stars in the night sky: red, yellow, orange, and blue. The photons of which star will carry the greatest amount of energy? a. blue b. orange c. red d. yellow 4. A lightbulb is wired to a variable resistor. What will happen to the color spectrum emitted by the bulb as the resistance of the circuit is increased? a. The bulb will emit greener light. b. The bulb will emit bluer light. c. The bulb will emit more ultraviolet light. d. The bulb will emit redder light. 21.2 Einstein and the Photoelectric Effect 5. Light is projected onto a semi-conductive surface. However, no electrons are ejected. What will happen when the light intensity is increased? a. An increase in light intensity decreases the number of photons. However, no electrons are ejected. Access for free at openstax.org. b. Increase in light intensity increases the number of photons, so electrons with higher kinetic energy are ejected. c. An increase in light intensity increases the number of photons, so electrons will be ejected. d. An increase in light intensity increases the number
of photons. However, no electrons are ejected. 6. True or false—The concept of a work function (or binding energy) is permissible under the classical wave model. a. b. false true 7. Can a single microwave photon cause cell damage? a. No, there is not enough energy associated with a single microwave photon to result in cell damage. b. No, there is zero energy associated with a single microwave photon, so it does not result in cell damage. c. Yes, a single microwave photon causes cell damage because it does not have high energy. d. Yes, a single microwave photon causes cell damage because it has enough energy. 21.3 The Dual Nature of Light 8. Why don’t we feel the momentum of sunlight when we are on the beach? a. The momentum of a singular photon is incredibly small. b. The momentum is not felt because very few photons strike us at any time, and not all have momentum. c. The momentum of a singular photon is large, but very few photons strike us at any time. d. A large number of photons strike us at any time, and so their combined momentum is incredibly large. 9. If a beam of helium atoms is projected through two slits and onto a screen, will an interference pattern emerge? a. No, an interference pattern will not emerge because helium atoms will strike a variety of locations on the screen. b. No, an interference pattern will not emerge because helium atoms will strike at certain locations on the screen. c. Yes, an interference pattern will emerge because helium atoms will strike a variety of locations on the screen. d. Yes, an interference pattern will emerge because helium atoms will strike at certain locations on the screen. Chapter 21 • Chapter Review 713 Critical Thinking Items 21.1 Planck and Quantum Nature of Light 13. Why is it assumed that a perfect absorber of light (like a blackbody) must also be a perfect emitter of light? a. To achieve electrostatic equilibrium with its 10. Explain why the frequency of a blackbody does not surroundings double when the temperature is doubled. a. Frequency is inversely proportional to temperature. b. Frequency is directly proportional to temperature. c. Frequency is directly proportional to the square of temperature. b. To achieve thermal equilibrium with its surroundings c. To achieve mechanical equilibrium with its surroundings d. To achieve chemical equilibrium with its d. Frequency is directly proportional to the fourth surroundings power of temperature. 11. Why does the intensity shown in the black
body radiation graph decrease after its peak frequency is achieved? a. Because after reaching the peak frequency, the photons created at a particular frequency are too many for energy intensity to continue to decrease. 21.2 Einstein and the Photoelectric Effect 14. Light is projected onto a semi-conductive surface. If the intensity is held constant but the frequency of light is increased, what will happen? a. As frequency is increased, electrons will stop being ejected from the surface. b. As frequency is increased, electrons will begin to be ejected from the surface. c. As frequency is increased, it will have no effect on the electrons being ejected as the intensity is the same. d. As frequency is increased, the rate at which the electrons are being ejected will increase. 15. Why is it important to consider what material to use when designing a light meter? Consider the worked example from Section 21-2 for assistance. a. A light meter should contain material that responds only to high frequency light. b. A light meter should contain material that responds b. Because after reaching the peak frequency, the to low frequency light. photons created at a particular frequency are too few for energy intensity to continue to decrease. c. A light meter should contain material that has high binding energy. c. Because after reaching the peak frequency, the d. A light meter should contain a material that does photons created at a particular frequency are too many for energy intensity to continue to increase. d. Because after reaching the peak frequency, the photons created at a particular frequency are too few for energy intensity to continue to increase. not show any photoelectric effect. 16. Why does overexposure to UV light often result in sunburn when overexposure to visible light does not? This is why you can get burnt even on a cloudy day. a. UV light carries less energy than visible light and 12. Shortly after the introduction of photography, it was can penetrate our body. found that photographic emulsions were more sensitive to blue and violet light than they were to red light. Explain why this was the case. a. Blue-violet light contains greater amount of energy than red light. b. UV light carries more energy than visible light, so it cannot break bonds at the cellular level. c. UV light carries more energy than visible light and can break bonds at the cellular level. d. UV light carries less energy than visible light and b. Blue-violet light contains lower amount of energy cannot penetrate the human body. than red light
. c. Both blue-violet light and red light have the same frequency but contain different amounts of energy. d. Blue-violet light frequency is lower than the frequency of red light. 17. If you pick up and shake a piece of metal that has electrons in it free to move as a current, no electrons fall out. Yet if you heat the metal, electrons can be boiled off. Explain both of these facts as they relate to the amount and distribution of energy involved with shaking the 714 Chapter 21 • Chapter Review object as compared with heating it. a. Thermal energy is added to the metal at a much higher rate than energy added due to shaking. b. Thermal energy is added to the metal at a much lower rate than energy added due to shaking. If the thermal energy added is below the binding energy of the electrons, they may be boiled off. If the mechanical energy added is below the binding energy of the electrons, they may be boiled off. d. c. 21.3 The Dual Nature of Light 18. In many macroscopic collisions, a significant amount of kinetic energy is converted to thermal energy. Explain why this is not a concern for Compton scattering. a. Because, photons and electrons do not exist on the molecular level, all energy of motion is considered kinetic energy. b. Because, photons exist on the molecular level while electrons do not exist on the molecular level, all energy of motion is considered kinetic energy. c. Because, electrons exist on the molecular level while photons do not exist on the molecular level, all energy of motion is considered kinetic energy. d. Because, photons and electrons exist on the molecular level, all energy of motion is considered kinetic energy. Problems 21.1 Planck and Quantum Nature of Light 22. How many X-ray photons per second are created by an X-ray tube that produces a flux of X-rays having a power of 1.00 W? Assume the average energy per photon is 75.0 keV. a. 8.33 × 1015 photons b. 9.1 × 107 photons c. 9.1 × 108 photons d. 8.33 × 1013 photons 23. What is the frequency of a photon produced in a CRT using a 25.0-kV accelerating potential? This is similar to the layout as in older color television sets. a. 6.04 × 10−48 Hz b. 2.77 × 10−48 Hz 3.02 × 1018 Hz c. d. 6.04
× 1018 Hz 21.2 Einstein and the Photoelectric Effect 24. What is the binding energy in eV of electrons in magnesium, if the longest-wavelength photon that can eject electrons is 337 nm? Access for free at openstax.org. 19. In what region of the electromagnetic spectrum will photons be most effective in accelerating a solar sail? a. ultraviolet rays b. infrared rays c. X-rays d. gamma rays 20. True or false—Electron microscopes can resolve images that are smaller than the images resolved by light microscopes. false a. true b. 21. How would observations of Compton scattering change if ultraviolet light were used in place of X-rays? a. Ultraviolet light carries less energy than X-rays. As a result, Compton scattering would be easier to detect. b. Ultraviolet light carries less energy than X-rays. As a result, Compton scattering would be more difficult to detect. c. Ultraviolet light carries more energy than X-rays. As a result, Compton scattering would be easier to detect. d. Ultraviolet light has higher energy than X-rays. As a result, Compton scattering would be more difficult to detect. a. b. c. d. 7.44 × 10−19 J 7.44 × 10−49 J 5.90 × 10−17 J 5.90 × 10−19 J 25. Photoelectrons from a material with a binding energy of 2.71 eV are ejected by 420-nm photons. Once ejected, how long does it take these electrons to travel 2.50 cm to a detection device? a. 8.5 × 10−6 s 3.5 × 10−7 s b. c. 43.5 × 10−9 s d. 8.5 × 10−8 s 21.3 The Dual Nature of Light 26. What is the momentum of a 0.0100-nm-wavelength photon that could detect details of an atom? a. 6.626 × 10−27 kg ⋅ m/s b. 6.626 × 10−32 kg ⋅ m/s c. 6.626 × 10−34 kg ⋅ m/s d. 6.626 × 10-23 kg ⋅ m/s 27. The momentum of light is exactly reversed when reflected straight back from a mirror, assuming negligible recoil of the mirror. Thus the change in Chapter 21 • Test Prep 715 momentum is twice the initial photon momentum.
Suppose light of intensity 1.00 kW/m2 reflectsfrom a mirror of area 2.00 m2 each second. Using the most general form of Newton’s second law, what is the force on the mirror? a. b. c. d. 1.33 × 10-5 N 1.33 × 10−6 N 1.33 × 10−7 N 1.33 × 10−8 N Performance Task 21.3 The Dual Nature of Light 28. Our scientific understanding of light has changed over time. There is evidence to support the wave model of light, just as there is evidence to support the particle model of light. 1. Construct a demonstration that supports the wave model of light. Note—One possible method is to use a piece of aluminum foil, razor blade, and laser to demonstrate wave interference. Can you arrange these materials to create an effective demonstration? In writing, explain how evidence TEST PREP Multiple Choice 21.1 Planck and Quantum Nature of Light 29. A perfect blackbody is a perfect absorber of energy transferred by what method? a. b. c. d. conduction convection induction radiation 30. Which of the following is a physical entity that is quantized? a. electric charge of an ion frequency of a sound b. speed of a car c. 31. Find the energy in joules of photons of radio waves that leave an FM station that has a 90.0-MHz broadcast frequency. a. b. c. d. 1.8 × 10−25 J 1.11 × 10−25 J 7.1 × 10−43 J 5.96 × 10-26 J 32. Which region of the electromagnetic spectrum will provide photons of the least energy? infrared light a. b. radio waves c. ultraviolet light d. X-rays 33. A hot, black coffee mug is sitting on a kitchen table in a dark room. Because it cannot be seen, one assumes that from your demonstration supports the wave model of light. 2. Construct a demonstration that supports the particle model of light. Note—One possible method is to use a negatively charged electroscope, zinc plate, and three light sources of different frequencies. A red laser, a desk lamp, and ultraviolet lamp are typically used. Can you arrange these materials to demonstrate the photoelectric effect? In writing, explain how evidence from your demonstration supports the particle model of light. it is not emitting energy in the form of light. Explain the fallacy in this logic. a. Not all
heat is in the form of light energy. b. Not all light energy falls in the visible portion of the electromagnetic spectrum. c. All heat is in the form of light energy. d. All light energy falls in the visible portion of the electromagnetic spectrum. 34. Given two stars of equivalent size, which will have a greater temperature: a red dwarf or a yellow dwarf? Explain. Note—Our sun is considered a yellow dwarf. a. a yellow dwarf, because yellow light has lower frequency b. a red dwarf, because red light has lower frequency c. a red dwarf, because red light has higher frequency d. a yellow dwarf, because yellow light has higher frequency 21.2 Einstein and the Photoelectric Effect 35. What is a quantum of light called? a. electron b. neutron c. photon d. proton 36. Which of the following observations from the photoelectric effect is not a violation of classical physics? a. Electrons are ejected immediately after impact from light. b. Light can eject electrons from a semi-conductive 716 Chapter 21 • Test Prep material. protons from a surface. c. Light intensity does not influence the kinetic d. UV, X-rays, and gamma rays are capable of ejecting energy of ejected electrons. electrons from a surface. d. No electrons are emitted if the light frequency is too low. 21.3 The Dual Nature of Light 37. If of energy is supplied to an electron with a 41. What two particles interact in Compton scattering? binding energy of the electron be launched? a. b. c. d., with what kinetic energy will a. photon and electron b. proton and electron c. neutron and electron d. proton and neutron 38. Which of the following terms translates to light- producing voltage? a. photoelectric b. quantum mechanics c. photoconductive d. photovoltaic 39. Why is high frequency EM radiation considered more dangerous than long wavelength EM radiation? a. Long wavelength EM radiation photons carry less energy and therefore have greater ability to disrupt materials through the photoelectric effect. b. Long wavelength EM radiation photons carry more energy and therefore have greater ability to disrupt materials through the photoelectric effect. c. High frequency EM radiation photons carry less energy and therefore have lower ability to disrupt materials through the photoelectric effect. d. High frequency EM radiation photons carry more energy and therefore have greater ability to disrupt materials through the photoelectric effect. 40. Why are UV, X-rays, and gamma rays considered
ionizing radiation? a. UV, X-rays, and gamma rays are capable of ejecting photons from a surface. b. UV, X-rays, and gamma rays are capable of ejecting neutrons from a surface. c. UV, X-rays, and gamma rays are capable of ejecting Short Answer 21.1 Planck and Quantum Nature of Light 42. What is the momentum of a 500-nm photon? a. 8.35 × 10−26 kg ⋅ m/s 3.31 × 10−40 kg ⋅ m/s b. 7.55 × 1026 kg ⋅ m/s c. 1.33 × 10-27 kg ⋅ m/s d. 43. The conservation of what fundamental physics principle charge is behind the technology of solar sails? a. b. mass c. momentum d. angular momentum 44. Terms like frequency, amplitude, and period are tied to what component of wave-particle duality? a. neither the particle nor the wave model of light b. both the particle and wave models of light c. d. the particle model of light the wave model of light 45. Why was it beneficial for Compton to scatter electrons using X-rays and not another region of light like microwaves? a. because X-rays are more penetrating than microwaves b. because X-rays have lower frequency than microwaves c. because microwaves have shorter wavelengths than X-rays d. because X-rays have shorter wavelength than microwaves elliptical path. c. The blackbody radiation curve would look like a vertical line. 46. Scientists once assumed that all frequencies of light d. The blackbody radiation curve would look like a were emitted with equal probability. Explain what the blackbody radiation curve would look like if this were the case. a. The blackbody radiation curve would look like a circular path. b. The blackbody radiation curve would look like an horizontal line. 47. Because there are more gradations to high frequency radiation than low frequency radiation, scientists also thought it possible that a curve titled the ultraviolet catastrophewould occur. Explain what the blackbody radiation curve would look like if this were the case. Access for free at openstax.org. a. The curve would steadily increase in intensity with increasing frequency. b. The curve would steadily decrease in intensity with increasing frequency. c. The curve would be much steeper than in the blackbody radiation graph. d. The curve would be much
flatter than in the blackbody radiation graph. 48. Energy provided by a light exists in the following quantities: 150 J, 225 J, 300 J. Define one possible quantum of energy and provide an energy state that cannot exist with this quantum. a. 65 J; 450 J cannot exist 70 J; 450 J cannot exist b. 75 J; 375 J cannot exist c. 75 J; 100 J cannot exist d. 49. Why is Planck’s recognition of quantum particles considered the dividing line between classical and modern physics? a. Planck recognized that energy is quantized, which was in sync with the classical physics concepts but not in agreement with modern physics concepts. b. Planck recognized that energy is quantized, which was in sync with modern physics concepts but not in agreement with classical physics concepts. Chapter 21 • Test Prep 717 a. b. c. d. radio, microwave, infrared, visible, ultraviolet, Xray, gamma radio, infrared, microwave, ultraviolet, visible, Xray, gamma radio, visible, microwave, infrared, ultraviolet, Xray, gamma radio, microwave, infrared, visible, ultraviolet, gamma, X-ray 53. Why are photons of gamma rays and X-rays able to penetrate objects more successfully than ultraviolet radiation? a. Photons of gamma rays and X-rays carry with them less energy. b. Photons of gamma rays and X-rays have longer wavelengths. c. Photons of gamma rays and X-rays have lower frequencies. d. Photons of gamma rays and X-rays carry with them more energy. 21.2 Einstein and the Photoelectric Effect 54. According to wave theory, what is necessary to eject electrons from a surface? a. Enough energy to overcome the binding energy of the electrons at the surface c. Prior to Planck’s hypothesis, all the classical b. A frequency that is higher than that of the electrons physics calculations were valid for subatomic particles, but quantum physics calculations were not valid. d. Prior to Planck’s hypothesis, all the classical physics calculations were not valid for macroscopic particles, but quantum physics calculations were valid. 50. How many 500-mm microwave photons are needed to supply the 8 kJ of energy necessary to heat a cup of water by 10 degrees Celsius? a. 8.05 × 1028 photons b. 8.05 × 1026 photons c. 2.01 × 1026 photons d. 2.01 × 1028 photons 51. What
is the efficiency of a 100-W, 550-nm lightbulb if a photometer finds that 1 × 1020 photons are emitted each second? a. b. c. d. 101 percent 72 percent 18 percent 36 percent 52. Rank the following regions of the electromagnetic spectrum by the amount of energy provided per photon: gamma, infrared, microwave, ultraviolet, radio, visible, X-ray. at the surface c. Energy that is lower than the binding energy of the electrons at the surface d. A very small number of photons 55. What is the wavelength of EM radiation that ejects 2.00-eV electrons from calcium metal, given that the binding energy is 2.71 eV? 16.1 × 105 m a. b. 6.21 × 10−5 m c. 9.94 × 10−26 m d. 2.63 × 10-7 m 56. Find the wavelength of photons that eject. electrons from potassium, given that the binding energy is a. b. c. d. 57. How do solar cells utilize the photoelectric effect? a. A solar cell converts all photons that it absorbs to electrical energy using the photoelectric effect. b. A solar cell converts all electrons that it absorbs to electrical energy using the photoelectric effect. c. A solar cell absorbs the photons with energy less 718 Chapter 21 • Test Prep than the energy gap of the material of the solar cell and converts it to electrical energy using the photoelectric effect. d. A solar cell absorbs the photons with energy greater than the energy gap of the material of the solar cell and converts it to electrical energy using the photoelectric effect. 58. Explain the advantages of the photoelectric effect to other forms of energy transformation. a. The photoelectric effect is able to work on the Sun’s natural energy. b. The photoelectric effect is able to work on energy generated by burning fossil fuels. c. The photoelectric effect can convert heat energy into electrical energy. d. The photoelectric effect can convert electrical energy into light energy. 21.3 The Dual Nature of Light 5.18 × 105 m/s c. d. 4.18 × 105 m/s 63. When a photon strikes a solar sail, what is the direction of impulse on the photon? a. parallel to the sail b. perpendicular to the sail c. tangential to the sail d. opposite to the sail 64. What is a fundamental difference between solar sails and sails that are used on sailboats?
a. Solar sails rely on disorganized strikes from light particles, while sailboats rely on disorganized strikes from air particles. b. Solar sails rely on disorganized strikes from air particles, while sailboats rely on disorganized strikes from light particles. c. Solar sails rely on organized strikes from air particles, while sailboats rely on organized strikes from light particles. 59. Upon collision, what happens to the frequency of a d. Solar sails rely on organized strikes from light photon? a. The frequency of the photon will drop to zero. b. The frequency of the photon will remain the same. c. The frequency of the photon will increase. d. The frequency of the photon will decrease. particles, while sailboats rely on organized strikes from air particles. 65. The wavelength of a particle is called the de Broglie wavelength, and it can be found with the equation. 60. How does the momentum of a photon compare to the momentum of an electron of identical energy? a. Momentum of the photon is greater than the momentum of an electron. b. Momentum of the photon is less than the momentum of an electron. c. Momentum of the photon is equal to the momentum of an electron. d. Momentum of the photon is zero due to zero rest mass but the momentum of an electron is finite. 61. A 500-nm photon strikes an electron and loses 20 percent of its energy. What is the new momentum of the photon? a. 4.24 × 10−27 kg ⋅ m/s 3.18 × 10−27 kg ⋅ m/s b. c. 2.12 × 10−27 kg ⋅ m/s 1.06 × 10−27 kg ⋅ m/s d. 62. A 500-nm photon strikes an electron and loses 20 percent of its energy. What is the speed of the recoiling electron? 7.18 × 105 m/s a. b. 6.18 × 105 m/s Yes or no—Can the wavelength of an electron match that of a proton? a. Yes, a slow-moving electron can achieve the same momentum as a slow-moving proton. b. No, a fast-moving electron cannot achieve the same momentum, and hence the same wavelength, as a proton. c. No, an electron can achieve the same momentum, and hence not the same wavelength, as a proton. d. Yes, a fast-moving electron can achieve the
same momentum, and hence have the same wavelength, as a slow-moving proton. 66. Large objects can move with great momentum. Why then is it difficult to see their wave-like nature? a. Their wavelength is equal to the object’s size. b. Their wavelength is very small compared to the object’s size. c. Their wavelength is very large compared to the object’s size. d. Their frequency is very small compared to the object’s size. Access for free at openstax.org. Extended Response 21.1 Planck and Quantum Nature of Light 67. Some television tubes are CRTs. They use an approximately 30-kV accelerating potential to send electrons to the screen, where the electrons stimulate phosphors to emit the light that forms the pictures we watch. Would you expect X-rays also to be created? Explain. a. No, because the full spectrum of EM radiation is not emitted at any temperature. b. No, because the full spectrum of EM radiation is not emitted at certain temperatures. c. Yes, because the full spectrum of EM radiation is emitted at any temperature. d. Yes, because the full spectrum of EM radiation is emitted at certain temperatures. 68. If Planck’s constant were large, say times greater than it is, we would observe macroscopic entities to be quantized. Describe the motion of a child’s swing under such circumstances. a. The child would not be able to swing with particular energies. b. The child could be released from any height. c. The child would be able to swing with constant velocity. d. The child could be released only from particular heights. 69. What is the accelerating voltage of an X-ray tube that produces X-rays with the shortest wavelength of 0.0103 nm? 1.21 × 1010 V a. b. 2.4 × 105 V c. d. 3.0 × 10−33 V 1.21 × 105 V 70. Patients in a doctor’s office are rightly concerned about receiving a chest X-ray. Yet visible light is also a form of electromagnetic radiation and they show little concern about sitting under the bright lights of the waiting room. Explain this discrepancy. a. X-ray photons carry considerably more energy so they can harm the patients. b. X-ray photons carry considerably less energy so they can harm the patients. c. X-ray photons have considerably longer wavelengths so they cannot harm
the patients. d. X-ray photons have considerably lower frequencies so they can harm the patients. 21.2 Einstein and the Photoelectric Effect 71. When increasing the intensity of light shining on a Chapter 21 • Test Prep 719 metallic surface, it is possible to increase the current created on that surface. Classical theorists would argue that this is evidence that intensity causes charge to move with a greater kinetic energy. Argue this logic from the perspective of a modern physicist. a. The increased intensity increases the number of ejected electrons. The increased current is due to the increase in the number of electrons. b. The increased intensity decreases the number of ejected electrons. The increased current is due to the decrease in the number of electrons ejected. c. The increased intensity does not alter the number of electrons ejected. The increased current is due to the increase in the kinetic energy of electrons. d. The increased intensity alters the number of electrons ejected, but an increase in the current is due to an increase in the kinetic energy of electrons. 72. What impact does the quantum nature of electromagnetic radiation have on the understanding of speed at the particle scale? a. Speed must also be quantized at the particle scale. b. Speed will not be quantized at the particle scale. c. Speed must be zero at the particle scale. d. Speed will be infinite at the particle scale. 73. A 500 nm photon of light strikes a semi-conductive surface with a binding energy of 2 eV. With what velocity will an electron be emitted from the semi-conductive surface? a. 8.38 × 105 m/s b. 9.33 × 105 m/s 3 × 108 m/s c. d. 4.11 × 105 m/s 74. True or false—Treating food with ionizing radiation helps keep it from spoiling. a. b. true false 21.3 The Dual Nature of Light 75. When testing atomic bombs, scientists at Los Alamos recognized that huge releases of energy resulted in problems with power and communications systems in the area surrounding the blast site. Explain the possible tie to Compton scattering. a. The release of light energy caused large-scale emission of electrons. b. The release of light energy caused large-scale emission of protons. c. The release of light energy caused large-scale emission of neutrons. d. The release of light energy caused large-scale 720 Chapter 21 • Test Prep emission of photons. 76. Sunlight above the Earth’s atmosphere has an
intensity of 1.30 kW/m2. If this is reflected straight back from a mirror that has only a small recoil, the light’s momentum is exactly reversed, giving the mirror twice the incident momentum. If the mirror were attached to a solar sail craft, how fast would the craft be moving after 24 hr? Note—The average mass per square meter of the craft is 0.100 kg. a. 8.67 × 10−5 m/s2 b. 8.67 × 10−6 m/s2 c. 94.2 m/s 7.49 m/s d. 77. Consider the counter-clockwise motion of LightSail-1 around Earth. When will the satellite move the fastest? a. point A b. point B c. point C d. point D 78. What will happen to the interference pattern created by electrons when their velocities are increased? a. There will be more zones of constructive interference and fewer zones of destructive interference. b. There will be more zones of destructive interference and fewer zones of constructive interference. c. There will be more zones of constructive and destructive interference. d. There will be fewer zones of constructive and destructive interference. Access for free at openstax.org. CHAPTER 22 The Atom Figure 22.1 Individual carbon atoms are visible in this image of a carbon nanotube made by a scanning tunneling electron microscope. (credit: Taner Yildirim, National Institute of Standards and Technology, Wikimedia Commons) Chapter Outline 22.1 The Structure of the Atom 22.2 Nuclear Forces and Radioactivity 22.3 Half Life and Radiometric Dating 22.4 Nuclear Fission and Fusion 22.5 Medical Applications of Radioactivity: Diagnostic Imaging and Radiation From childhood on, we learn that atoms are a substructure of all things around us, from the air we breathe to INTRODUCTION the autumn leaves that blanket a forest trail. Invisible to the eye, the atoms have properties that are used to explain many phenomena—a theme found throughout this text. In this chapter, we discuss the discovery of atoms and their own substructures. We will then learn about the forces that keep them together and the tremendous energy they release when we break them apart. Finally, we will see how the knowledge and manipulation of atoms allows us to better understand geology, biology, and the world around us. 22.1 The Structure of the Atom Section Learning Objectives By the end of this section, you will be
able to do the following: • Describe Rutherford’s experiment and his model of the atom • Describe emission and absorption spectra of atoms • Describe the Bohr model of the atom • Calculate the energy of electrons when they change energy levels • Calculate the frequency and wavelength of emitted photons when electrons change energy levels • Describe the quantum model of the atom 722 Chapter 22 • The Atom Section Key Terms energy-level diagram excited state Fraunhofer lines ground state Heisenberg Uncertainty Principle hydrogen-like atoms planetary model of the atom Rutherford scattering Rydberg constant How do we know that atoms are really there if we cannot see them with our own eyes? While often taken for granted, our knowledge of the existence and structure of atoms is the result of centuries of contemplation and experimentation. The earliest known speculation on the atom dates back to the fifth century B.C., when Greek philosophers Leucippus and Democritus contemplated whether a substance could be divided without limit into ever smaller pieces. Since then, scientists such as John Dalton (1766–1844), Amadeo Avogadro (1776–1856), and Dmitri Mendeleev (1834–1907) helped to discover the properties of that fundamental structure of matter. While much could be written about any number of important scientific philosophers, this section will focus on the role played by Ernest Rutherford (1871–1937). Though his understanding of our most elemental matter is rooted in the success of countless prior investigations, his surprising discovery about the interior of the atom is most fundamental in explaining so many well-known phenomena. Rutherford’s Experiment In the early 1900’s, the plum puddingmodel was the accepted model of the atom. Proposed in 1904 by J. J. Thomson, the model suggested that the atom was a spherical ball of positive charge, with negatively charged electrons scattered evenly throughout. In that model, the positive charges made up the pudding, while the electrons acted as isolated plums. During its short life, the model could be used to explain why most particles were neutral, although with an unbalanced number of plums, electrically charged atoms could exist. When Ernest Rutherford began his gold foil experiment in 1909, it is unlikely that anyone would have expected that the plum pudding model would be challenged. However, using a radioactive source, a thin sheet of gold foil, and a phosphorescent screen, Rutherford would uncover something so great that he would later call it “the most incredible
event that has ever happened to me in my life”[James, L. K. (1993). Nobel Laureates in Chemistry, 1901–1992. Washington, DC: American Chemical Society.] The experiment that Rutherford designed is shown in Figure 22.2. As you can see in, a radioactive source was placed in a lead container with a hole in one side to produce a beam of positively charged helium particles, called alpha particles. Then, a thin gold foil sheet was placed in the beam. When the high-energy alpha particles passed through the gold foil, they were scattered. The scattering was observed from the bright spots they produced when they struck the phosphor screen. Figure 22.2 Rutherford’s experiment gave direct evidence for the size and mass of the nucleus by scattering alpha particles from a thin gold foil. The scattering of particles suggests that the gold nuclei are very small and contain nearly all of the gold atom’s mass. Particularly significant in showing the size of the nucleus are alpha particles that scatter to very large angles, much like a soccer ball bouncing off a goalie’s head. The expectation of the plum pudding model was that the high-energy alpha particles would be scattered only slightly by the presence of the gold sheet. Because the energy of the alpha particles was much higher than those typically associated with atoms, the alpha particles should have passed through the thin foil much like a supersonic bowling ball would crash through a Access for free at openstax.org. 22.1 • The Structure of the Atom 723 few dozen rows of bowling pins. Any deflection was expected to be minor, and due primarily to the electrostatic Coulomb force between the alpha particles and the foil’s interior electric charges. However, the true result was nothing of the sort. While the majority of alpha particles passed through the foil unobstructed, Rutherford and his collaborators Hans Geiger and Ernest Marsden found that alpha particles occasionally were scattered to large angles, and some even came back in the direction from which they came! The result, called Rutherford scattering, implied that the gold nuclei were actually very small when compared with the size of the gold atom. As shown in Figure 22.3, the dense nucleus is surrounded by mostly empty space of the atom, an idea verified by the fact that only 1 in 8,000 particles was scattered backward. Figure 22.3 An expanded view of the atoms in the gold foil in Rutherford’s experiment. Circles represent the atoms that are about 10−10
m in diameter, while the dots represent the nuclei that are about 10−15 m in diameter. To be visible, the dots are much larger than scale—if the nuclei were actually the size of the dots, each atom would have a diameter of about five meters! Most alpha particles crash through but are relatively unaffected because of their high energy and the electron’s small mass. Some, however, strike a nucleus and are scattered straight back. A detailed analysis of their interaction gives the size and mass of the nucleus. Although the results of the experiment were published by his colleagues in 1909, it took Rutherford two years to convince himself of their meaning. Rutherford later wrote: “It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you. On consideration, I realized that this scattering backwards... [meant]... the greatest part of the mass of the atom was concentrated in a tiny nucleus.” In 1911, Rutherford published his analysis together with a proposed model of the atom, which was in part based on Geiger’s work from the previous year. As a result of the paper, the size of the nucleus was determined to be about cm3, much greater than any macroscopic matter. m, or 100,000 times smaller than the atom. That implies a huge density, on the order of g/ Based on the size and mass of the nucleus revealed by his experiment, as well as the mass of electrons, Rutherford proposed the planetary model of the atom. The planetary model of the atom pictures low-mass electrons orbiting a large-mass nucleus. The sizes of the electron orbits are large compared with the size of the nucleus, and most of the atom is a vacuum. The model is analogous to how low-mass planets in our solar system orbit the large-mass Sun. In the atom, the attractive Coulomb force is analogous to gravitation in the planetary system (see Figure 22.4). Figure 22.4 Rutherford’s planetary model of the atom incorporates the characteristics of the nucleus, electrons, and the size of the atom. The model was the first to recognize the structure of atoms, in which low-mass electrons orbit a very small, massive nucleus in orbits much larger than the nucleus. The atom is mostly empty and is analogous to our planetary system. 724 Chapter 22 • The Atom Virtual Physics Rutherford Scattering Click to view content (https://www.openstax.org/l/28
rutherford) How did Rutherford figure out the structure of the atom without being able to see it? Explore the answer through this simulation of the famous experiment in which he disproved the plum pudding model by observing alpha particles bouncing off atoms and determining that they must have a small core. TIPS FOR SUCCESS As you progress through the model of the atom, consider the effect that experimentation has on the scientific process. Ask yourself the following: What would our model of the atom be without Rutherford’s gold foil experiment? What further understanding of the atom would not have been gained? How would that affect our current technologies? Though often confusing, experiments taking place today to further understand composition of the atom could perhaps have a similar effect. Absorption and Emission Spectra In 1900, Max Planck recognized that all energy radiated from a source is emitted by atoms in quantum states. How would that radical idea relate to the interior of an atom? The answer was first found by investigating the spectrum of light or emission spectrum produced when a gas is highly energized. Figure 22.5 shows how to isolate the emission spectrum of one such gas. The gas is placed in the discharge tube at the left, where it is energized to the point at which it begins to radiate energy or emit light. The radiated light is channeled by a thin slit and then passed through a diffraction grating, which will separate the light into its constituent wavelengths. The separated light will then strike the photographic film on the right. The line spectrum shown in part (b) of Figure 22.5 is the output shown on the film for excited iron. Note that this spectrum is not continuous but discrete. In other words, only particular wavelengths are emitted by the iron source. Why would that be the case? Figure 22.5 Part (a) shows, from left to right, a discharge tube, slit, and diffraction grating producing a line spectrum. Part (b) shows the emission spectrum for iron. The discrete lines imply quantized energy states for the atoms that produce them. The line spectrum for each element is unique, providing a powerful and much-used analytical tool, and many line spectra were well known for many years before they could be explained with physics. (credit:(b) Yttrium91, Wikimedia Commons) The spectrum of light created by excited iron shows a variety of discrete wavelengths emitted within the visible spectrum. Each element, when excited to the appropriate degree, will create a discrete emission spectrum as in part (b
) of Figure 22.5. However, the wavelengths emitted will vary from element to element. The emission spectrum for iron was chosen for Figure 22.5 solely Access for free at openstax.org. because a substantial portion of its emission spectrum is within the visible spectrum. Figure 22.6 shows the emission spectrum for hydrogen. Note that, while discrete, a large portion of hydrogen emission takes place in the ultraviolet and infrared regions. 22.1 • The Structure of the Atom 725 Figure 22.6 A schematic of the hydrogen spectrum shows several series named for those who contributed most to their determination. Part of the Balmer series is in the visible spectrum, while the Lyman series is entirely in the ultraviolet, and the Paschen series and others are in the infrared. Values of nf and ni are shown for some of the lines. Their importance will be described shortly. Just as an emission spectrum shows all discrete wavelengths emitted by a gas, an absorption spectrum will show all light that is absorbed by a gas. Black lines exist where the wavelengths are absorbed, with the remainder of the spectrum lit by light is free to pass through. What relationship do you think exists between the black lines of a gas’s absorption spectrum and the colored lines of its emission spectrum? Figure 22.7 shows the absorption spectrum of the Sun. The black lines are called Fraunhofer lines, and they correspond to the wavelengths absorbed by gases in the Sun’s exterior. Figure 22.7 The absorption spectrum of the Sun. The black lines appear at wavelengths absorbed by the Sun’s gas exterior. The energetic photons emitted from the Sun’s interior are absorbed by gas in its exterior and reemitted in directions away from the observer. That results in dark lines within the absorption spectrum. The lines are called Fraunhofer lines, in honor of the German physicist who discovered them. Lines similar to those are used to determine the chemical composition of stars well outside our solar system. Bohr’s Explanation of the Hydrogen Spectrum To tie the unique signatures of emission spectra to the composition of the atom itself would require clever thinking. Niels Bohr (1885–1962), a Danish physicist, did just that, by making immediate use of Rutherford’s planetary model of the atom. Bohr, shown in Figure 22.8, became convinced of its validity and spent part of 1912 at Rutherford’s laboratory. In 1913, after returning to Copenhagen, he began publishing his theory of
the simplest atom, hydrogen, based on Rutherford’s planetary model. Figure 22.8 Niels Bohr, Danish physicist, used the planetary model of the atom to explain the atomic spectrum and size of the hydrogen 726 Chapter 22 • The Atom atom. His many contributions to the development of atomic physics and quantum mechanics, his personal influence on many students and colleagues, and his personal integrity, especially in the face of Nazi oppression, earned him a prominent place in history. (credit: Unknown Author, Wikimedia Commons) Bohr was able to derive the formula for the hydrogen spectrum using basic physics, the planetary model of the atom, and some very important new conjectures. His first conjecture was that only certain orbits are allowed: In other words, in an atom, the orbits of electrons are quantized. Each quantized orbit has a different distinct energy, and electrons can move to a higher orbit by absorbing energy or drop to a lower orbit by emitting energy. Because of the quantized orbits, the amount of energy emitted or absorbed must also be quantized, producing the discrete spectra seen in Figure 22.5 and Figure 22.7. In equation form, the amount of energy absorbed or emitted can be found as refers to the energy of the initial quantized orbit, and where wavelength emitted can be found using the equation refers to the energy of the final orbits. Furthermore, the and relating the wavelength to the frequency found using the equation, where vcorresponds to the speed of light. It makes sense that energy is involved in changing orbits. For example, a burst of energy is required for a satellite to climb to a higher orbit. What is not expected is that atomic orbits should be quantized. Quantization is not observed for satellites or planets, which can have any orbit, given the proper energy (see Figure 22.9). 22.2 22.1 Figure 22.9 The planetary model of the atom, as modified by Bohr, has the orbits of the electrons quantized. Only certain orbits are allowed, explaining why atomic spectra are discrete or quantized. The energy carried away from an atom by a photon comes from the electron dropping from one allowed orbit to another and is thus quantized. The same is true for atomic absorption of photons. Figure 22.10 shows an energy-level diagram, a convenient way to display energy states. Each of the horizontal lines corresponds to the energy of an electron in a different orbital. Energy is plotted vertically with the lowest or ground state at the bottom and with
excited states above. The vertical arrow downwards shows energy being emitted out of the atom due to an electron dropping from one excited state to another. That would correspond to a line shown on the atom’s emission spectrum. The Lyman series shown in Figure 22.6 results from electrons dropping to the ground state, while the Balmer and Paschen series result to electrons dropping to the n = 2and n = 3states, respectively. Access for free at openstax.org. 22.1 • The Structure of the Atom 727 Figure 22.10 An energy-level diagram plots energy vertically and is useful in visualizing the energy states of a system and the transitions between them. This diagram is for the hydrogen-atom electrons, showing a transition between two orbits having energies and. The energy transition results in a Balmer series line in an emission spectrum. Energy and Wavelength of Emitted Hydrogen Spectra The energy associated with a particular orbital of a hydrogen atom can be found using the equation 22.3 where ncorresponds to the orbital value from the atom’s nucleus. The negative value in the equation is based upon a baseline energy of zero when the electron is infinitely far from the atom. As a result, the negative value shows that energy is necessary to free the electron from its orbital state. The minimum energy to free the electron is also referred to as its binding energy. The equation is only valid for atoms with single electrons in their orbital shells (like hydrogen). For ionized atoms similar to hydrogen, the following formula may be used. 22.4 corresponds to –13.6 eV, as mentioned earlier. Additionally, Please note that studied. The atomic number is the number of protons in the nucleus—it is different for each element. The above equation is derived from some basic physics principles, namely conservation of energy, conservation of angular momentum, Coulomb’s law, and centripetal force. There are three derivations that result in the orbital energy equations, and they are shown below. While you can use the energy equations without understanding the derivations, they will help to remind you of just how valuable those fundamental concepts are. refers to the atomic number of the element Derivation 1 (Finding the Radius of an Orbital) One primary difference between the planetary model of the solar system and the planetary model of the atom is the cause of the circular motion. While gravitation causes the motion of orbiting planets around an interior star, the Coulomb force is responsible for the circular shape
of the electron’s orbit. The magnitude of the centripetal force is, while the magnitude of the Coulomb force is. The assumption here is that the nucleus is more massive than the stationary electron, and the electron orbits about it. That is consistent with the planetary model of the atom. Equating the Coulomb force and the centripetal force, 22.5 which yields 728 Chapter 22 • The Atom Derivation 2 (Finding the Velocity of the Orbiting Electron) Bohr was clever enough to find a way to calculate the electron orbital energies in hydrogen. That was an important first step that has been improved upon, but it is well worth repeating here, because it does correctly describe many characteristics of hydrogen. Assuming circular orbits, Bohr proposed that the angular momentum Lof an electron in its orbit is also quantized, that is, it has only specific, discrete values. The value for Lis given by the formula 22.7 22.6 where Lis the angular momentum, meis the electron’s mass, rnis the radius of the nth orbit, and his Planck’s constant. Note that angular momentum is. For a small object at a radius r,, so that, and Bohr himself did not know why angular momentum should be quantized, but by using that assumption, he was able to calculate the energies in the hydrogen spectrum, something no one else had done at the time. Quantization says that the value of mvrcan only be equal to h/ 2, 2h/ 2, 3h/ 2, etc. At the time, Derivation 3 (Finding the Energy of the Orbiting Electron) To get the electron orbital energies, we start by noting that the electron energy is the sum of its kinetic and potential energy. 22.8 Kinetic energy is the familiar electron is electrical, or, assuming the electron is not moving at a relativistic speed. Potential energy for the, where Vis the potential due to the nucleus, which looks like a point charge. The nucleus has a positive charge ; thus,, recalling an earlier equation for the potential due to a point charge from the chapter on Electricity and Magnetism. Since the electron’s charge is negative, we see that Substituting the expressions for KE and PE, Now we solve for rnand vusing the equation for angular momentum, giving and Substituting the expression for rnand vinto the above expressions for energy (KE and PE), and performing algebra
ic manipulation, yields 22.9 22.10 22.11 22.12 for the orbital energies of hydrogen-like atoms. Here, Eois the ground-state energy (n= 1) for hydrogen (Z= 1) and is given by Thus, for hydrogen, The relationship between orbital energies and orbital states for the hydrogen atom can be seen in Figure 22.11. 22.13 22.14 Access for free at openstax.org. 22.1 • The Structure of the Atom 729 Figure 22.11 Energy-level diagram for hydrogen showing the Lyman, Balmer, and Paschen series of transitions. The orbital energies are calculated using the above equation, first derived by Bohr. WORKED EXAMPLE A hydrogen atom is struck by a photon. How much energy must be absorbed from the photon to raise the electron of the hydrogen atom from its ground state to its second orbital? Strategy The hydrogen atom has an atomic number of Z= 1. Raising the electron from the ground state to its second orbital will increase its orbital level from n= 1 to n= 2. The energy determined will be measured in electron-volts. Solution The amount of energy necessary to cause the change in electron state is the difference between the final and initial energies of the electron. The final energy state of the electron can be found using Knowing the nand Zvalues for the hydrogen atom, and knowing that Eo= –13.6 eV, the result is The original amount of energy associated with the electron is equivalent to the ground state orbital, or 22.15 22.16 22.17 730 Chapter 22 • The Atom The amount of energy necessary to change the orbital state of the electron can be found by determining the electron’s change in energy. 22.18 Discussion The energy required to change the orbital state of the electron is positive. That means that for the electron to move to a state with greater energy, energy must be added to the atom. Should the electron drop back down to its original energy state, a change of –10.2 eV would take place, and 10.2 eV of energy would be emitted from the atom. Just as only quantum amounts of energy may be absorbed by the atom, only quantum amounts of energy can be emitted from the atom. That helps to explain many of the quantum light effects that you have learned about previously. WORKED EXAMPLE Characteristic X-Ray Energy Calculate the approximate energy of an X-ray emitted for an n
= 2 to n= 1 transition in a tungsten anode in an X-ray tube. Strategy How do we calculate energies in a multiple-electron atom? In the case of characteristic X-rays, the following approximate calculation is reasonable. Characteristic X-rays are produced when an inner-shell vacancy is filled. Inner-shell electrons are nearer the nucleus than others in an atom and thus feel little net effect from the others. That is similar to what happens inside a charged conductor, where its excess charge is distributed over the surface so that it produces no electric field inside. It is reasonable to assume the inner-shell electrons have hydrogen-like energies, as given by For tungsten, Z= 74, so that the effective charge is 73. Solution The amount of energy given off as an X-ray is found using where and Thus, 22.19 22.20 22.21 22.22 22.23 Discussion This large photon energy is typical of characteristic X-rays from heavy elements. It is large compared with other atomic emissions because it is produced when an inner-shell vacancy is filled, and inner-shell electrons are tightly bound. Characteristic X-ray energies become progressively larger for heavier elements because their energy increases approximately as Z2. Significant accelerating voltage is needed to create such inner-shell vacancies, because other shells are filled and you cannot simply bump one electron to a higher filled shell. You must remove it from the atom completely. In the case of tungsten, at least 72.5 kV is needed. Tungsten is a common anode material in X-ray tubes; so much of the energy of the impinging electrons is absorbed, raising its temperature, that a high-melting-point material like tungsten is required. The wavelength of light emitted by an atom can also be determined through basic derivations. Let us consider the energy of a photon emitted from a hydrogen atom in a downward transition, given by the equation Access for free at openstax.org. Substituting, we get Dividing both sides of the equation by hcgives us an expression for, It can be shown that where Ris the Rydberg constant. Simplified, the formula for determining emitted wavelength can now be written as 22.1 • The Structure of the Atom 731 22.24 22.25 22.26 22.27 22.28 WORKED EXAMPLE What wavelength of light is emitted by an electron dropping from the third orbital
to the ground state of a hydrogen atom? Strategy The ground state of a hydrogen atom is considered the first orbital of the atom. As a result, nf= 1 and ni= 3. The Rydberg constant has already been determined and will be constant regardless of atom chosen. Solution For the equation above, calculate wavelength based on the known energy states. Rearranging the equation for wavelength yields 22.29 22.30 22.31 Discussion This wavelength corresponds to light in the ultraviolet spectrum. As a result, we would not be able to see the photon of light emitted when an electron drops from its third to first energy state. However, it is worth noting that by supplying light of wavelength precisely 102.6 nm, we can cause the electron in hydrogen to move from its first to its third orbital state. Limits of Bohr’s Theory and the Quantum Model of the Atom There are limits to Bohr’s theory. It does not account for the interaction of bound electrons, so it cannot be fully applied to multielectron atoms, even one as simple as the two-electron helium atom. Bohr’s model is what we call semiclassical. The orbits are quantized (nonclassical) but are assumed to be simple circular paths (classical). As quantum mechanics was developed, it became clear that there are no well-defined orbits; rather, there are clouds of probability. Additionally, Bohr’s theory did not explain that some spectral lines are doublets or split into two when examined closely. While we shall examine a few of those aspects of quantum mechanics in more detail, it should be kept in mind that Bohr did not fail. Rather, he made very important steps along the path to greater knowledge and laid the foundation for all of atomic physics that has since evolved. 732 Chapter 22 • The Atom DeBroglie’s Waves Following Bohr’s initial work on the hydrogen atom, a decade was to pass before Louis de Broglie proposed that matter has wave properties. The wave-like properties of matter were subsequently confirmed by observations of electron interference when scattered from crystals. Electrons can exist only in locations where they interfere constructively. How does that affect electrons in atomic orbits? When an electron is bound to an atom, its wavelength must fit into a small space, something like a standing wave on a string (see Figure 22.12). Orbits in which an electron can constructively interfere with itself are allowed. All orbits in which constructive
interference cannot occur are not able to exist. Thus, only certain orbits are allowed. The wave nature of an electron, according to de Broglie, is why the orbits are quantized! Figure 22.12 (a) Standing waves on a string have a wavelength related to the length of the string, allowing them to interfere constructively. (b) If we imagine the string formed into a closed circle, we get a rough idea of how electrons in circular orbits can interfere constructively. (c) If the wavelength does not fit into the circumference, the electron interferes destructively; it cannot exist in such an orbit. For a circular orbit, constructive interference occurs when the electron’s wavelength fits neatly into the circumference, so that wave crests always align with crests and wave troughs align with troughs, as shown in Figure 22.12(b). More precisely, when an integral multiple of the electron’s wavelength equals the circumference of the orbit, constructive interference is obtained. In equation form, the condition for constructive interference and an allowed electron orbit is where of a hydrogen atom. is the electron’s wavelength and rnis the radius of that circular orbit. Figure 22.13 shows the third and fourth orbitals 22.32 Figure 22.13 The third and fourth allowed circular orbits have three and four wavelengths, respectively, in their circumferences. Access for free at openstax.org. 22.1 • The Structure of the Atom 733 Heisenberg Uncertainty How does determining the location of an electron change its trajectory? The answer is fundamentally important—measurement affects the system being observed. It is impossible to measure a physical quantity exactly, and greater precision in measuring one quantity produces less precision in measuring a related quantity. It was Werner Heisenberg who first stated that limit to knowledge in 1929 as a result of his work on quantum mechanics and the wave characteristics of all particles (see Figure 22.14). Figure 22.14 Werner Heisenberg was the physicist who developed the first version of true quantum mechanics. Not only did his work give a description of nature on the very small scale, it also changed our view of the availability of knowledge. Although he is universally recognized for the importance of his work by receiving the Nobel Prize in 1932, for example, Heisenberg remained in Germany during World War II and headed the German effort to build a nuclear bomb, permanently alienating himself from most of the scientific community. (credit: Unknown Author, Wikimedia Commons) For
example, you can measure the position of a moving electron by scattering light or other electrons from it. However, by doing so, you are giving the electron energy, and therefore imparting momentum to it. As a result, the momentum of the electron is affected and cannot be determined precisely. This change in momentum could be anywhere from close to zero up to the relative momentum of the electron ( particle. ). Note that, in this case, the particle is an electron, but the principle applies to any Viewing the electron through the model of wave-particle duality, Heisenberg recognized that, because a wave is not located at one fixed point in space, there is an uncertainty associated with any electron’s position. That uncertainty in position, approximately equal to the wavelength of the particle. That is, momentum. The uncertainty in an electron’s position can be reduced by using a shorter-wavelength electron, since But shortening the wavelength increases the uncertainty in momentum, since momentum can be reduced by using a longer-wavelength electron, but that increases the uncertainty in position. Mathematically, you can express the trade-off by multiplying the uncertainties. The wavelength cancels, leaving. There is an interesting trade-off between position and. Conversely, the uncertainty in,is. Therefore, if one uncertainty is reduced, the other must increase so that their product is mathematics, Heisenberg showed that the best that can be done in a simultaneous measurement of position and momentum is.With the use of advanced 22.33 That relationship is known as the Heisenberg uncertainty principle. The Quantum Model of the Atom Because of the wave characteristic of matter, the idea of well-defined orbits gives way to a model in which there is a cloud of probability, consistent with Heisenberg’s uncertainty principle. Figure 22.15 shows how the principle applies to the ground state of hydrogen. If you try to follow the electron in some well-defined orbit using a probe that has a wavelength small enough to 734 Chapter 22 • The Atom measure position accurately, you will instead knock the electron out of its orbit. Each measurement of the electron’s position will find it to be in a definite location somewhere near the nucleus. Repeated measurements reveal a cloud of probability like that in the figure, with each speck the location determined by a single measurement. There is not a well-defined, circular-orbit type of distribution. Nature again proves to be different on a small scale than on a macroscopic scale.
Figure 22.15 The ground state of a hydrogen atom has a probability cloud describing the position of its electron. The probability of finding the electron is proportional to the darkness of the cloud. The electron can be closer or farther than the Bohr radius, but it is very unlikely to be a great distance from the nucleus. Virtual Physics Models of the Hydrogen Atom Click to view content (https://www.openstax.org/l/28atom_model) How did scientists figure out the structure of atoms without looking at them? Try out different models by shooting light at the atom. Use this simulation to see how the prediction of the model matches the experimental results. Check Your Understanding 1. Alpha particles are positively charged. What influence did their charge have on the gold foil experiment? a. The positively charged alpha particles were attracted by the attractive electrostatic force from the positive nuclei of the gold atoms. b. The positively charged alpha particles were scattered by the attractive electrostatic force from the positive nuclei of the gold atoms. c. The positively charged alpha particles were scattered by the repulsive electrostatic force from the positive nuclei of the gold atoms. d. The positively charged alpha particles were attracted by the repulsive electrostatic force from the positive nuclei of the gold atoms. 22.2 Nuclear Forces and Radioactivity Section Learning Objectives By the end of this section, you will be able to do the following: • Describe the structure and forces present within the nucleus • Explain the three types of radiation • Write nuclear equations associated with the various types of radioactive decay Section Key Terms alpha decay atomic number beta decay gamma decay Geiger tube isotope mass number nucleons radioactive radioactive decay Access for free at openstax.org. 22.2 • Nuclear Forces and Radioactivity 735 radioactivity scintillator strong nuclear force transmutation There is an ongoing quest to find the substructures of matter. At one time, it was thought that atoms would be the ultimate substructure. However, just when the first direct evidence of atoms was obtained, it became clear that they have a substructure and a tiny nucleus. The nucleus itself has spectacular characteristics. For example, certain nuclei are unstable, and their decay emits radiations with energies millions of times greater than atomic energies. Some of the mysteries of nature, such as why the core of Earth remains molten and how the Sun produces its energy, are explained by nuclear phenomena. The exploration of radioactivity and the nucleus has revealed new fundamental particles, forces, and conservation
laws. That exploration has evolved into a search for further underlying structures, such as quarks. In this section, we will explore the fundamentals of the nucleus and nuclear radioactivity. The Structure of the Nucleus At this point, you are likely familiar with the neutron and proton, the two fundamental particles that make up the nucleus of an atom. Those two particles, collectively called nucleons, make up the small interior portion of the atom. Both particles have nearly the same mass, although the neutron is about two parts in 1,000 more massive. The mass of a proton is equivalent to 1,836 electrons, while the mass of a neutron is equivalent to that of 1,839 electrons. That said, each of the particles is significantly more massive than the electron. When describing the mass of objects on the scale of nucleons and atoms, it is most reasonable to measure their mass in terms of atoms. The atomic mass unit (u) was originally defined so that a neutral carbon atom would have a mass of exactly 12 u. Given that protons and neutrons are approximately the same mass, that there are six protons and six neutrons in a carbon atom, and that the mass of an electron is minuscule in comparison, measuring this way allows for both protons and neutrons to have masses close to 1 u. Table 22.1 shows the mass of protons, neutrons, and electrons on the new scale. TIPS FOR SUCCESS For most conceptual situations, the difference in mass between the proton and neutron is insubstantial. In fact, for calculations that require fewer than four significant digits, both the proton and neutron masses may be considered equivalent to one atomic mass unit. However, when determining the amount of energy released in a nuclear reaction, as in Equation 22.40, the difference in mass cannot be ignored. Another other useful mass unit on the atomic scale is the one uses the equation, as will be addressed later in this text.. While rarely used in most contexts, it is convenient when Proton Mass Neutron Mass Electron Mass Kilograms (kg) Atomic mass units (u) Table 22.1 Atomic Masses for Multiple Units To more completely characterize nuclei, let us also consider two other important quantities: the atomic number and the mass number. The atomic number, Z, represents the number of protons within a nucleus. That value determines the elemental quality of each atom. Every carbon atom, for instance, has a Zvalue of 6
, whereas every oxygen atom has a Zvalue of 8. For clarification, only oxygen atoms may have a Zvalue of 8. If the Zvalue is not 8, the atom cannot be oxygen. The mass number, A, represents the total number of protons and neutrons, or nucleons, within an atom. For an ordinary carbon atom the mass number would be 12, as there are typically six neutrons accompanying the six protons within the atom. In the case of carbon, the mass would be exactly 12 u. For oxygen, with a mass number of 16, the atomic mass is 15.994915 u. Of course, the difference is minor and can be ignored for most scenarios. Again, because the mass of an electron is so small compared to the nucleons, the mass number and the atomic mass can be essentially equivalent. Figure 22.16 shows an example of Lithium-7, which has an atomic number of 3 and a mass number of 7. How does the mass number help to differentiate one atom from another? If each atom of carbon has an atomic number of 6, 736 Chapter 22 • The Atom then what is the value of including the mass number at all? The intent of the mass number is to differentiate between various isotopes of an atom. The term isotope refers to the variation of atoms based upon the number of neutrons within their nucleus. While it is most common for there to be six neutrons accompanying the six protons within a carbon atom, it is possible to find carbon atoms with seven neutrons or eight neutrons. Those carbon atoms are respectively referred to as carbon-13 and carbon-14 atoms, with their mass numbers being their primary distinction. The isotope distinction is an important one to make, as the number of neutrons within an atom can affect a number of its properties, not the least of which is nuclear stability. Figure 22.16 Lithium-7 has three protons and four neutrons within its nucleus. As a result, its mass number is 7, while its atomic number is 3. The actual mass of the atom is 7.016 u. Lithium 7 is an isotope of lithium. To more easily identify various atoms, their atomic number and mass number are typically written in a form of representation called the nuclide. The nuclide form appears as follows: neutrons., where Xis the atomic symbol and Nrepresents the number of Let us look at a few examples of nuclides expressed in the notation.
The nucleus of the simplest atom, hydrogen, is a single (the zero for no neutrons is often omitted). To check the symbol, refer to the periodic table—you see that the proton, or atomic number Zof hydrogen is 1. Since you are given that there are no neutrons, the mass number Ais also 1. There is a scarce form of hydrogen found in nature called deuterium; its nucleus has one proton and one neutron and, hence, twice the mass of common hydrogen. The symbol for deuterium is, thus, tritium, since it has a single proton and two neutrons, and it is written identical chemistries, but the nuclei differ greatly in mass, stability, and other characteristics. Again, the different nuclei are referred to as isotopes of the same element.. An even rarer—and radioactive—form of hydrogen is called. The three varieties of hydrogen have nearly There is some redundancy in the symbols A, X, Z, and N. If the element Xis known, then Zcan be found in a periodic table. If both Aand Xare known, then Ncan also be determined by first finding Z; then, N= A– Z. Thus the simpler notation for nuclides is 22.34 which is sufficient and is most commonly used. For example, in this simpler notation, the three isotopes of hydrogen are,, and. For, should we need to know, we can determine that Z= 92 for uranium from the periodic table, and thus, N = 238 − 92 = 146. Radioactivity and Nuclear Forces In 1896, the French physicist Antoine Henri Becquerel (1852–1908) noticed something strange. When a uranium-rich mineral called pitchblende was placed on a completely opaque envelope containing a photographic plate, it darkened spots on the photographic plate.. Becquerel reasoned that the pitchblende must emit invisible rays capable of penetrating the opaque material. Stranger still was that no light was shining on the pitchblende, which means that the pitchblende was emitting the invisible rays continuously without having any energy input! There is an apparent violation of the law of conservation of energy, one that scientists can now explain using Einstein’s famous equation originate in the nuclei of the atoms and have other unique characteristics. It was soon evident that Becquerel’s rays To this point, most reactions you have studied have been chemical reactions, which are reactions involving the electrons
Access for free at openstax.org. 22.2 • Nuclear Forces and Radioactivity 737 surrounding the atoms. However, two types of experimental evidence implied that Becquerel’s rays did not originate with electrons, but instead within the nucleus of an atom. First, the radiation is found to be only associated with certain elements, such as uranium. Whether uranium was in the form of an element or compound was irrelevant to its radiation. In addition, the presence of radiation does not vary with temperature, pressure, or ionization state of the uranium atom. Since all of those factors affect electrons in an atom, the radiation cannot come from electron transitions, as atomic spectra do. The huge energy emitted during each event is the second piece of evidence that the radiation cannot be atomic. Nuclear radiation has energies on the order of 106 eV per event, which is much greater than typical atomic energies that are a few eV, such as those observed in spectra and chemical reactions, and more than ten times as high as the most energetic X-rays. But why would reactions within the nucleus take place? And what would cause an apparently stable structure to begin emitting energy? Was there something special about Becquerel’s uranium-rich pitchblende? To answer those questions, it is necessary to look into the structure of the nucleus. Though it is perhaps surprising, you will find that many of the same principles that we observe on a macroscopic level still apply to the nucleus. Nuclear Stability A variety of experiments indicate that a nucleus behaves something like a tightly packed ball of nucleons, as illustrated in Figure 22.17. Those nucleons have large kinetic energies and, thus, move rapidly in very close contact. Nucleons can be separated by a large force, such as in a collision with another nucleus, but strongly resist being pushed closer together. The most compelling evidence that nucleons are closely packed in a nucleus is that the radius of a nucleus, r, is found to be approximately where 1.2 femtometer (fm) and Ais the mass number of the nucleus. Note that. Since many nuclei are spherical, and the volume of a sphere is, we see that —that is, the volume of a nucleus is proportional to the number of nucleons in it. That is what you expect if you pack nucleons so close that there is no empty space between them. 22.35 Figure 22.17 Nucleons are held together by nuclear forces and resist both being pulled apart
and pushed inside one another. The volume of the nucleus is the sum of the volumes of the nucleons in it, here shown in different colors to represent protons and neutrons. So what forces hold a nucleus together? After all, the nucleus is very small and its protons, being positive, should exert tremendous repulsive forces on one another. Considering that, it seems that the nucleus would be forced apart, not together! The answer is that a previously unknown force holds the nucleus together and makes it into a tightly packed ball of nucleons. This force is known as the strong nuclear force. The strong force has such a short range that it quickly fall to zero over a distance of only 10–15 meters. However, like glue, it is very strong when the nucleons get close to one another. The balancing of the electromagnetic force with the nuclear forces is what allows the nucleus to maintain its spherical shape. If, for any reason, the electromagnetic force should overcome the nuclear force, components of the nucleus would be projected outward, creating the very radiation that Becquerel discovered! Understanding why the nucleus would break apart can be partially explained using Table 22.2. The balance between the strong nuclear force and the electromagnetic force is a tenuous one. Recall that the attractive strong nuclear force exists between any two nucleons and acts over a very short range while the weaker repulsive electromagnetic force only acts between protons, although over a larger range. Considering the interactions, an imperfect balance between neutrons and protons can result in a nuclear reaction, with the result of regaining equilibrium. 738 Chapter 22 • The Atom Range of Force Direction Nucleon Interaction Magnitude of Force Electromagnetic Force Long range, though decreasing by 1/r2 Repulsive Proton –proton repulsion Relatively small Strong Nuclear Force Very short range, essentially zero at 1 femtometer Attractive Attraction between any two nucleons 100 times greater than the electromagnetic force Table 22.2 Comparing the Electromagnetic and Strong Forces The radiation discovered by Becquerel was due to the large number of protons present in his uranium-rich pitchblende. In short, the large number of protons caused the electromagnetic force to be greater than the strong nuclear force. To regain stability, the nucleus needed to undergo a nuclear reaction called alpha (α) decay. The Three Types of Radiation Radioactivity refers to the act of emitting particles or energy from the nucleus. When the uranium nucleus emits energetic nucleons in Becquere
l’s experiment, the radioactive process causes the nucleus to alter in structure. The alteration is called radioactive decay. Any substance that undergoes radioactive decay is said to be radioactive. That those terms share a root with the term radiationshould not be too surprising, as they all relate to the transmission of energy. Alpha Decay Alpha decay refers to the type of decay that takes place when too many protons exist in the nucleus. It is the most common type of decay and causes the nucleus to regain equilibrium between its two competing internal forces. During alpha decay, the nucleus ejects two protons and two neutrons, allowing the strong nuclear force to regain balance with the repulsive electromagnetic force. The nuclear equation for an alpha decay process can be shown as follows. 22.36 Figure 22.18 A nucleus undergoes alpha decay. The alpha particle can be seen as made up of two neutrons and two protons, which constitute a helium-4 atom. Three things to note as a result of the above equation: 1. By ejecting an alpha particle, the original nuclide decreases in atomic number. That means that Becquerel’s uranium nucleus, upon decaying, is actually transformed into thorium, two atomic numbers lower on the periodic table! The process of changing elemental composition is called transmutation. 2. Note that the two protons and two neutrons ejected from the nucleus combine to form a helium nucleus. Shortly after decay, the ejected helium ion typically acquires two electrons to become a stable helium atom. 3. Finally, it is important to see that, despite the elemental change, physical conservation still takes place. The mass number of the new element and the alpha particle together equal the mass number of the original element. Also, the net charge of all particles involved remains the same before and after the transmutation. Beta Decay Like alpha decay, beta ( For beta decay, however, a neutron is transformed into a proton and electron or vice versa. The transformation allows for the total mass number of the atom to remain the same, although the atomic number will increase by one (or decrease by one). Once again, the transformation of the neutron allows for a rebalancing of the strong nuclear and electromagnetic forces. The nuclear ) decay also takes place when there is an imbalance between neutrons and protons within the nucleus. Access for free at openstax.org. 22.2 • Nuclear Forces and Radioactivity 739 equation for a beta decay process is shown below. in the equation above
stands for a high-energy particle called the neutrino. A nucleus may also emit a positron, The symbol and in that case Zdecreases and Nincreases. It is beyond the scope of this section and will be discussed in further detail in the chapter on particles. It is worth noting, however, that the mass number and charge in all beta-decay reactions are conserved. Figure 22.19 A nucleus undergoes beta decay. The neutron splits into a proton, electron, and neutrino. This particular decay is called decay. Gamma Decay Gamma decay is a unique form of radiation that does not involve balancing forces within the nucleus. Gamma decay occurs when a nucleus drops from an excited state to the ground state. Recall that such a change in energy state will release energy from the nucleus in the form of a photon. The energy associated with the photon emitted is so great that its wavelength is shorter than that of an X-ray. Its nuclear equation is as follows. 22.37 Figure 22.20 A nucleus undergoes gamma decay. The nucleus drops in energy state, releasing a gamma ray. WORKED EXAMPLE Creating a Decay Equation Write the complete decay equation in Strategy Beta decay results in an increase in atomic number. As a result, the original (or parent) nucleus, must have an atomic number of one fewer proton.. Refer to the periodic table for values of Z. notation for beta decay producing Solution The equation for beta decay is as follows Considering that barium is the product (or daughter) nucleus and has an atomic number of 56, the original nucleus must be of an atomic number of 55. That corresponds to cesium, or Cs. 22.38 22.39 740 Chapter 22 • The Atom The number of neutrons in the parent cesium and daughter barium can be determined by subtracting the atomic number from the mass number (137 – 55 for cesium, 137 – 56 for barium). Substitute those values for the Nand N – 1subscripts in the above equation. 22.40 Discussion The terms parentand daughternucleus refer to the reactants and products of a nuclear reaction. The terminology is not just used in this example, but in all nuclear reaction examples. The cesium-137 nuclear reaction poses a significant health risk, as its chemistry is similar to that of potassium and sodium, and so it can easily be concentrated in your cells if ingested. WORKED EXAMPLE Alpha Decay Energy Found
from Nuclear Masses Find the energy emitted in the decay of 239Pu. Strategy Nuclear reaction energy, such as released in decay, can be found using the equation difference in mass between the parent nucleus and the products of the decay. The mass of pertinent particles is as follows. We must first find, the 239Pu: 239.052157 u 235U: 235.043924 u 4He: 4.002602 u. Solution The decay equation for 239Pu is Determine the amount of mass lost between the parent and daughter nuclei. Now we can find Eby entering into the equation. And knowing that, we can find that 22.41 22.42 22.43 22.44 Discussion The energy released in this decay is in the MeV range, about 106 times as great as typical chemical reaction energies, consistent with previous discussions. Most of the energy becomes kinetic energy of the particle (or 4He nucleus), which moves away at high speed. The energy carried away by the recoil of the 235U nucleus is much smaller, in order to conserve momentum. The 235U nucleus can be left in an excited state to later emit photons ( rays). The decay is spontaneous and releases energy, because the products have less mass than the parent nucleus. Properties of Radiation The charges of the three radiated particles differ. Alpha particles, with two protons, carry a net charge of +2. Beta particles, with one electron, carry a net charge of –1. Meanwhile, gamma rays are solely photons, or light, and carry no charge. The difference Access for free at openstax.org. 22.2 • Nuclear Forces and Radioactivity 741 in charge plays an important role in how the three radiations affect surrounding substances. Alpha particles, being highly charged, will quickly interact with ions in the air and electrons within metals. As a result, they have a short range and short penetrating distance in most materials. Beta particles, being slightly less charged, have a larger range and larger penetrating distance. Gamma rays, on the other hand, have little electric interaction with particles and travel much farther. Two diagrams below show the importance of difference in penetration. Table 22.3 shows the distance of radiation penetration, and Figure 22.21 shows the influence various factors have on radiation penetration distance. Type of Radiation Range particles A sheet of paper, a few cm of air, fractions of a millimeter of tissue particles A thin aluminum plate, tens of cm of tissue rays Several cm of lead, meters of
concrete Table 22.3 Comparing Ranges of Radioactive Decay Figure 22.21 The penetration or range of radiation depends on its energy, the material it encounters, and the type of radiation. (a) Greater energy means greater range. (b) Radiation has a smaller range in materials with high electron density. (c) Alphas have the smallest range, betas have a greater range, and gammas have the greatest range. LINKS TO PHYSICS Radiation Detectors The first direct detection of radiation was Becquerel’s darkened photographic plate. Photographic film is still the most common detector of ionizing radiation, being used routinely in medical and dental X-rays. Nuclear radiation can also be captured on film, as seen in Figure 22.22. The mechanism for film exposure by radiation is similar to that by photons. A quantum of energy from a radioactive particle interacts with the emulsion and alters it chemically, thus exposing the film. Provided the radiation has more than the few eV of energy needed to induce the chemical change, the chemical alteration will occur. The amount of film darkening is related to the type of radiation and amount of exposure. The process is not 100 percent efficient, since not all incident radiation interacts and not all interactions produce the chemical change. 742 Chapter 22 • The Atom Figure 22.22 Film badges contain film similar to that used in this dental X-ray film. It is sandwiched between various absorbers to determine the penetrating ability of the radiation as well as the amount. Film badges are worn to determine radiation exposure. (credit: Werneuchen, Wikimedia Commons) Another very common radiation detector is the Geiger tube. The clicking and buzzing sound we hear in dramatizations and documentaries, as well as in our own physics labs, is usually an audio output of events detected by a Geiger counter. These relatively inexpensive radiation detectors are based on the simple and sturdy Geiger tube, shown schematically in Figure 22.23. A conducting cylinder with a wire along its axis is filled with an insulating gas so that a voltage applied between the cylinder and wire produces almost no current. Ionizing radiation passing through the tube produces free ion pairs that are attracted to the wire and cylinder, forming a current that is detected as a count. Not every particle is detected, since some radiation can pass through without producing enough ionization. However, Geiger counters are very useful in producing a prompt output that reveals the existence and relative intensity of ionizing radiation. Figure 22.23
(a) Geiger counters such as this one are used for prompt monitoring of radiation levels, generally giving only relative intensity and not identifying the type or energy of the radiation. (credit: Tim Vickers, Wikimedia Commons) (b) Voltage applied between the cylinder and wire in a Geiger tube affects ions and electrons produced by radiation passing through the gas-filled cylinder. Ions move toward the cylinder and electrons toward the wire. The resulting current is detected and registered as a count. Another radiation detection method records light produced when radiation interacts with materials. The energy of the radiation is sufficient to excite atoms in a material that may fluoresce, such as the phosphor used by Rutherford’s group. Materials called scintillators use a more complex process to convert radiation energy into light. Scintillators may be liquid or solid, and they can Access for free at openstax.org. 22.3 • Half Life and Radiometric Dating 743 be very efficient. Their light output can provide information about the energy, charge, and type of radiation. Scintillator light flashes are very brief in duration, allowing the detection of a huge number of particles in short periods of time. Scintillation detectors are used in a variety of research and diagnostic applications. Among those are the detection of the radiation from distant galaxies using satellite-mounted equipment and the detection of exotic particles in accelerator laboratories. Virtual Physics Beta Decay Click to view content (https://www.openstax.org/l/21betadecayvid) Watch beta decay occur for a collection of nuclei or for an individual nucleus. With this applet, individuals or groups of students can compare half-lives! Check Your Understanding 2. What leads scientists to infer that the nuclear strong force exists? a. A strong force must hold all the electrons outside the nucleus of an atom. b. A strong force must counteract the highly attractive Coulomb force in the nucleus. c. A strong force must hold all the neutrons together inside the nucleus. d. A strong force must counteract the highly repulsive Coulomb force between protons in the nucleus. 22.3 Half Life and Radiometric Dating Section Learning Objectives By the end of this section, you will be able to do the following: • Explain radioactive half-life and its role in radiometric dating • Calculate radioactive half-life and solve problems associated with radiometric dating Section Key Terms activity becquerel carbon-14 dating decay constant half-life
radioactive dating Half-Life and the Rate of Radioactive Decay Unstable nuclei decay. However, some nuclides decay faster than others. For example, radium and polonium, discovered by Marie and Pierre Curie, decay faster than uranium. That means they have shorter lifetimes, producing a greater rate of decay. Here we will explore half-life and activity, the quantitative terms for lifetime and rate of decay. Why do we use the term like half-liferather than lifetime? The answer can be found by examining Figure 22.24, which shows how the number of radioactive nuclei in a sample decreases with time. The time in which half of the original number of nuclei decay is defined as the half-life,. After one half-life passes, half of the remaining nuclei will decay in the next half-life. Then, half of that amount in turn decays in the following half-life. Therefore, the number of radioactive nuclei decreases from Nto N/ 2 in one half-life, to N/ 4 in the next, to N/ 8 in the next, and so on. Nuclear decay is an example of a purely statistical process. TIPS FOR SUCCESS A more precise definition of half-life is that each nucleus has a 50 percent chance of surviving for a time equal to one halflife. If an individual nucleus survives through that time, it still has a 50 percent chance of surviving through another half-life. Even if it happens to survive hundreds of half-lives, it still has a 50 percent chance of surviving through one more. Therefore, the decay of a nucleus is like random coin flipping. The chance of heads is 50 percent, no matter what has happened before. The probability concept aligns with the traditional definition of half-life. Provided the number of nuclei is reasonably large, half of the original nuclei should decay during one half-life period. 744 Chapter 22 • The Atom Figure 22.24 Radioactive decay reduces the number of radioactive nuclei over time. In one half-life ( ), the number decreases to half of its original value. Half of what remains decays in the next half-life, and half of that in the next, and so on. This is exponential decay, as seen in the graph of the number of nuclei present as a function of time. The following equation gives the quantitative relationship between the original number of nuclei present at time zero the number at a later time t and 22.
45 where e= 2.71828... is the base of the natural logarithm, and is the decay constant for the nuclide. The shorter the half-life, the larger is the value of equation decreases with time. The decay constant can be found with the, and the faster the exponential Activity, the Rate of Decay What do we mean when we say a source is highly radioactive? Generally, it means the number of decays per unit time is very high. We define activity Rto be the rate of decay expressed in decays per unit time. In equation form, this is 22.46 where is the number of decays that occur in time. Activity can also be determined through the equation 22.47 22.48 which shows that as the amount of radiative material (N) decreases, the rate of decay decreases as well. The SI unit for activity is one decay per second and it is given the name becquerel (Bq) in honor of the discoverer of radioactivity. That is, Activity Ris often expressed in other units, such as decays per minute or decays per year. One of the most common units for activity is the curie (Ci), defined to be the activity of 1 g of 226Ra, in honor of Marie Curie’s work with radium. The definition of the curie is 22.49 Access for free at openstax.org. or decays per second. Radiometric Dating 22.3 • Half Life and Radiometric Dating 745 Radioactive dating or radiometric dating is a clever use of naturally occurring radioactivity. Its most familiar application is carbon-14 dating. Carbon-14 is an isotope of carbon that is produced when solar neutrinos strike atmosphere. Radioactive carbon has the same chemistry as stable carbon, and so it mixes into the biosphere, where it is consumed and becomes part of every living organism. Carbon-14 has an abundance of 1.3 parts per trillion of normal carbon, so if you know the number of carbon nuclei in an object (perhaps determined by mass and Avogadro’s number), you can multiply that number by back to with the environment ceases, and wrappings, with the normal abundance in living tissue, it is possible to determine the artifact’s age (or time since death). Carbon-14 dating can be used for biological tissues as old as 50 or 60 thousand years, but is most accurate for younger samples,
since the abundance of with a half-life of 5,730 years (note that this is an example of beta decay). When an organism dies, carbon exchange nuclei within the object. Over time, carbon-14 will naturally decay is not replenished. By comparing the abundance of in an artifact, such as mummy nuclei in them is greater. to find the number of particles within the One of the most famous cases of carbon-14 dating involves the Shroud of Turin, a long piece of fabric purported to be the burial shroud of Jesus (see Figure 22.25). This relic was first displayed in Turin in 1354 and was denounced as a fraud at that time by a French bishop. Its remarkable negative imprint of an apparently crucified body resembles the then-accepted image of Jesus. As a result, the relic has been remained controversial throughout the centuries. Carbon-14 dating was not performed on the shroud until 1988, when the process had been refined to the point where only a small amount of material needed to be destroyed. Samples were tested at three independent laboratories, each being given four pieces of cloth, with only one unidentified piece found in from the shroud, to avoid prejudice. All three laboratories found samples of the shroud contain 92 percent of the living tissues, allowing the shroud to be dated (see Equation 22.57). Figure 22.25 Part of the Shroud of Turin, which shows a remarkable negative imprint likeness of Jesus complete with evidence of crucifixion wounds. The shroud first surfaced in the 14th century and was only recently carbon-14 dated. It has not been determined how the image was placed on the material. (credit: Butko, Wikimedia Commons) WORKED EXAMPLE Carbon-11 Decay Carbon-11 has a half-life of 20.334 min. (a) What is the decay constant for carbon-11? If 1 kg of carbon-11 sample exists at the beginning of an hour, (b) how much material will remain at the end of the hour and (c) what will be the decay activity at that time? Strategy Since refers to the amount of carbon-11 at the start, then after one half-life, the amount of carbon-11 remaining will be The decay constant is equivalent to the probability that a nucleus will decay each second. As a result, the half-life will need to be converted to seconds. Solution (a) Since half of the carbon-11 remains after one half-life,. 22.50 22.51 746
Chapter 22 • The Atom Take the natural logarithm of each side to isolate the decay constant. Convert the 20.334 min to seconds. (b) The amount of material after one hour can be found by using the equation with tconverted into seconds and NOwritten as 1,000 g (c) The decay activity after one hour can be found by using the equation for the mass value after one hour. 22.52 22.53 22.54 22.55 22.56 22.57 Discussion (a) The decay constant shows that 0.0568 percent of the nuclei in a carbon-11 sample will decay each second. Another way of considering the decay constant is that a given carbon-11 nuclei has a 0.0568 percent probability of decaying each second. The decay of carbon-11 allows it to be used in positron emission topography (PET) scans; however, its 20.334 min half-life does pose challenges for its administration. (b) One hour is nearly three full half-lives of the carbon-11 nucleus. As a result, one would expect the amount of sample remaining to be approximately one eighth of the original amount. The 129.4 g remaining is just a bit larger than one-eighth, which is sensible given a half-life of just over 20 min. (c) Label analysis shows that the unit of Becquerel is sensible, as there are 0.0735 g of carbon-11 decaying each second. That is smaller amount than at the beginning of the hour, when g of carbon-11 were decaying each second. WORKED EXAMPLE How Old is the Shroud of Turin? Calculate the age of the Shroud of Turin given that the amount of Strategy Because 92 percent of the know that the half-life of assume that the decrease in remains, is 5,730 years, and so once is solely due to nuclear decay. found in it is 92 percent of that in living tissue.. Therefore, the equation can be used to find. We also is known, we can find and then find tas requested. Here, we Solution Solving the equation for gives Access for free at openstax.org. 22.58 22.4 • Nuclear Fission and Fusion 747 Thus, Taking the natural logarithm of both sides of the equation yields so that Rearranging to isolate tgives Now, the equation can be used to find for. Solving for and substituting the known half-
life gives We enter that value into the previous equation to find t. 22.59 22.60 22.61 22.62 22.63 22.64 Discussion This dates the material in the shroud to 1988–690 = 1300. Our calculation is only accurate to two digits, so that the year is rounded to 1300. The values obtained at the three independent laboratories gave a weighted average date of 1320 ± 60. That uncertainty is typical of carbon-14 dating and is due to the small amount of 14 C in living tissues, the amount of material available, and experimental uncertainties (reduced by having three independent measurements). That said, is it notable that the carbon-14 date is consistent with the first record of the shroud’s existence and certainly inconsistent with the period in which Jesus lived. There are other noncarbon forms of radioactive dating. Rocks, for example, can sometimes be dated based on the decay of The decay series for been since the rock solidified. Knowledge of the years ago. about, so the ratio of those nuclides in a rock can be used an indication of how long it has half-life has shown, for example, that the oldest rocks on Earth solidified ends with Virtual Physics Radioactive Dating Game Click to view content (https://www.openstax.org/l/02radioactive_dating_game) Learn about different types of radiometric dating, such as carbon dating. Understand how decay and half-life work to enable radiometric dating to work. Play a game that tests your ability to match the percentage of the dating element that remains to the age of the object. 22.4 Nuclear Fission and Fusion Section Learning Objectives By the end of this section, you will be able to do the following: • Explain nuclear fission • Explain nuclear fusion • Describe how the processes of fission and fusion work in nuclear weapons and in generating nuclear power 748 Chapter 22 • The Atom Section Key Terms chain reaction critical mass liquid drop model nuclear fission nuclear fusion proton-proton cycle The previous section dealt with naturally occurring nuclear decay. Without human intervention, some nuclei will change composition in order to achieve a stable equilibrium. This section delves into a less-natural process. Knowing that energy can be emitted in various forms of nuclear change, is it possible to create a nuclear reaction through our own intervention? The answer to this question is yes. Through two distinct methods, humankind has discovered multiple ways of manipulating the atom to release its internal energy. Nuclear Fission In
simplest terms, nuclear fission is the splitting of an atomic bond. Given that it requires great energy separate two nucleons, it may come as a surprise to learn that splitting a nucleus can releasevast potential energy. And although it is true that huge amounts of energy can be released, considerable effort is needed to do so in practice. An unstable atom will naturally decay, but it may take millions of years to do so. As a result, a physical catalyst is necessary to produce useful energy through nuclear fission. The catalyst typically occurs in the form of a free neutron, projected directly at the nucleus of a high-mass atom. As shown in Figure 22.26, a neutron strike can cause the nucleus to elongate, much like a drop of liquid water. This is why the model is known as the liquid drop model. As the nucleus elongates, nucleons are no longer so tightly packed, and the repulsive electromagnetic force can overcome the short-range strong nuclear force. The imbalance of forces can result in the two ends of the drop flying apart, with some of the nuclear binding energy released to the surroundings. Figure 22.26 Neutron-induced fission is shown. First, energy is put into a large nucleus when it absorbs a neutron. Acting like a struck liquid drop, the nucleus deforms and begins to narrow in the middle. Since fewer nucleons are in contact, the repulsive Coulomb force is able to break the nucleus into two parts with some neutrons also flying away. As you can imagine, the consequences of the nuclei splitting are substantial. When a nucleus is split, it is not only energy that is released, but a small number of neutrons as well. Those neutrons have the potential to cause further fission in other nuclei, especially if they are directed back toward the other nuclei by a dense shield or neutron reflector (see part (d) of Figure 22.26). Access for free at openstax.org. 22.4 • Nuclear Fission and Fusion 749 However, not every neutron produced by fission induces further fission. Some neutrons escape the fissionable material, while others interact with a nucleus without making it split. We can enhance the number of fissions produced by neutrons by having a large amount of fissionable material as well as a neutron reflector. The minimum amount necessary for self-sustained fission of a given nuclide is called its critical mass. Some nuclides, such as 239
Pu, produce more neutrons per fission than others, such as 235U. Additionally, some nuclides are easier to make fission than others. In particular, 235U and 239Pu are easier to fission than the much more abundant 238U. Both factors affect critical mass, which is smallest for 239Pu. The self-sustained fission of nuclei is commonly referred to as a chain reaction, as shown in Figure 22.27. Figure 22.27 A chain reaction can produce self-sustained fission if each fission produces enough neutrons to induce at least one more fission. This depends on several factors, including how many neutrons are produced in an average fission and how easy it is to make a particular type of nuclide fission. A chain reaction can have runaway results. If each atomic split results in two nuclei producing a new fission, the number of nuclear reactions will increase exponentially. One fission will produce two atoms, the next round of fission will create four atoms, the third round eight atoms, and so on. Of course, each time fission occurs, more energy will be emitted, further increasing the power of the atomic reaction. And that is just if two neutrons create fission reactions each round. Perhaps you can now see why so many people consider atomic energy to be an exciting energy source! To make a self-sustained nuclear fission reactor with 235U, it is necessary to slow down the neutrons. Water is very effective at this, since neutrons collide with protons in water molecules and lose energy. Figure 22.28 shows a schematic of a reactor design called the pressurized water reactor. 750 Chapter 22 • The Atom Figure 22.28 A pressurized water reactor is cleverly designed to control the fission of large amounts of 235U, while using the heat produced in the fission reaction to create steam for generating electrical energy. Control rods adjust neutron flux so that it is self-sustaining. In case the reactor overheats and boils the water away, the chain reaction terminates, because water is needed to slow down the neutrons. This inherent safety feature can be overwhelmed in extreme circumstances. Control rods containing nuclides that very strongly absorb neutrons are used to adjust neutron flux. To produce large amounts of power, reactors contain hundreds to thousands of critical masses, and the chain reaction easily becomes self-sustaining. Neutron flux must be carefully regulated to avoid
an out-of-control exponential increase in the rate of fission. Control rods help prevent overheating, perhaps even a meltdown or explosive disassembly. The water that is used to slow down neutrons, necessary to get them to induce fission in 235U, and achieve criticality, provides a negative feedback for temperature increase. In case the reactor overheats and boils the water to steam or is breached, the absence of water kills the chain reaction. Considerable heat, however, can still be generated by the reactor’s radioactive fission products. Other safety features, thus, need to be incorporated in the event of a loss of coolant accident, including auxiliary cooling water and pumps. Energies in Nuclear Fission The following are two interesting facts to consider: • The average fission reaction produces 200 MeV of energy. • If you were to measure the mass of the products of a nuclear reaction, you would find that their mass was slightly less than the mass of the original nucleus. How are those things possible? Doesn’t the fission reaction’s production of energy violate the conservation of energy? Furthermore, doesn’t the loss in mass in the reaction violate the conservation of mass? Those are important questions, and they can both be answered with one of the most famous equations in scientific history. 22.65 Recall that, according to Einstein’s theory, energy and mass are essentially the same thing. In the case of fission, the mass of the products is less than that of the reactants because the missing mass appears in the form of the energy released in the reaction, with a constant value of c2 Joules of energy converted for each kilogram of material. The value of c2 is substantial—from Einstein’s equation, the amount of energy in just 1 gram of mass would be enough to support the average U.S. citizen for more than 270 years! The example below will show you how a mass-energy transformation of this type takes place. Access for free at openstax.org. 22.4 • Nuclear Fission and Fusion 751 WORKED EXAMPLE Calculating Energy from a Kilogram of Fissionable Fuel Calculate the amount of energy produced by the fission of 1.00 kg of, given the average fission reaction of 22.66 Strategy The total energy produced is the number of number of atoms in 1.00 kg. atoms times the given energy per fission. We should therefore find the Solution The number of
g; thus, there are atoms in 1.00 kg is Avogadro’s number times the number of moles. One mole of has a mass of 235.04. The number of atoms is therefore So the total energy released is Discussion The result is another impressively large amount of energy, equivalent to about 14,000 barrels of crude oil or 600,000 gallons of gasoline. But, it is only one fourth the energy produced by the fusion of a kilogram of a mixture of deuterium and tritium. Even though each fission reaction yields about ten times the energy of a fusion reaction, the energy per kilogram of fission fuel is less, because there are far fewer moles per kilogram of the heavy nuclides. Fission fuel is also much scarcer than fusion fuel, and less than 1 percent of uranium (the 235 U) is readily usable. 22.67 Virtual Physics Nuclear Fission Click to view content (https://www.openstax.org/l/16fission) Start a chain reaction, or introduce nonradioactive isotopes to prevent one. Use the applet to control energy production in a nuclear reactor! Nuclear Fusion Nuclear fusion is defined as the combining, or fusing, of two nuclei and, the combining of nuclei also results in an emission of energy. For many, the concept is counterintuitive. After all, if energy is released when a nucleus is split, how can it also be released when nucleons are combined together? The difference between fission and fusion, which results from the size of the nuclei involved, will be addressed next. Remember that the structure of a nucleus is based on the interplay of the compressive nuclear strong force and the repulsive electromagnetic force. For nuclei that are less massive than iron, the nuclear force is actually stronger than that of the Coulomb force. As a result, when a low-mass nucleus absorbs nucleons, the added neutrons and protons bind the nucleus more tightly. The increased nuclear strong force does work on the nucleus, and energy is released. Once the size of the created nucleus exceeds that of iron, the short-ranging nuclear force does not have the ability to bind a nucleus more tightly, and the emission of energy ceases. In fact, for fusion to occur for elements of greater mass than iron, energy must be added to the system! Figure 22.29 shows an energy-mass curve commonly used to describe nuclear reactions. Notice the location of iron (
Fe) on the graph. All low-mass nuclei to the left of iron release energy through fusion, while all highmass particles to the right of iron produce energy through fission. 752 Chapter 22 • The Atom Figure 22.29 Fusion of light nuclei to form medium-mass nuclei converts mass to energy, because binding energy per nucleon ( ) is greater for the product nuclei. The larger is, the less mass per nucleon, and so mass is converted to energy and released in such fusion reactions. TIPS FOR SUCCESS Just as it is not possible for the elements to the left of iron in the figure to naturally fission, it is not possible for elements to the right of iron to naturally undergo fusion, as that process would require the addition of energy to occur. Furthermore, notice that elements commonly discussed in fission and fusion are elements that can provide the greatest change in binding energy, such as uranium and hydrogen. Iron’s location on the energy-mass curve is important, and explains a number of its characteristics, including its role as an elemental endpoint in fusion reactions in stars. The major obstruction to fusion is the Coulomb repulsion force between nuclei. Since the attractive nuclear force that can fuse nuclei together is short ranged, the repulsion of like positive charges must be overcome in order to get nuclei close enough to induce fusion. Figure 22.30 shows an approximate graph of the potential energy between two nuclei as a function of the distance between their centers. The graph resembles a hill with a well in its center. A ball rolled to the left must have enough kinetic energy to get over the hump before it falls into the deeper well with a net gain in energy. So it is with fusion. If the nuclei are given enough kinetic energy to overcome the electric potential energy due to repulsion, then they can combine, release energy, and fall into a deep well. One way to accomplish that end is to heat fusion fuel to high temperatures so that the kinetic energy of thermal motion is sufficient to get the nuclei together. Figure 22.30 Potential energy between two light nuclei graphed as a function of distance between them. If the nuclei have enough kinetic energy to get over the Coulomb repulsion hump, they combine, release energy, and drop into a deep attractive well. You might think that, in our Sun, nuclei are constantly coming into contact and fusing. However, this is only partially true. Only at the Sun’s core are
the particles close enough and the temperature high enough for fusion to occur! In the series of reactions below, the Sun produces energy by fusing protons, or hydrogen nuclei (, by far the Sun’s most Access for free at openstax.org. 22.4 • Nuclear Fission and Fusion 753 abundant nuclide) into helium nuclei cycle. The principal sequence of fusion reactions forms what is called the proton-proton 22.68 stands for a positron and where the first two reactions must occur twice for the third to be possible, so the cycle consumes six protons ( Furthermore, the two positrons produced will find two electrons and annihilate to form four more overall cycle is thus is an electron neutrino. The energy in parentheses is releasedby the reaction. Note that ) but gives back two. rays, for a total of six. The where the 26.7 MeV includes the annihilation energy of the positrons and electrons and is distributed among all the reaction products. The solar interior is dense, and the reactions occur deep in the Sun where temperatures are highest. It takes about 32,000 years for the energy to diffuse to the surface and radiate away. However, the neutrinos can carry their energy out of the Sun in less than two seconds, because they interact so weakly with other matter. Negative feedback in the Sun acts as a thermostat to regulate the overall energy output. For instance, if the interior of the Sun becomes hotter than normal, the reaction rate increases, producing energy that expands the interior. The expansion cools it and lowers the reaction rate. Conversely, if the interior becomes too cool, it contracts, increasing the temperature and therefore the reaction rate (see Figure 22.31). Stars like the Sun are stable for billions of years, until a significant fraction of their hydrogen has been depleted. Figure 22.31 Nuclear fusion in the Sun converts hydrogen nuclei into helium; fusion occurs primarily at the boundary of the helium core, where the temperature is highest and sufficient hydrogen remains. Energy released diffuses slowly to the surface, with the exception of neutrinos, which escape immediately. Energy production remains stable because of negative-feedback effects. Nuclear Weapons and Nuclear Power The world was in political turmoil when fission was discovered in 1938. Compounding the troubles, the possibility of a selfsustained chain reaction was immediately recognized by leading scientists the world over. The enormous energy known to be in nuclei, but considered inaccessible,
now seemed to be available on a large scale. Within months after the announcement of the discovery of fission, Adolf Hitler banned the export of uranium from newly occupied Czechoslovakia. It seemed that the possible military value of uranium had been recognized in Nazi Germany, and that a serious effort to build a nuclear bomb had begun. Alarmed scientists, many of whom fled Nazi Germany, decided to take action. None was more famous or revered than Einstein. It was felt that his help was needed to get the American government to make a serious effort at constructing nuclear weapons as a matter of survival. Leo Szilard, a Hungarian physicist who had emigrated to America, took a draft of a letter to Einstein, who, although a pacifist, signed the final version. The letter was for President Franklin Roosevelt, warning of the German potential to build extremely powerful bombs of a new type. It was sent in August of 1939, just before the German invasion of Poland that marked the start of World War II. It was not until December 6, 1941, the day before the Japanese attack on Pearl Harbor, that the United States made a massive commitment to building a nuclear bomb. The top secret Manhattan Project was a crash program aimed at beating the Germans. 754 Chapter 22 • The Atom It was carried out in remote locations, such as Los Alamos, New Mexico, whenever possible, and eventually came to cost billions of dollars and employ the efforts of more than 100,000 people. J. Robert Oppenheimer (1904–1967), a talented physicist, was chosen to head the project. The first major step was made by Enrico Fermi and his group in December 1942, when they completed the first self-sustaining nuclear reactor. This first atomic pile, built in a squash court at the University of Chicago, proved that a fission chain reaction was possible. Plutonium was recognized as easier to fission with neutrons and, hence, a superior fission material very early in the Manhattan Project. Plutonium availability was uncertain, and so a uranium bomb was developed simultaneously. Figure 22.32 shows a guntype bomb, which takes two subcritical uranium masses and shoots them together. To get an appreciable yield, the critical mass must be held together by the explosive charges inside the cannon barrel for a few microseconds. Since the buildup of the uranium chain reaction is relatively slow, the device to bring the critical mass together can be relatively simple. Owing to the fact that the rate of spontaneous fission is low,
a neutron source is at the center the assembled critical mass. Figure 22.32 A gun-type fission bomb for utilizes two subcritical masses forced together by explosive charges inside a cannon barrel. The energy yield depends on the amount of uranium and the time it can be held together before it disassembles itself. Plutonium’s special properties necessitated a more sophisticated critical mass assembly, shown schematically in Figure 22.33. A spherical mass of plutonium is surrounded by shaped charges (high explosives that focus their blast) that implode the plutonium, crushing it into a smaller volume to form a critical mass. The implosion technique is faster and more effective, because it compresses three-dimensionally rather than one-dimensionally as in the gun-type bomb. Again, a neutron source is included to initiate the chain reaction. Figure 22.33 An implosion created by high explosives compresses a sphere of 239Pu into a critical mass. The superior fissionability of plutonium has made it the preferred bomb material. Access for free at openstax.org. Owing to its complexity, the plutonium bomb needed to be tested before there could be any attempt to use it. On July 16, 1945, the test named Trinity was conducted in the isolated Alamogordo Desert in New Mexico, about 200 miles south of Los Alamos (see Figure 22.34). A new age had begun. The yield of the Trinity device was about 10 kilotons (kT), the equivalent of 5,000 of the largest conventional bombs. 22.4 • Nuclear Fission and Fusion 755 Figure 22.34 Trinity test (1945), the first nuclear bomb (credit: U.S. Department of Energy) Although Germany surrendered on May 7, 1945, Japan had been steadfastly refusing to surrender for many months, resulting large numbers of civilian and military casualties. Invasion plans by the Allies estimated a million casualties of their own and untold losses of Japanese lives. The bomb was viewed as a way to end the war. The first bomb used was a gun-type uranium bomb dropped on Hiroshima on August 6 by the United States. Its yield of about 15 kT destroyed the city and killed an estimated 80,000 people, with 100,000 more being seriously injured. The second bomb was an implosion-type plutonium bomb dropped on Nagasaki only three days later. Its 20-kT yield killed at least 50,000 people, something less than Hiroshima because of the hilly terrain and the fact that it was a
few kilometers off target. The Japanese were told that one bomb a week would be dropped until they surrendered unconditionally, which they did on August 14. In actuality, the United States had only enough plutonium for one more bomb, as yet unassembled. Knowing that fusion produces several times more energy per kilogram of fuel than fission, some scientists pursued the idea of constructing a fusion bomb. The first such bomb was detonated by the United States several years after the first fission bombs, on October 31, 1952, at Eniwetok Atoll in the Pacific Ocean. It had a yield of 10 megatons (MT), about 670 times that of the fission bomb that destroyed Hiroshima. The Soviet Union followed with a fusion device of its own in August 1953, and a weapons race, beyond the aim of this text to discuss, continued until the end of the Cold War. Figure 22.35 shows a simple diagram of how a thermonuclear bomb is constructed. A fission bomb is exploded next to fusion fuel in the solid form of lithium deuteride. Before the shock wave blows it apart, create tritium through the reaction. Additional fusion and fission fuels are enclosed in a dense shell of rays heat and compress the fuel, and neutrons. At the same time that the uranium shell reflects the neutrons back into the fuel to enhance its fusion, the fast-moving neutrons cause the plentiful and inexpensive to fission, part of what allows thermonuclear bombs to be so large. 756 Chapter 22 • The Atom Figure 22.35 This schematic of a fusion bomb (H-bomb) gives some idea of how the fission trigger is used to ignite fusion fuel. Neutrons and γrays transmit energy to the fusion fuel, create tritium from deuterium, and heat and compress the fusion fuel. The outer shell of serves to reflect some neutrons back into the fuel, causing more fusion, and it boosts the energy output by fissioning itself when neutron energies become high enough. Of course, not all applications of nuclear physics are as destructive as the weapons described above. Hundreds of nuclear fission power plants around the world attest to the fact that controlled fission is both practical and economical. Given growing concerns over global warming, nuclear power is often seen as a viable alternative to energy derived from fossil fuels. BOUNDLESS PHYSICS Fusion Reactors For decades, fusion reactors have been deemed the energy of the future. A safer, cleaner, and more abundant potential source of
energy than its fission counterpart, images of the fusion reactor have been conjured up each time the need for a renewable, environmentally friendly resource is discussed. Now, after more than half a century of speculating, some scientists believe that fusion reactors are nearly here. In creating energy by combining atomic nuclei, the fusion reaction holds many advantages over fission. First, fusion reactions are more efficient, releasing 3 to 4 times more energy than fission per gram of fuel. Furthermore, unlike fission reactions that require heavyelements like uranium that are difficult to obtain, fusion requires lightelements that are abundant in nature. The greatest advantage of the fusion reaction, however, is in its ability to be controlled. While traditional nuclear reactors create worries about meltdowns and radioactive waste, neither is a substantial concern with the fusion reaction. Consider that fusion reactions require a large amount of energy to overcome the repulsive Coulomb force and that the byproducts of a fusion reaction are largely limited to helium nuclei. In order for fusion to occur, hydrogen isotopes of deuterium and tritium must be acquired. While deuterium can easily be gathered from ocean water, tritium is slightly more difficult to come by, though it can be manufactured from Earth’s abundant lithium. Once acquired, the hydrogen isotopes are injected into an empty vessel and subjected to temperature and pressure great enough to mimic the conditions at the core of our Sun. Using carefully controlled high-frequency radio waves, the hydrogen isotopes are broken into plasma and further controlled through an electromagnetic field. As the electromagnetic field continues to exert pressure on the hydrogen plasma, enough energy is supplied to cause the hydrogen plasma to fuse into helium. Access for free at openstax.org. 22.5 • Medical Applications of Radioactivity: Diagnostic Imaging and Radiation 757 Figure 22.36 Tokamak confinement of nuclear fusion plasma. The magnetic field lines are used to confine the high-temperature plasma (purple). Research is currently being done to increase the efficiency of the tokamak confinement model. Once the plasma fuses, high-velocity neutrons are ejected from the newly formed helium atoms. Those high velocity neutrons, carrying the excess energy stored within bonds of the original hydrogen, are able to travel unaffected by the applied magnetic field. In doing so, they strike a barrier around the nuclear reactor, transforming their excess energy to heat. The heat is then harvested to make steam that drives turbines. Hydrogen’s tremendous power
is now usable! The historical concern with nuclear fusion reactors is that the energy required to control the electromagnetic field is greater than the energy harvested from the hydrogen atoms. However, recent research by both Lockheed Martin engineers and scientists at the Lawrence Livermore National Laboratory has yielded exciting theoretical improvements in efficiency. At the time of this writing, a test facility called ITER (International Thermonuclear Experimental Reactor) is being constructed in southern France. A joint venture of the European Union, the United States, Japan, Russia, China, South Korea, and India, ITER is designed for further study into the future of nuclear fusion energy production. 22.5 Medical Applications of Radioactivity: Diagnostic Imaging and Radiation Section Learning Objectives By the end of this section, you will be able to do the following: • Describe how nuclear imaging works (e.g., radioisotope imaging, PET) • Describe the ionizing effects of radiation and how they can be used for medical treatment Section Key Terms Anger camera rad radiopharmaceutical therapeutic ratio relative biological effectiveness (RBE) roentgen equivalent man (rem) tagged Medical Applications of Nuclear Physics Applications of nuclear physics have become an integral part of modern life. From the bone scan that detects one cancer to the radioiodine treatment that cures another, nuclear radiation has diagnostic and therapeutic effects on medicine. Medical Imaging A host of medical imaging techniques employ nuclear radiation. What makes nuclear radiation so useful? First, radiation can 758 Chapter 22 • The Atom easily penetrate tissue; hence, it is a useful probe to monitor conditions inside the body. Second, nuclear radiation depends on the nuclide and not on the chemical compound it is in, so that a radioactive nuclide can be put into a compound designed for specific purposes. When that is done, the compound is said to be tagged. A tagged compound used for medical purposes is called a radiopharmaceutical. Radiation detectors external to the body can determine the location and concentration of a radiopharmaceutical to yield medically useful information. For example, certain drugs are concentrated in inflamed regions of the body, and their locations can aid diagnosis and treatment as seen in Figure 22.37. Another application utilizes a radiopharmaceutical that the body sends to bone cells, particularly those that are most active, to detect cancerous tumors or healing points. Images can then be produced of such bone scans. Clever use of radioisotopes determines the functioning of body organs, such as blood flow, heart muscle activity, and iodine uptake in
the thyroid gland. For instance, a radioactive form of iodine can be used to monitor the thyroid, a radioactive thallium salt can be used to follow the blood stream, and radioactive gallium can be used for cancer imaging. Figure 22.37 A radiopharmaceutical was used to produce this brain image of a patient with Alzheimer’s disease. Certain features are computer enhanced. (credit: National Institutes of Health) Once a radioactive compound has been ingested, a device like that shown in Figure 22.38 is used to monitor nuclear activity. The device, called an Anger camera or gamma camera uses a piece of lead with holes bored through it. The gamma rays are redirected through the collimator to narrow their beam, and are then interpreted using a device called a scintillator. The computer analysis of detector signals produces an image. One of the disadvantages of this detection method is that there is no depth information (i.e., it provides a two-dimensional view of the tumor as opposed to a three-dimensional view), because radiation from any location under that detector produces a signal. Figure 22.38 An Anger or gamma camera consists of a lead collimator and an array of detectors. Gamma rays produce light flashes in the scintillators. The light output is converted to an electrical signal by the photomultipliers. A computer constructs an image from the detector output. Single-photon-emission computer tomography (SPECT) used in conjunction with a CT scanner improves on the process carried out by the gamma camera. Figure 22.39 shows a patient in a circular array of SPECT detectors that may be stationary or rotated, with detector output used by a computer to construct a detailed image. The spatial resolution of this technique is poor, but the three-dimensional image created results in a marked improvement in contrast. Access for free at openstax.org. 22.5 • Medical Applications of Radioactivity: Diagnostic Imaging and Radiation 759 Figure 22.39 SPECT uses a rotating camera to form an image of the concentration of a radiopharmaceutical compound. (credit: Woldo, Wikimedia Commons) Positron emission tomography (or PET) scans utilize images produced by encounters an electron, mutual annihilation occurs, producing two γrays. Those energy comes from the destruction of an electron or positron mass) and they move directly away from each other, allowing detectors to determine their point of origin accurately (as shown in Figure 22.40). It requires detectors on opposite
sides to simultaneously (i.e., at the same time) detect photons of 0.511 MeV energy and utilizes computer imaging techniques similar to those in SPECT and CT scans. PET is used extensively for diagnosing brain disorders. It can note decreased metabolism in certain regions that accompany Alzheimer’s disease. PET can also locate regions in the brain that become active when a person carries out specific activities, such as speaking, closing his or her eyes, and so on. rays have identical 0.511 MeV energies (the emitters. When the emitted positron β+ Figure 22.40 A PET system takes advantage of the two identical -ray photons produced by positron-electron annihilation. The rays are emitted in opposite directions, so that the line along which each pair is emitted is determined. Various events detected by several pairs of detectors are then analyzed by the computer to form an accurate image. Ionizing Radiation on the Body We hear many seemingly contradictory things about the biological effects of ionizing radiation. It can cause cancer, burns, and hair loss, and yet it is used to treat and even cure cancer. How do we understand such effects? Once again, there is an underlying simplicity in nature, even in complicated biological organisms. All the effects of ionizing radiation on biological tissue can be understood by knowing that ionizing radiation affects molecules within cells, particularly DNA molecules. Let us take a brief look at molecules within cells and how cells operate. Cells have long, double-helical DNA molecules containing chemical patterns called genetic codes that govern the function and processes undertaken by the cells. Damage to DNA consists of breaks in chemical bonds or other changes in the structural features of the DNA chain, leading to changes in the genetic code. In human cells, we can have as many as a million individual instances of damage to DNA per cell per day. The repair ability of DNA is vital for maintaining the integrity of the genetic code and for the normal functioning of the entire organism. A cell with a damaged ability to repair DNA, which could have been induced by ionizing radiation, can do one of the following: • The cell can go into an irreversible state of dormancy, known as senescence. • The cell can commit suicide, known as programmed cell death. • The cell can go into unregulated cell division, leading to tumors and cancers. 760 Chapter 22 • The Atom Since ionizing radiation damages the DNA, ionizing radiation has its greatest effect on cells that rapidly reproduce, including most types of cancer. Thus, cancer cells are
more sensitive to radiation than normal cells and can be killed by it easily. Cancer is characterized by a malfunction of cell reproduction, and can also be caused by ionizing radiation. There is no contradiction to say that ionizing radiation can be both a cure and a cause. Radiotherapy Radiotherapy is effective against cancer because cancer cells reproduce rapidly and, consequently, are more sensitive to radiation. The central problem in radiotherapy is to make the dose for cancer cells as high as possible while limiting the dose for normal cells. The ratio of abnormal cells killed to normal cells killed is called the therapeutic ratio, and all radiotherapy techniques are designed to enhance that ratio. Radiation can be concentrated in cancerous tissue by a number of techniques. One of the most prevalent techniques for well-defined tumors is a geometric technique shown in Figure 22.41. A narrow beam of radiation is passed through the patient from a variety of directions with a common crossing point in the tumor. The technique concentrates the dose in the tumor while spreading it out over a large volume of normal tissue. Figure 22.41 The 60Co source of -radiation is rotated around the patient so that the common crossing point is in the tumor, concentrating the dose there. This geometric technique works for well-defined tumors. Another use of radiation therapy is through radiopharmaceuticals. Cleverly, radiopharmaceuticals are used in cancer therapy by tagging antibodies with radioisotopes. Those antibodies are extracted from the patient, cultured, loaded with a radioisotope, and then returned to the patient. The antibodies are then concentrated almost entirely in the tissue they developed to fight, thus localizing the radiation in abnormal tissue. This method is used with radioactive iodine to fight thyroid cancer. While the therapeutic ratio can be quite high for such short-range radiation, there can be a significant dose for organs that eliminate radiopharmaceuticals from the body, such as the liver, kidneys, and bladder. As with most radiotherapy, the technique is limited by the tolerable amount of damage to the normal tissue. Radiation Dosage To quantitatively discuss the biological effects of ionizing radiation, we need a radiation dose unit that is directly related to those effects. To do define such a unit, it is important to consider both the biological organism and the radiation itself. Knowing that the amount of ionization is proportional to the amount of deposited energy, we define a radiation dose unit called the rad. It 1/100 of a joule of ionizing
energy deposited per kilogram of tissue, which is For example, if a 50.0-kg person is exposed to ionizing radiation over her entire body and she absorbs 1.00 J, then her wholebody radiation dose is 22.69 If the same 1.00 J of ionizing energy were absorbed in her 2.00-kg forearm alone, then the dose to the forearm would be 22.70 22.71 and the unaffected tissue would have a zero rad dose. When calculating radiation doses, you divide the energy absorbed by the mass of affected tissue. You must specify the affected region, such as the whole body or forearm in addition to giving the numerical dose in rads. Although the energy per kilogram in 1 rad is small, it can still have significant effects. Since only a few eV cause ionization, just 0.01 J of ionizing energy can create a huge number of ion pairs and have an effect at the cellular level. The effects of ionizing radiation may be directly proportional to the dose in rads, but they also depend on the type of radiation and the type of tissue. That is, for a given dose in rads, the effects depend on whether the radiation is, X-ray, or some,, Access for free at openstax.org. 22.5 • Medical Applications of Radioactivity: Diagnostic Imaging and Radiation 761 other type of ionizing radiation. The relative biological effectiveness (RBE) relates to the amount of biological damage that can occur from a given type of radiation and is given in Table 22.4 for several types of ionizing radiation. Type and energy of radiation RBE X-rays rays rays greater than 32 keV rays less than 32 keV 1 1 1 1.7 Neutrons, thermal to slow (< 20 keV) 2–5 Neutrons, fast (1–10 MeV) 10 (body), 32 (eyes) Protons (1–10 MeV) 10 (body), 32 (eyes) rays from radioactive decay 10–20 Heavy ions from accelerators 10–20 Table 22.4 Relative Biological Effectiveness TIPS FOR SUCCESS The RBEs given in Table 22.4 are approximate, but they yield certain valuable insights. • The eyes are more sensitive to radiation, because the cells of the lens do not repair themselves. • Though both are neutral and have large ranges, neutrons cause more damage than rays because neutrons often cause secondary radiation when they are captured. • Short-range
particles such as rays have a severely damaging effect to internal anatomy, as their damage is concentrated and more difficult for the biological organism to repair. However, the skin can usually block alpha particles from entering the body. Can you think of any other insights from the table? A final dose unit more closely related to the effect of radiation on biological tissue is called the roentgen equivalent man, or rem. A combination of all factors mentioned previously, the roentgen equivalent man is defined to be the dose in rads multiplied by the relative biological effectiveness. The large-scale effects of radiation on humans can be divided into two categories: immediate effects and long-term effects. Table 22.5 gives the immediate effects of whole-body exposures received in less than one day. If the radiation exposure is spread out over more time, greater doses are needed to cause the effects listed. Any dose less than 10 rem is called a low dose, a dose 10 to 100 rem is called a moderate dose, and anything greater than 100 rem is called a high dose. 22.72 Dose (rem) Effect 0–10 No observable effect 10–100 Slight to moderate decrease in white blood cell counts Table 22.5 Immediate Effects of Radiation (Adults, Whole Body, Single Exposure) 762 Chapter 22 • The Atom Dose (rem) Effect 50 Temporary sterility 100–200 Significant reduction in blood cell counts, brief nausea, and vomiting; rarely fatal 200–500 Nausea, vomiting, hair loss, severe blood damage, hemorrhage, fatalities 450 LD50/32; lethal to 50% of the population within 32 days after exposure if untreated 500–2,000 Worst effects due to malfunction of small intestine and blood systems; limited survival > 2,000 Fatal within hours due to collapse of central nervous system Table 22.5 Immediate Effects of Radiation (Adults, Whole Body, Single Exposure) WORK IN PHYSICS Health Physicist Are you interested in learning more about radiation? Are you curious about studying radiation dosage levels and ensuring the safety of the environment and people that are most closely affected by it? If so, you may be interested in becoming a health physicist. The field of health physics draws from a variety of science disciplines with the central aim of mitigating radiation concerns. Those that work as health physicists have a diverse array of potential jobs available to them, including those in research, industry, education, environmental protection, and governmental regulation. Furthermore, while the term health physicistmay lead many to think of the medical field,
there are plenty of applications within the military, industrial, and energy fields as well. As a researcher, a health physicist can further environmental studies on the effects of radiation, design instruments for more accurate measurements, and assist in establishing valuable radiation standards. Within the energy field, a health physicsist often acts as a manager, closely tied to all operations at all levels, from procuring appropirate equipment to monitoring health data. Within industry, the health physicist acts as a consultant, assisting industry management in important decisions, designing facilities, and choosing appropriate detection tools. The health physicist possesses a unique knowledge base that allows him or her to operate in a wide variety of interesting disciplines! To become a health physicist, it is necessary to have a background in the physical sciences. Understanding the fields of biology, physiology, biochemistry, and genetics are all important as well. The ability to analyze and solve new problems is critical, and a natural aptitude for science and mathematics will assist in the continued necessary training. There are two possible certifications for health physicists: from the American Board of Health Physicists (ABHP) and the National Registry of Radiation Protection Technologists (NRRPT). Access for free at openstax.org. Chapter 22 • Key Terms 763 KEY TERMS activity rate of decay for radioactive nuclides alpha decay type of radioactive decay in which an atomic nucleus emits an alpha particle anger camera common medical imaging device that uses a scintillator connected to a series of photomultipliers atomic number number of protons in a nucleus becquerel SI unit for rate of decay of a radioactive material beta decay type of radioactive decay in which an atomic nucleus emits a beta particle nuclear fusion reaction in which two nuclei are combined, or fused, to form a larger nucleus nucleons particles found inside nuclei planetary model of the atom model of the atom that shows electrons orbiting like planets about a Sun-like nucleus proton-proton cycle combined reactions and that begins with carbon-14 dating radioactive dating technique based on the hydrogen and ends with helium radioactivity of carbon-14 rad amount of ionizing energy deposited per kilogram of chain reaction self-sustaining sequence of events, tissue exemplified by the self-sustaining nature of a fission reaction at critical mass radioactive substance or object that emits nuclear radiation critical mass minimum amount necessary for self- radioactive dating application of radioactive decay in sustained fission of a given nuclide decay constant quantity that is inversely proportional to the half-life and that is
absorbed as the electrons decrease and increase in orbital energy. • The energy emitted or absorbed by an electron as it changes energy state can be determined with the equation, where 764 Chapter 22 • Key Equations. • The wavelength of energy absorbed or emitted by an electron as it changes energy state can be determined by the equation, where. • Described as an electron cloud, the quantum model of the atom is the result of de Broglie waves and Heisenberg’s uncertainty principle. 22.2 Nuclear Forces and Radioactivity • The structure of the nucleus is defined by its two nucleons, the neutron and proton. • Atomic numbers and mass numbers are used to differentiate between various atoms and isotopes. Those numbers can be combined into an easily recognizable form called a nuclide. • The size and stability of the nucleus is based upon two forces: the electromagnetic force and strong nuclear force. • Radioactive decay is the alteration of the nucleus through the emission of particles or energy. • Alpha decay occurs when too many protons exist in the nucleus. It results in the ejection of an alpha particle, as described in the equation. • Beta decay occurs when too many neutrons (or protons) exist in the nucleus. It results in the transmutation of a neutron into a proton, electron, and neutrino. The decay is expressed through the equation. (Beta decay may also transform a proton into a neutron.) • Gamma decay occurs when a nucleus in an excited state move to a more stable state, resulting in the release of a photon. Gamma decay is represented with the equation. • The penetration distance of radiation depends on its energy, charge, and type of material it encounters. 22.3 Half Life and Radiometric Dating • Radioactive half-life is the time it takes a sample of nuclei to decay to half of its original amount. • The rate of radioactive decay is defined as the sample’s KEY EQUATIONS 22.1 The Structure of the Atom energy of hydrogen electron in an orbital Access for free at openstax.org. activity, represented by the equation. • Knowing the half-life of a radioactive isotope allows for the process of radioactive dating to determine the age of a material. If the half-life of a material is known, the age of the material can be found using the equation. • • The age of organic material can be determined using the decay of the carbon-14 isotope, while the age of rocks can be determined using the
decay of uranium-238. 22.4 Nuclear Fission and Fusion • Nuclear fission is the splitting of an atomic bond, releasing a large amount of potential energy previously holding the atom together. The amount of energy released can be determined through the equation. • Nuclear fusion is the combining, or fusing together, of two nuclei. Energy is also released in nuclear fusion as the combined nuclei are closer together, resulting in a decreased strong nuclear force. • Fission was used in two nuclear weapons at the conclusion of World War II: the gun-type uranium bomb and the implosion-type plutonium bomb. • While fission has been used in both nuclear weapons and nuclear reactors, fusion is capable of releasing more energy per reaction. As a result, fusion is a wellresearched, if not yet well-controlled, energy source. 22.5 Medical Applications of Radioactivity: Diagnostic Imaging and Radiation • Medical imaging occurs when a radiopharmaceutical placed in the body provides information to an array of radiation detectors outside the body. • Devices utilizing medical imaging include the Anger • camera, SPECT detector, and PET scan. Ionizing radiation can both cure and cause cancer through the manipulation of DNA molecules. • Radiation dosage and its effect on the body can be measured using the quantities radiation dose unit (rad), relative biological effectiveness (RBE), and the roentgen equivalent man (rem). energy of any hydrogen-like electron in orbital wavelength of light emitted by an electron changing states wavelength of an orbital heisenberg’s uncertainty principle 22.2 Nuclear Forces and Radioactivity alpha decay equation beta decay equation gamma decay equation CHAPTER REVIEW Concept Items 22.1 The Structure of the Atom 1. A star emits light from its core. One observer views the emission unobstructed while a second observer views the emission while obstructed by a cloud of hydrogen gas. Describe the difference between their observations. a. b. c. Frequencies will be absorbed from the spectrum. d. Frequencies will be added to the spectrum. Intensity of the light in the spectrum will increase. Intensity of the light in the spectrum will decrease. 2. How does the orbital energy of a hydrogen-like atom change as it increases in atomic number? Critical Thinking Items 22.1 The Structure of the Atom 4. How would the gold foil experiment have changed if electrons were used in place of alpha particles, assuming that the electrons hit the gold foil with the same force as the alpha particles? a. Being less massive,
the electrons might have been scattered to a greater degree than the alpha particles. b. Being less massive, the electrons might have been scattered to a lesser degree than the alpha particles. Chapter 22 • Chapter Review 765 22.3 Half Life and Radiometric Dating radioactive half-life 22.4 Nuclear Fission and Fusion energy–mass conversion protonproton chain 22.5 Medical Applications of Radioactivity: Diagnostic Imaging and Radiation roentgen equivalent man a. The orbital energy will increase. b. The orbital energy will decrease. c. The orbital energy will remain constant. d. The orbital energy will be halved. 22.4 Nuclear Fission and Fusion 3. Aside from energy yield, why are nuclear fusion reactors more desirable than nuclear fission reactors? a. Nuclear fusion reactors have a low installation cost. b. Radioactive waste is greater for a fusion reactor. c. Nuclear fusion reactors are easy to design and build. d. A fusion reactor produces less radioactive waste. c. Being more massive, the electrons would have been scattered to a greater degree than the alpha particles. d. Being more massive, the electrons would have been scattered to a lesser degree than the alpha particles. 5. Why does the emission spectrum of an isolated gas differ from the emission spectrum created by a white light? a. White light and an emission spectrum are different varieties of continuous distribution of frequencies. b. White light and an emission spectrum are different series of discrete frequencies. 766 Chapter 22 • Chapter Review c. White light is a continuous distribution of frequencies, and an emission spectrum is a series of discrete frequencies. d. White light is a series of discrete frequencies, and an emission spectrum is a continuous distribution of frequencies. 6. Why would it most likely be difficult to observe quantized orbital states for satellites orbiting the earth? a. On a macroscopic level, the orbital states do exist for satellites orbiting Earth but are too closely spaced for us to see. d. While the alpha particle has a greater charge than a beta particle, the electron density in lead is much higher than that in air. 10. What influence does the strong nuclear force have on the electrons in an atom? a. b. c. The strong force makes electrons revolve around It attracts them toward the nucleus. It repels them away from the nucleus. the nucleus. It does not have any influence. d. b. On a macroscopic level, the orbital states do not 22.3 Half Life and Radiometric Dating exist for satellites
orbiting Earth. c. On a macroscopic level, we cannot control the amount of energy that we give to an artificial satellite and thus control its orbital altitude. d. On a macroscopic level, we cannot control the amount of energy that we give to an artificial satellite but we can control its orbital altitude. 7. Do standing waves explain why electron orbitals are quantized? a. no b. yes 8. Some terms referring to the observation of light include emission spectrumand absorption spectrum. Based on these definitions, what would a reflection spectrum describe? a. The reflection spectrum would describe when incident waves are selectively reflected by a substance. b. The reflection spectrum would describe when incident waves are completely reflected by a substance. c. The reflection spectrum would describe when incident waves are not absorbed by a substance. d. The reflection spectrum would describe when incident waves are completely absorbed by a substance. 22.2 Nuclear Forces and Radioactivity 9. Explain why an alpha particle can have a greater range in air than a beta particle in lead. a. While the alpha particle has a lesser charge than a beta particle, the electron density in lead is much less than that in air. b. While the alpha particle has a greater charge than a beta particle, the electron density in lead is much lower than that in air. c. While the alpha particle has a lesser charge than a beta particle, the electron density in lead is much greater than that in air. Access for free at openstax.org. 11. Provide an example of something that decreases in a manner similar to radioactive decay. a. The potential energy of an object falling under the influence of gravity b. The kinetic energy of a ball that is dropped from a building to the ground c. Theh charge transfer from an ebonite rod to fur d. The heat transfer from a hot to a cold object 12. A sample of radioactive material has a decay constant of 0.05 s–1. Why is it wrong to presume that the sample will take just 20 seconds to fully decay? a. The decay constant varies with the mass of the sample. b. The decay constant results vary with the amount of the sample. c. The decay constant represents a percentage of the sample that cannot decay. d. The decay constant represents only the fraction of a sample that decays in a unit of time, not the decay of the entire sample. 22.4 Nuclear Fission and Fusion 13. What is the atomic number of the most strongly bound nuclide? a
. b. c. d. 14. Why are large electromagnets necessary in nuclear fusion reactors? a. Electromagnets are used to slow down the movement of charge hydrogen plasma. b. Electromagnets are used to decrease the temperature of hydrogen plasma. c. Electromagnets are used to confine the hydrogen plasma. d. Electromagnets are used to stabilize the temperature of the hydrogen plasma. 22.5 Medical Applications of Radioactivity: Diagnostic Imaging and Radiation 15. Why are different radiopharmaceuticals used to image different parts of the body? a. The different radiopharmaceuticals travel through different blood vessels. b. The different radiopharmaceuticals travel to different parts of the body. c. The different radiopharmaceuticals are used to treat different diseases of the body. d. The different radiopharmaceuticals produce different amounts of ionizing radiation. 16. Why do people think carefully about whether to receive a diagnostic test such as a CT scan? a. The radiation from a CT scan is capable of creating cancerous cells. Performance Task 22.5 Medical Applications of Radioactivity: Diagnostic Imaging and Radiation 18. On the Environmental Protection Agency’s website, a helpful tool exists to allow you to determine your average annual radiation dose. Use the tool to determine whether the radiation level you have been exposed to is dangerous and to compare your radiation dosage to other radiative events. 1. Visit the webpage (http://www.openstax.org/l/ 28calculate) and answer the series of questions provided to determine the average annual radiation dosage that you receive. 2. Table 22.5 shows the immediate effects of a radiation dosage. Using the table, explain what you would experience if your yearly dosage of radiation was received all over the course of one day. Also, determine whether your dosage is considered a low, moderate, or high. 3. Using the information input into the webpage, what percentage of your dosage comes from TEST PREP Multiple Choice 22.1 The Structure of the Atom 19. If electrons are negatively charged and the nucleus is positively charged, why do they not attract and collide with each other? Chapter 22 • Test Prep 767 b. The radiation from a CT scan is capable of destroying cancerous cells. c. The radiation from a CT scan is capable of creating diabetic cells. d. The radiation from a CT scan is capable of destroying diabetic cells. 17. Sometimes it is necessary to take a PET scan
very soon after ingesting a radiopharmaceutical. Why is that the case? a. The radiopharmaceutical may have a short half-life. b. The radiopharmaceutical may have a long half-life. c. The radiopharmaceutical quickly passes through the digestive system. d. The radiopharmaceutical can become lodged in the digestive system. natural sources? The average percentage of radiation from natural sources for an individual is around 85 percent. 4. Research radiation dosages for evacuees from events like the Chernobyl and Fukushima meltdowns. How does your annual radiation exposure rate compare to the net dosage for evacuees of each event. Use numbers to support your answer. 5. The U.S. Department of Labor limits the amount of radiation that a given worker may receive in a 12 month period. a. Research the present maximum value and compare your annual exposure rate to that of a radiation worker. Use numbers to support your answer. b. What types of work are likely to cause an increase in the radiation exposure of a particular worker? Provide one question based upon the information gathered on the EPA website. a. The pull from the nucleus provides a centrifugal force, which is not strong enough to draw the electrons into the nucleus. b. The pull from the nucleus provides a centripetal force, which is not strong enough to draw the electrons into the nucleus. 768 Chapter 22 • Test Prep c. The pull from the nucleus provides a helical motion. d. The pull from the nucleus provides a cycloid b. Coulomb force between protons only c. Strong nuclear force between all nucleons and motion. 22.4 Nuclear Fission and Fusion 20. If a nucleus elongates due to a neutron strike, which of the following forces will decrease? a. Nuclear force between neutrons only Coulomb force between protons, but the strong force will decrease more d. Strong nuclear force between neutrons and Coulomb force between protons, but Coulomb force will decrease more Short Answer 22.1 The Structure of the Atom a. The radioactive activity decreases exponentially. b. The radioactive activity undergoes linear decay. c. The radioactive activity undergoes logarithmic 21. Why do Bohr’s calculations for electron energies not decay. work for all atoms? a. In atoms with more than one electron is an atomic shell, the electrons will interact. That requires a more complex formula than Bohr’s calculations accounted for. In atoms with 10 or more
electorns in an atomic shell, the electrons will interact. That requires a more complex formula than Bohr’s calculations accounted for. In atoms with more than one electron in an atomic shell, the electrons will not interact. That requires a more complex formula than Bohr’s calculations accounted for. In atoms with 10 or more electrons in an atomic shell, the electrons will not interact. That requires a more complex formula than Bohr’s calculations accounted for. b. c. d. 22.2 Nuclear Forces and Radioactivity 22. Does transmutation occur within chemical reactions? a. no b. yes 22.3 Half Life and Radiometric Dating 23. How does the radioactive activity of a sample change with time? Extended Response 22.1 The Structure of the Atom 26. Compare the standing wavelength of an orbital to the standing wavelength of an a. The standing wavelength of an orbital. orbital is greater than the standing wavelength of an orbital. b. The standing wavelength of an than the standing wavelength of an orbital is less orbital. Access for free at openstax.org. d. The radioactive activity will not change with time. 22.4 Nuclear Fission and Fusion 24. Why does fission of heavy nuclei result in the release of neutrons? a. Heavy nuclei require more neutrons to achieve stability. b. Heavy nuclei require more neutrons to balance charge. c. Light nuclei require more neutrons to achieve stability. d. Light nuclei require more neutrons to balance charge. 22.5 Medical Applications of Radioactivity: Diagnostic Imaging and Radiation 25. Why is radioactive iodine used to monitor the thyroid? a. Radioactive iodine can be used by the thyroid while absorbing information about the thyroid. b. Radioactive iodine can be used by the thyroid while emitting information about the thyroid. c. Radioactive iodine can be secreted by the thyroid while absorbing information about the thyroid. d. Radioactive iodine can be secreted by the thyroid while emitting information about the thyroid. c. There is no relation between the standing wavelength of an wavelength of an orbital and the standing orbital. d. The standing wavelength of an orbital is the same as the standing wavelength of an orbital. 27. Describe the shape of the electron cloud, based on total energy levels, for an atom with electrons in multiple orbital states. a. There are multiple regions of high electron Chapter 22 • Test Prep 769 probability of various shapes surrounding the nucleus. b. There is a single solid spherical
region of high electron probability surrounding the nucleus. brought together and then rapidly increase once a minimum is reached. b. The potential energy will decrease as the nuclei are brought together. c. There are multiple concentric shells of high c. The potential energy will increase as the nuclei are electron probability surrounding the nucleus. brought together. d. There is a single spherical shell of high electron probability surrounding the nucleus. 22.2 Nuclear Forces and Radioactivity 28. How did Becquerel’s observations of pitchblende imply the existence of radioactivity? a. A chemical reaction occurred on the photographic plate without any external source of energy. b. Bright spots appeared on the photographic plate due to an external source of energy. c. Energy from the Sun was absorbed by the pitchblende and reflected onto the photographic plate. d. Dark spots appeared on the photographic plate due to an external source of energy. 22.4 Nuclear Fission and Fusion 29. Describe the potential energy of two nuclei as they approach each other. a. The potential energy will decrease as the nuclei are d. The potential energy will increase as the nuclei are brought together and then rapidly decrease once a maximum is reached. 22.5 Medical Applications of Radioactivity: Diagnostic Imaging and Radiation 30. Why do X-rays and gamma rays have equivalent RBE values if they provide different amounts of energy to the body? a. The penetration distance, which depends on energy, is short for both X-rays and gamma rays. b. The penetration distance, which depends on energy, is long for both X-rays and gamma rays. c. The penetration distance, as determined by their high mass, is different for both X-rays and gamma rays. d. The penetration distance, as determined by their low mass, is the same for both X-rays and gamma rays. 770 Chapter 22 • Test Prep Access for free at openstax.org. CHAPTER 23 Particle Physics Figure 23.1 Part of the Large Hadron Collider (LHC) at CERN, on the border of Switzerland and France. The LHC is a particle accelerator, designed to study fundamental particles. (credit: Image Editor, Flickr) Chapter Outline 23.1 The Four Fundamental Forces 23.2 Quarks 23.3 The Unification of Forces Following ideas remarkably similar to those of the ancient Greeks, we continue to look for smaller and INTRODUCTION smaller structures in nature, hoping ultimately to find and understand the most fundamental
building blocks that exist. Atomic physics deals with the smallest units of elements and compounds. In its study, we have found a relatively small number of atoms with systematic properties, and these properties have explained a tremendous range of phenomena. Nuclear physics is concerned with the nuclei of atoms and their substructures. Here, a smaller number of components—the proton and neutron—make up all nuclei. Exploring the systematic behavior of their interactions has revealed even more about matter, forces, and energy. Particle physics deals with the substructures of atoms and nuclei and is particularly aimed at finding those truly fundamental particles that have no further substructure. Just as in atomic and nuclear physics, we have found a complex array of particles and properties with systematic characteristics analogous to the periodic table and the chart of nuclides. An underlying structure is apparent, and there is some reason to think that we arefinding particles that have no substructure. Of course, we have been in similar situations before. For example, atoms were once thought to be the ultimate substructures. It is possible that we could continue to find deeper and deeper structures without ever discovering the ultimate substructure—in science there is never complete certainty. See Figure 23.2. The properties of matter are based on substructures called molecules and atoms. Each atom has the substructure of a nucleus surrounded by electrons, and their interactions explain atomic properties. Protons and neutrons—and the interactions between them—explain the stability and abundance of elements and form the substructure of nuclei. Protons and neutrons are not fundamental—they are composed of quarks. Like electrons and a few other particles, quarks may be the fundamental building blocks of all matter, lacking any further substructure. But the story is not complete because quarks and electrons may have substructures smaller than details that are presently observable. 772 Chapter 23 • Particle Physics Figure 23.2 A solid, a molecule, an atom, a nucleus, a nucleon (a particle that makes up the nucleus—either a proton or a neutron), and a quark. This chapter covers the basics of particle physics as we know it today. An amazing convergence of topics is evolving in particle physics. We find that some particles are intimately related to forces and that nature on the smallest scale may have its greatest influence on the large scale character of the universe. It is an adventure exceeding the best science fiction because it is not only fantastic but also real. 23.1 The Four Fundamental
Forces Section Learning Objectives By the end of the section, you will be able to do the following: • Define, describe, and differentiate the four fundamental forces • Describe the carrier particles and explain how their exchange transmits force • Explain how particle accelerators work to gather evidence about particle physics Section Key Terms carrier particle colliding beam cyclotron Feynman diagram graviton particle physics pion quantum electrodynamics synchrotron boson boson weak nuclear force boson Despite the apparent complexity within the universe, there remain just four basic forces. These forces are responsible for all interactions known to science: from the very small to the very large to those that we experience in our day-to-day lives. These forces describe the movement of galaxies, the chemical reactions in our laboratories, the structure within atomic nuclei, and the cause of radioactive decay. They describe the true cause behind familiar terms like friction and the normal force. These four basic forces are known as fundamental because they alone are responsible for all observations of forces in nature. The four fundamental forces are gravity, electromagnetism, weak nuclear force, and strong nuclear force. Understanding the Four Forces The gravitational force is most familiar to us because it describes so many of our common observations. It explains why a dropped ball falls to the ground and why our planet orbits the Sun. It gives us the property of weight and determines much about the motion of objects in our daily lives. Because gravitational force acts between all objects of mass and has the ability to act over large distances, the gravitational force can be used to explain much of what we observe and can even describe the motion of objects on astronomical scales! That said, gravity is incredibly weak compared to the other fundamental forces and is the weakest of all of the fundamental forces. Consider this: The entire mass of Earth is needed to hold an iron nail to the ground. Yet with a simple magnet, the force of gravity can be overcome, allowing the nail to accelerate upward through space. The electromagnetic force is responsible for both electrostatic interactions and the magnetic force seen between bar magnets. When focusing on the electrostatic relationship between two charged particles, the electromagnetic force is known as the coulomb force. The electromagnetic force is an important force in the chemical and biological sciences, as it is responsible for molecular connections like ionic bonding and hydrogen bonding. Additionally, the electromagnetic force is behind the common physics forces of friction and the normal force. Like the gravitational force, the electromagnetic force is an inverse square law. However, the electromagnetic
force does not exist between any two objects of mass, only those that are charged. When considering the structure of an atom, the electromagnetic force is somewhat apparent. After all, the electrons are held in place by an attractive force from the nucleus. But what causes the nucleus to remain intact? After all, if all protons are positive, it Access for free at openstax.org. 23.1 • The Four Fundamental Forces 773 makes sense that the coulomb force between the protons would repel the nucleus apart immediately. Scientists theorized that another force must exist within the nucleus to keep it together. They further theorized that this nuclear force must be significantly stronger than gravity, which has been observed and measured for centuries, and also stronger than the electromagnetic force, which would cause the protons to want to accelerate away from each other. The strong nuclear force is an attractive force that exists between all nucleons. This force, which acts equally between protonproton connections, proton-neutron connections, and neutron-neutron connections, is the strongest of all forces at short ranges. However, at a distance of 10–13 cm, or the diameter of a single proton, the force dissipates to zero. If the nucleus is large (it has many nucleons), then the distance between each nucleon could be much larger than the diameter of a single proton. The weak nuclear force is responsible for beta decay, as seen in the equation decay is when a beta particle is ejected from an atom. In order to accelerate away from the nucleus, the particle must be acted on by a force. Enrico Fermi was the first to envision this type of force. While this force is appropriately labeled, it remains stronger than the gravitational force. However, its range is even smaller than that of the strong force, as can be seen in Table 23.1. The weak nuclear force is more important than it may appear at this time, as will be addressed when we discuss quarks. Recall that beta Force Approximate Relative Strength[1] Range Gravity Weak Electromagnetic Strong 1 ∞ ∞ [1]Relative strength is based on the strong force felt by a proton–proton pair. Table 23.1 Relative strength and range of the four fundamental forces Transmitting the Four Fundamental Forces Just as it troubled Einstein prior to formulating the gravitational field theory, the concept of forces acting over a distance had greatly troubled particle physicists. That is, how does one pro
ton knowthat another exists? Furthermore, what causes one proton to make a second proton repel? Or, for that matter, what is it about a proton that causes a neutron to attract? These mysterious interactions were first considered by Hideki Yukawa in 1935 and laid the foundation for much of what we now understand about particle physics. Hideki Yukawa’s focus was on the strong nuclear force and, in particular, its incredibly short range. His idea was a blend of particles, relativity, and quantum mechanics that was applicable to all four forces. Yukawa proposed that the nuclear force is actually transmitted by the exchange of particles, called carrier particles, and that what we commonly refer to as the force’s field consists of these carrier particles. Specifically for the strong nuclear force, Yukawa proposed that a previously unknown particle, called a pion, is exchanged between nucleons, transmitting the force between them. Figure 23.3 illustrates how a pion would carry a force between a proton and a neutron. Figure 23.3 The strong nuclear force is transmitted between a proton and neutron by the creation and exchange of a pion. The pion, created through a temporary violation of conservation of mass-energy, travels from the proton to the neutron and is recaptured. It is not 774 Chapter 23 • Particle Physics directly observable and is called a virtual particle. Note that the proton and neutron change identity in the process. The range of the force is limited by the fact that the pion can exist for only the short time allowed by the Heisenberg uncertainty principle. Yukawa used the finite range of the strong nuclear force to estimate the mass of the pion; the shorter the range, the larger the mass of the carrier particle. In Yukawa’s strong force, the carrier particle is assumed to be transmitted at the speed of light and is continually transferred between the two nucleons shown. The particle that Yukawa predicted was finally discovered within cosmic rays in 1947. Its name, the pion, stands for pi meson, where meson means medium mass; it’s a medium mass because it is smaller than a nucleon but larger than an electron. Yukawa launched the field that is now called quantum chromodynamics, and the carrier particles are now called gluons due to their strong binding power. The reason for the change in the particle name will be explained when quarks are discussed later in this section. As you may assume, the strong
force is not the only force with a carrier particle. Nuclear decay from the weak force also requires a particle transfer. In the weak force are the following three: the weak negative carrier, W–; the weak positive carrier, W+; and the zero charge carrier, Z0. As we will see, Fermi inferred that these particles must carry mass, as the total mass of the products of nuclear decay is slightly larger than the total mass of all reactants after nuclear decay. The carrier particle for the electromagnetic force is, not surprisingly, the photon. After all, just as a lightbulb can emit photons from a charged tungsten filament, the photon can be used to transfer information from one electrically charged particle to another. Finally, the graviton is the proposed carrier particle for gravity. While it has not yet been found, scientists are currently looking for evidence of its existence (see Boundless Physics: Searching for the Graviton). So how does a carrier particle transmit a fundamental force? Figure 23.4 shows a virtual photon transmitted from one positively charged particle to another. The transmitted photon is referred to as a virtual particle because it cannot be directly observed while transmitting the force. Figure 23.5 shows a way of graphing the exchange of a virtual photon between the two positively charged particles. This graph of time versus position is called a Feynman diagram, after the brilliant American physicist Richard Feynman (1918–1988), who developed it. Figure 23.4 The image in part (a) shows the exchange of a virtual photon transmitting the electromagnetic force between charges, just as virtual pion exchange carries the strong nuclear force between nucleons. The image in part (b) shows that the photon cannot be directly observed in its passage because this would disrupt it and alter the force. In this case, the photon does not reach the other charge. The Feynman diagram should be read from the bottom up to show the movement of particles over time. In it, you can see that the left proton is propelled leftward from the photon emission, while the right proton feels an impulse to the right when the photon is received. In addition to the Feynman diagram, Richard Feynman was one of the theorists who developed the field of quantum electrodynamics (QED), which further describes electromagnetic interactions on the submicroscopic scale. For this work, he shared the 1965 Nobel Prize with Julian Schwinger and S.I. Tomonaga. A Feynman diagram explaining
the strong force interaction hypothesized by Yukawa can be seen in Figure 23.6. Here, you can see the change in particle type due to the exchange of the pi meson. Access for free at openstax.org. 23.1 • The Four Fundamental Forces 775 Figure 23.5 The Feynman diagram for the exchange of a virtual photon between two positively charged particles illustrates how electromagnetic force is transmitted on a quantum mechanical scale. Time is graphed vertically, while the distance is graphed horizontally. The two positively charged particles are seen to repel each other by the photon exchange. Figure 23.6 The image shows a Feynman diagram for the exchange of a π+ (pion) between a proton and a neutron, carrying the strong nuclear force between them. This diagram represents the situation shown more pictorially in Figure 23.3. The relative masses of the listed carrier particles describe something valuable about the four fundamental forces, as can be seen bosons) and Z bosons ( in Table 23.2. W bosons (consisting of nearly 1,000 times more massive than pions, carriers of the strong nuclear force. Simultaneously, the distance that the weak nuclear force can be transmitted is approximately times the strong force transmission distance. Unlike carrier particles, bosons), carriers of the weak nuclear force, are and which have a limited range, the photon is a massless particle that has no limit to the transmission distance of the electromagnetic force. This relationship leads scientists to understand that the yet-unfound graviton is likely massless as well. Force Carrier Particle Range Relative Strength[1] Gravity Graviton (theorized) Weak W and Z bosons Electromagnetic Photon ∞ ∞ Strong Pi mesons or pions (now known as gluons) 1 [1]Relative strength is based on the strong force felt by a proton-proton pair. Table 23.2 Carrier particles and their relative masses compared to pions for the four fundamental forces 776 Chapter 23 • Particle Physics BOUNDLESS PHYSICS Searching for the Graviton From Newton’s Universal Law of Gravitation to Einstein’s field equations, gravitation has held the focus of scientists for centuries. Given the discovery of carrier particles during the twentieth century, the importance of understanding gravitation has yet again gained the interest of prominent physicists everywhere. With carrier particles discovered for three of the four fundamental forces, it is sensible to scientists that a similar particle
, titled the graviton, must exist for the gravitational force. While evidence of this particle is yet to be uncovered, scientists are working diligently to discover its existence. So what do scientists think about the unfound particle? For starters, the graviton (like the photon) should be a massless particle traveling at the speed of light. This is assumed because, like the electromagnetic force, gravity is an inverse square law,. Scientists also theorize that the graviton is an electrically neutral particle, as an empty space within the influence of gravity is chargeless. However, because gravity is such a weak force, searching for the graviton has resulted in some unique methods. LIGO, the Laser Interferometer Gravitational-Wave Observatory, is one tool currently being utilized (see Figure 23.7). While searching for a gravitational wave to find a carrier particle may seem counterintuitive, it is similar to the approach taken by Planck and Einstein to learn more about the photon. According to wave-particle duality, if a gravitational wave can be found, the graviton should be present along with it. Predicted by Einstein’s theory of general relativity, scientists have been monitoring binary star systems for evidence of these gravitational waves. Figure 23.7 In searching for gravitational waves, scientists are using the Laser Interferometer Gravitational-Wave Observatory (LIGO). Here we see the control room of LIGO in Hanford, Washington. Particle accelerators like the Large Hadron Collider (LHC) are being used to search for the graviton through high-energy collisions. While scientists at the LHC speculate that the particle may not exist long enough to be seen, evidence of its prior existence, like footprints in the sand, can be found through gaps in projected energy and momentum. Some scientists are even searching the remnants of the Big Bang in an attempt to find the graviton. By observing the cosmic background radiation, they are looking for anomalies in gravitational waves that would provide information about the gravity particles that existed at the start of our universe. Regardless of the method used, scientists should know the graviton once they find it. A massless, chargeless particle with a spin of 2 and traveling at the speed of light—there is no other particle like it. Should it be found, its discovery would surely be considered by future generations to be on par with those of Newton and Einstein. GRASP CHECK Why are binary star systems used by LIGO to find gravitational waves?
a. Binary star systems have high temperature. b. Binary star systems have low density. c. Binary star systems contain a large amount of mass, but because they are orbiting each other, the gravitational field between the two is much less. d. Binary star systems contain a large amount of mass. As a result, the gravitational field between the two is great. Access for free at openstax.org. 23.1 • The Four Fundamental Forces 777 Accelerators Create Matter From Energy Before looking at all the particles that make up our universe, let us first examine some of the machines that create them. The fundamental process in creating unknown particles is to accelerate known particles, such as protons or electrons, and direct a beam of them toward a target. Collisions with target nuclei provide a wealth of information, such as information obtained by Rutherford in the gold foil experiment. If the energy of the incoming particles is large enough, new matter can even be created in the collision. The more energy input or ΔE, the more matter mcan be created, according to mass energy equivalence. Limitations are placed on what can occur by known conservation laws, such as conservation of mass-energy, momentum, and charge. Even more interesting are the unknown limitations provided by nature. While some expected reactions do occur, others do not, and still other unexpected reactions may appear. New laws are revealed, and the vast majority of what we know about particle physics has come from accelerator laboratories. It is the particle physicist’s favorite indoor sport. Our earliest model of a particle accelerator comes from the Van de Graaff generator. The relatively simple device, which you have likely seen in physics demonstrations, can be manipulated to produce potentials as great as 50 million volts. While these machines do not have energies large enough to produce new particles, analysis of their accelerated ions was instrumental in exploring several aspects of the nucleus. Another equally famous early accelerator is the cyclotron, invented in 1930 by the American physicist, E.O. Lawrence (1901–1958). Figure 23.8 is a visual representation with more detail. Cyclotrons use fixed-frequency alternating electric fields to accelerate particles. The particles spiral outward in a magnetic field, making increasingly larger radius orbits during acceleration. This clever arrangement allows the successive addition of electric potential energy with each loop. As a result, greater particle energies are possible than in a Van de Graaff generator. Figure 23.8 On the left is an artist’s rendition of the popular physics demonstration
tool, the Van de Graaff generator. A battery (A) supplies excess positive charge to a pointed conductor, the points of which spray the charge onto a moving insulating belt near the bottom. The pointed conductor (B) on top in the large sphere picks up the charge. (The induced electric field at the points is so large that it removes the charge from the belt.) This can be done because the charge does not remain inside the conducting sphere but moves to its outer surface. An ion source inside the sphere produces positive ions, which are accelerated away from the positive sphere to high velocities. On the right is a cyclotron. Cyclotrons use a magnetic field to cause particles to move in circular orbits. As the particles pass between the plates of the Dees, the voltage across the gap is oscillated to accelerate them twice in each orbit. A synchrotron is a modification of the cyclotron in which particles continually travel in a fixed-radius orbit, increasing speed each time. Accelerating voltages are synchronized with the particles to accelerate them, hence the name. Additionally, magnetic field strength is increased to keep the orbital radius constant as energy increases. A ring of magnets and accelerating tubes, as shown in Figure 23.9, are the major components of synchrotrons. High-energy particles require strong magnetic fields to steer 778 Chapter 23 • Particle Physics them, so superconducting magnets are commonly employed. Still limited by achievable magnetic field strengths, synchrotrons need to be very large at very high energies since the radius of a high-energy particle’s orbit is very large. To further probe the nucleus, physicists need accelerators of greater energy and detectors of shorter wavelength. To do so requires not only greater funding but greater ingenuity as well. Colliding beams used at both the Fermi National Accelerator Laboratory (Fermilab; see Figure 23.11) near Chicago and the LHC in Switzerland are designed to reduce energy loss in particle collisions. Typical stationary particle detectors lose a large amount of energy to the recoiling target struck by the accelerating particle. By providing head-on collisions between particles moving in opposite directions, colliding beams make it possible to create particles with momenta and kinetic energies near zero. This allows for particles of greater energy and mass to be created. Figure 23.10 is a schematic representation of this effect. In addition to circular accelerators, linear accelerators can be used to reduce energy radiation losses. The Stanford Linear
Accelerator Center (now called the SLAC National Accelerator Laboratory) in California is home to the largest such accelerator in the world. Figure 23.9 (a) A synchrotron has a ring of magnets and accelerating tubes. The frequency of the accelerating voltages is increased to cause the beam particles to travel the same distance in a shorter time. The magnetic field should also be increased to keep each beam burst traveling in a fixed-radius path. Limits on magnetic field strength require these machines to be very large in order to accelerate particles to very high energies. (b) A positively charged particle is shown in the gap between accelerating tubes. (c) While the particle passes through the tube, the potentials are reversed so that there is another acceleration at the next gap. The frequency of the reversals needs to be varied as the particle is accelerated to achieve successive accelerations in each gap. Figure 23.10 This schematic shows the two rings of Fermilab’s accelerator and the scheme for colliding protons and antiprotons (not to scale). Figure 23.11 The Fermi National Accelerator Laboratory, near Batavia, Illinois, was a subatomic particle collider that accelerated protons and antiprotons to attain energies up to 1 Tev (a trillion electronvolts). The circular ponds near the rings were built to dissipate waste heat. This accelerator was shut down in September 2011. (credit: Fermilab, Reidar Hahn) Check Your Understanding 1. Which of the four forces is responsible for radioactive decay? a. the electromagnetic force Access for free at openstax.org. 23.2 • Quarks 779 b. c. d. the gravitational force the strong nuclear force the weak nuclear force 2. What force or forces exist between an electron and a proton? a. b. c. d. the strong nuclear force, the electromagnetic force, and gravity the weak nuclear force, the strong nuclear force, and gravity the weak nuclear force, the strong nuclear force, and the electromagnetic force the weak nuclear force, the electromagnetic force, and gravity 3. What is the proposed carrier particle for the gravitational force? a. boson b. graviton c. gluon d. photon 4. What is the relationship between the mass and range of a carrier particle? a. Range of a carrier particle is inversely proportional to its mass. b. Range of a carrier particle is inversely proportional to square of its mass. c. Range of a
carrier particle is directly proportional to its mass. d. Range of a carrier particle is directly proportional to square of its mass. 5. What type of particle accelerator uses fixed-frequency oscillating electric fields to accelerate particles? cyclotron synchrotron a. b. c. betatron d. Van de Graaff accelerator 6. How does the expanding radius of the cyclotron provide evidence of particle acceleration? a. A constant magnetic force is exerted on particles at all radii. As the radius increases, the velocity of the particle must increase to maintain this constant force. b. A constant centripetal force is exerted on particles at all radii. As the radius increases, the velocity of the particle must decrease to maintain this constant force. c. A constant magnetic force is exerted on particles at all radii. As the radius increases, the velocity of the particle must decrease to maintain this constant force. d. A constant centripetal force is exerted on particles at all radii. As the radius increases, the velocity of the particle must increase to maintain this constant force. 7. Which of the four forces is responsible for the structure of galaxies? a. electromagnetic force b. gravity c. d. weak nuclear force strong nuclear force 23.2 Quarks Section Learning Objectives By the end of the section, you will be able to do the following: • Describe quarks and their relationship to other particles • Distinguish hadrons from leptons • Distinguish matter from antimatter • Describe the standard model of the atom • Define a Higgs boson and its importance to particle physics 780 Chapter 23 • Particle Physics Section Key Terms annihilation antimatter baryon bottom quark charmed quark color down quark flavor gluon hadron Higgs boson Higgs field lepton meson pair production positron quantum chromodynamics quark Standard Model strange quark top quark up quark Quarks “The first principles of the universe are atoms and empty space. Everything else is merely thought to exist…” “… Further, the atoms are unlimited in size and number, and they are borne along with the whole universe in a vortex, and thereby generate all composite things—fire, water, air, earth. For even these are conglomerations of given atoms. And it because of their solidity that these atoms are impassive and unalterable.” —Diogenes Laertius (summarizing the views of Democrit
us, circa 460–370 B.C.) The search for fundamental particles is nothing new. Atomists of the Greek and Indian empires, like Democritus of fifth century B.C., openly wondered about the most finite components of our universe. Though dormant for centuries, curiosity about the atomic nature of matter was reinvigorated by Rutherford’s gold foil experiment and the discovery of the nucleus. By the early 1930s, scientists believed they had fully determined the tiniest constituents of matter—in the form of the proton, neutron, and electron. This would be only partially true. At present, scientists know that there are hundreds of particles not unlike our electron and nucleons, all making up what some have termed the particle zoo. While we are confident that the electron remains fundamental, it is surrounded by a plethora of similar sounding terms, like leptons, hadrons, baryons, and mesons. Even though not every particle is considered fundamental, they all play a vital role in understanding the intricate structure of our universe. A fundamental particle is defined as a particle with no substructure and no finite size. According to the Standard Model, there are three types of fundamental particles: leptons, quarks, and carrier particles. As you may recall, carrier particles are responsible for transmitting fundamental forces between their interacting masses. Leptons are a group of six particles not bound by the strong nuclear force, of which the electron is one. As for quarks, they are the fundamental building blocks of a group of particles called hadrons, a group that includes both the proton and the neutron. Now for a brief history of quarks. Quarks were first proposed independently by American physicists Murray Gell-Mann and George Zweig in 1963. Originally, three quark types—or flavors—were proposed with the names up (u), down (d), and strange (s). At first, physicists expected that, with sufficient energy, we should be able to free quarks and observe them directly. However, this has not proved possible, as the current understanding is that the force holding quarks together is incredibly great and, much like a spring, increases in magnitude as the quarks are separated. As a result, when large energies are put into collisions, other particles are created—but no quarks emerge. With that in mind, there is compelling evidence for the existence of quarks. By 1967, experiments at the SLAC National Accelerator Laboratory scattering 20-GeV electrons

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