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sihnvASQFSw-022|So how long is a half-life? |
WitrD0bFL24-000|We know electrons can absorb energy and change energy levels, go from one orbital to another, in an atom. |
WitrD0bFL24-009|We're talking about electronic transitions, going from n equal 1 to n equal 2. |
WitrD0bFL24-010|Which of these possible species can do that? |
WitrD0bFL24-011|Let's look at their electronic configurations. |
WitrD0bFL24-012|Here's fluorine, 1s2 2s2 2p5. |
WitrD0bFL24-013|Fluorine minus has an extra electron, one more electron than fluorine, so it has electronic configuration 1s2 2s2 2p6. |
WitrD0bFL24-016|So 1s2 2s2 2p6. |
WitrD0bFL24-017|Notice fluorine minus and neon have the same electronic configuration, plus it has an electron in the three s orbital. |
Zfx54dYEDBs-007|We're looking at the oxidation of NO2 minus to NO3 minus and looking at how the formal charge on nitrogen changes in that process. |
Zfx54dYEDBs-012|And it'll share one of these and two of those. |
Zfx54dYEDBs-013|So one, two, three, four, five electrons in a sharing mode for nitrogen in that molecule. |
Zfx54dYEDBs-014|This molecule, nitrogen shares all the electrons around it. |
Zfx54dYEDBs-015|It shares one from this bond, one from this bond, and two from this bond. |
Zfx54dYEDBs-016|One, two, three, four. |
Zfx54dYEDBs-017|Four electrons. |
Zfx54dYEDBs-018|Nitrogen as an atom has five electrons. |
4blDQFbly-Y-000|Let's look at the Gibbs free energy for a system. |
4blDQFbly-Y-001|The Gibbs function is defined as the enthalpy minus the temperature times the entropy. |
4blDQFbly-Y-004|You can talk about that for any system in general or you can talk about specifically the standard states of the system. |
4blDQFbly-Y-014|But the free energy is always a predictor of the direction of the chemical reaction. |
4blDQFbly-Y-019|So I have predictive power now, whether a reaction is likely to go based on the free energy of the system. |
TPQjDIPwJ-A-000|Let's do a calculation involving the photoelectric effect, ejected photo electrons and photons of a certain energy hitting a metal. |
TPQjDIPwJ-A-002|So this work function for chromium metal, 4.37 electron volts, is how strongly chromium holds on to its electrons. |
TPQjDIPwJ-A-003|So we have to say, well, what is the photoelectric effect? |
TPQjDIPwJ-A-004|The photoelectric effect, we have to balance the energies. |
TPQjDIPwJ-A-006|So the metal holds on to the electron, but if we bring in a high enough energy photon we can eject an electron with a certain kinetic energy. |
TPQjDIPwJ-A-007|So that's what we have to find. |
TPQjDIPwJ-A-010|Now, we do that to keep all the things we multiply together consistent unit wise. |
TPQjDIPwJ-A-011|We're dimensionally consistent. |
TPQjDIPwJ-A-012|We always use kilograms for mass, so don't just put your masses into your equations with grams or pounds or some random mass unit. |
TPQjDIPwJ-A-013|If you come across a mass, convert it to the kilograms. |
TPQjDIPwJ-A-014|If you come across a distance, a length, convert that to meters. |
TPQjDIPwJ-A-015|A time, seconds. |
TPQjDIPwJ-A-016|An energy, joules. |
TPQjDIPwJ-A-022|That's going to be an energy, a kinetic energy, so this will be joules. |
TPQjDIPwJ-A-023|In fact, that's how I remember the SI units of joules. |
TPQjDIPwJ-A-030|Tiny, tiny, tiny number of joules. |
TPQjDIPwJ-A-031|Of course, it's an electron. |
TPQjDIPwJ-A-035|Now electron volts in a unit of energy, it's the kinetic energy that an electron gains as you accelerate it across the potential of 1 volt. |
TPQjDIPwJ-A-037|So we can do that product, so the work function in terms of joules, 7.04 times 10 to the minus 19th joules. |
TPQjDIPwJ-A-046|And that's what we want. |
TPQjDIPwJ-A-047|It's always good to check, do the units that we have left makes sense for the quantity that we're solving for? |
TPQjDIPwJ-A-048|We're solving for wave length, a length. |
TPQjDIPwJ-A-049|Do I have meters? |
TPQjDIPwJ-A-050|In this case, I do. |
TPQjDIPwJ-A-053|109 nanometers is, as we recall, in the UV, the ultraviolet region. |
TPQjDIPwJ-A-054|We know visible went from 700 down to 400 nanometers. |
wA_IvsdRjpM-000|Let's look at the root mean squared velocity for bromine, at 300 Kelvin in a sample of gas, in our Nuts and Boltz section. |
wA_IvsdRjpM-001|And in honor of Maxwell Boltzmann distribution, we'll spell bolts with a z this time. |
wA_IvsdRjpM-006|And I can solve for that. |
wA_IvsdRjpM-007|I get 216 meters per second. |
wA_IvsdRjpM-008|Now, 216 meters per second, that's about 800 feet per second. |
wA_IvsdRjpM-011|That's amazing speeds for gas particles at modest temperatures. |
ktySzUONBUU-000|So how do we determine the exact mass of an atom? |
ktySzUONBUU-001|Well, one way is to use a mass spectrometer. |
ktySzUONBUU-003|A mass spectrometer, you take a sample and you ionize it. |
ktySzUONBUU-004|And by ionization, we mean stripping off electrons. |
ktySzUONBUU-005|In this case, we strip off all of the electrons. |
ktySzUONBUU-006|The sample is then accelerated towards an electric field and directed through a magnetic field. |
ktySzUONBUU-007|Now when charged particles hit a magnetic field, they are deflected by the magnetic field. |
ktySzUONBUU-008|And they're deflected according to their mass. |
ktySzUONBUU-009|Heavier particles are not as deflected as lighter particles. |
ktySzUONBUU-010|So you have this stream of particles and it hits that magnetic field, and it's fanned out based on their mass. |
ktySzUONBUU-015|Let's look at a larger mass coming through the same system. |
ktySzUONBUU-016|We're going to ionize it, it's not deflected as much. |
ktySzUONBUU-017|So you have a distribution of masses that you can resolve using a mass spectrometer. |
ktySzUONBUU-019|And the vertical axis will tell us about how many particles there were, and that's very useful. |
ktySzUONBUU-020|Remember when we had our carbon, we said that one in hundred of these particles was a carbon-13. |
ktySzUONBUU-021|The other 99 were carbon-12. |
ktySzUONBUU-022|That's a difference of one mass unit. |
ktySzUONBUU-024|So, let's look at a few masses. |
ktySzUONBUU-025|We have, for instance, what would show up at mass 1? |
ktySzUONBUU-029|So you can tell by the mass spectrometer you can get the mass. |
ktySzUONBUU-030|But remember, mass does not determine the identity of the element-- certainly a clue, but it does not determine the identity. |
ktySzUONBUU-031|We need to know the number of protons in the nucleus to get the identity of an element. |
ktySzUONBUU-032|So mass three, that could be tritium or it could be a molecule of deuterium and hydrogen hooked together to form a hydrogen molecule. |
ktySzUONBUU-036|And you can go on down the line. |
ktySzUONBUU-037|Here is oxygen 16. |
ktySzUONBUU-038|At mass 18, you could have a water molecule with oxygen 16 and two protons, two hydrogen atoms. |
ktySzUONBUU-044|So a mass spectrometer, exquisitely sensitive makes. |
ktySzUONBUU-046|The e equals mc squared mass loss can be determined by modern mass spectrometer. |
ktySzUONBUU-047|That's how accurate they are. |
ktySzUONBUU-048|Incredibly tiny mass losses can be determined by this method. |
ktySzUONBUU-049|It's a beautiful instrument and it's very, very, very powerful for measuring atomic masses. |
ksZoOde-7zY-000|We noticed that the trend in ionization energies is to kind of increase across a row in the periodic table. |
ksZoOde-7zY-001|Let's look at that in more detail. |
ksZoOde-7zY-003|That's a 1s electron. |
ksZoOde-7zY-004|For helium, about 2,400 kilojoules per mole for that ionization. |
ksZoOde-7zY-006|So increasing that positive charge on the nucleus was much more dramatic than adding another electron. |
ksZoOde-7zY-007|It's much harder to pull off an electron from helium than hydrogen. |
ksZoOde-7zY-011|And lithium is remarkably easy to ionize. |
ksZoOde-7zY-018|It's a general trend. |
ksZoOde-7zY-019|And we can kind of rationalize that. |
ksZoOde-7zY-020|Because what we're doing is we're adding more nuclear charge, but we're staying in the same principle quantum level. |
ksZoOde-7zY-021|So electrons in about the same region of space but more positive charge holding them. |
ksZoOde-7zY-022|Of course, those will be slightly more difficult to ionize. |
ksZoOde-7zY-023|As you continue across, there's a slight dip here at oxygen. |
ksZoOde-7zY-024|And I think we can understand that too in terms of the fact that nitrogen now is that half-filled state. |
ksZoOde-7zY-025|Oxygen has that paired electron. |
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