query stringlengths 17 1.94k | response stringlengths 21 6.24k | subject stringclasses 14
values |
|---|---|---|
Determine the term used to describe the total number of wave cycles that pass a fixed reference point within a time interval of one second. | **Final Answer:** Frequency $f = \frac{n}{t}$ $f = \frac{n}{1 \text{ s}} = n$ $\text{Result} = \text{Frequency}$ | Chemistry 10th JEE |
Consider the following types of rays: X-rays, cathode rays, and canal rays. Identify which of these is classified as electromagnetic radiation and explain the fundamental difference in nature between electromagnetic radiation and the other two types mentioned. | **Final Answer:** X-rays are classified as electromagnetic radiation. The fundamental difference is that X-rays are waves (photons) with zero rest mass ($m_{0} = 0$), whereas cathode rays and canal rays are particulate in nature with finite rest mass. $\text{Classification: X-rays} \in \text{Electromagnetic Radiation}$... | Chemistry 10th JEE |
Within the visible light spectrum (VIBGYOR), different colors correspond to different energy levels. Compare the relative energies of Red, Blue, Green, and Yellow light, and identify which specific color is associated with the highest energy. | **Final Answer:** Blue light has the highest energy among the given colors. Using Planck's quantum theory formula: $E = \frac{hc}{\lambda}$ Relationship: $E \propto \frac{1}{\lambda}$ Wavelength order: $\lambda_{Blue} < \lambda_{Green} < \lambda_{Yellow} < \lambda_{Red}$ Energy order: $E_{Blue} > E_{Green} > E_{Yellow}... | Chemistry 10th JEE |
State the specific Greek symbol used to denote the frequency of an electromagnetic wave and write down the formula that relates this frequency to the velocity of light ($c$) and wavelength ($\lambda$). | **Final Answer:** $\nu \text{ and } \nu = \frac{c}{\lambda}$ $\text{Frequency symbol} = \nu$ $$c = \nu \lambda$$ $$\nu = \frac{c}{\lambda}$$ | Chemistry 10th JEE |
The wavelength of light is often measured in various units such as meters ($m$), centimeters ($cm$), Angstroms ($\mathring{A}$), and nanometers ($nm$). Determine the conversion factor to express $1\text{ nm}$ in terms of meters ($m$). | **Final Answer:** $1\text{ nm} = 10^{-9}\text{ m}$ Prefix definition: $\text{nano (n)} = 10^{-9}$ $1\text{ nm} = 1 \times 10^{-9}\text{ m} = 10^{-9}\text{ m}$ | Chemistry 10th JEE |
Express the length of $1\text{ nanometer}$ ($nm$) in meters ($m$) using scientific notation. | **Final Answer:** $1 \times 10^{-9}\text{ m}$ $1\text{ nano (n)} = 10^{-9}$ $1\text{ nm} = 1 \times 10^{-9}\text{ m}$ | Chemistry 10th JEE |
The velocity of electromagnetic radiation in a vacuum is a universal constant. Determine the value of this velocity and express it in the following two units:
(i) Meters per second ($m/s$)
(ii) Centimeters per second ($cm/s$)
Show the calculation step used to convert between these two units. | **Final Answer:** (i) $3 \times 10^{8} \, \text{m/s}$, (ii) $3 \times 10^{10} \, \text{cm/s}$ Given the velocity of light in vacuum: $c = 3 \times 10^{8} \, \text{m/s}$ Using conversion factor: $1 \, \text{m} = 10^{2} \, \text{cm}$ $$c = 3 \times 10^{8} \times 10^{2} = 3 \times 10^{10} \, \text{cm/s}$$ | Chemistry 10th JEE |
Calculate the wavelength ($\lambda$) in meters for an electromagnetic wave belonging to the X-ray region, given that its frequency is $1.5 \times 10^{18} \text{ cycles per second}$ ($\text{Hz}$). Use the value of the speed of light $c = 3 \times 10^8 \text{ m s}^{-1}$. | **Final Answer:** $\lambda = 2 \times 10^{-10} \, \text{m}$ Given: $c = 3 \times 10^{8} \, \text{m s}^{-1}, \nu = 1.5 \times 10^{18} \, \text{Hz}$ Using $$\lambda = \frac{c}{\nu}$$ $$\lambda = \frac{3 \times 10^{8}}{1.5 \times 10^{18}} = 2 \times 10^{-10} \, \text{m}$$ | Chemistry 10th JEE |
Analyze the spectrum produced when white light is passed through a dispersive medium like a prism. State whether this spectrum is classified as continuous or emission, and justify your answer based on the presence or absence of gaps between colors. | **Final Answer:** Continuous Emission Spectrum $\lambda_{white} \in [400\text{ nm}, 700\text{ nm}]$ $n(\lambda) = A + \frac{B}{\lambda^{2}}$ $\delta = (n-1)A$ $$\delta_{V} > \delta_{R} \implies \text{Continuous distribution without gaps}$$ | Chemistry 10th JEE |
Explain whether a line spectrum (discontinuous spectrum) is considered a characteristic property of individual atoms or of more complex molecular structures. | **Final Answer:** A line spectrum is a characteristic property of individual atoms (atomic spectra). $\Delta E = E_{n_{2}} - E_{n_{1}}$ $\Delta E = h \nu = \frac{hc}{\lambda}$ $\lambda = \frac{hc}{E_{n_{2}} - E_{n_{1}}}$ $n_{1}, n_{2} \in \text{integers} \implies \text{discrete } \lambda \implies \text{Individual Atoms... | Chemistry 10th JEE |
According to Sommerfeld's modification of the Bohr model, determine the number of sub-orbits that are elliptical in shape for an electron residing in the second principal energy level ($n=2$). | **Final Answer:** 1 Given: $n = 2$ Azimuthal quantum number $k \in \{1, 2, \dots, n\} \implies k = 1, 2$ Ratio of semi-major axis $a$ to semi-minor axis $b$: $a/b = n/k$ If $k=1, a/b = 2/1 = 2 \implies a \neq b$ (Elliptical) If $k=2, a/b = 2/2 = 1 \implies a = b$ (Circular) | Chemistry 10th JEE |
In Bohr's theory of the atom, energy levels are designated by the principal quantum number $n$. Provide the corresponding alphabetical letter symbols used to represent the orbits where $n = 1, 2, 3,$ and $4$. | **Final Answer:** $n=1: K, n=2: L, n=3: M, n=4: N$ $n = 1 \implies K$ $n = 2 \implies L$ $n = 3 \implies M$ $n = 4 \implies N$ | Chemistry 10th JEE |
According to Bohr's postulates for the hydrogen atom, the angular momentum of an electron is quantized. Write the general expression for the orbital angular momentum ($mvr$) of an electron in the $n^{th}$ orbit and use it to calculate the angular momentum of an electron in the $3^{rd}$ stationary orbit. | **Final Answer:** $L = 1.5 \frac{h}{\pi}$ Using Bohr's quantization condition $mvr = \frac{nh}{2\pi}$ $mvr = \frac{3 \times h}{2 \times \pi}$ $mvr = 1.5 \frac{h}{\pi}$ | Chemistry 10th JEE |
Calculate the radius of the $L$ shell ($n=2$) for a $Be^{3+}$ ion in Angstroms ($Å$), given that the Bohr radius for the first orbit of hydrogen is $0.529\text{ }Å$. | **Final Answer:** $0.529 \, \text{\AA}$ Given: $n = 2$ (L shell), $Z = 4$ ($\text{Be}^{3+}$), $a_{0} = 0.529 \, \text{\AA}$ Using $$r_{n} = a_{0} \frac{n^{2}}{Z}$$ $$r_{2} = 0.529 \times \frac{2^{2}}{4} = 0.529 \, \text{\AA}$$ | Chemistry 10th JEE |
Suppose an electron is characterized by a principal quantum number $n = 3$. Determine and list all the permissible sets of the remaining three quantum numbers: the azimuthal quantum number ($l$), the magnetic quantum number ($m_l$), and the spin quantum number ($m_s$) that this electron can have according to the quantu... | **Final Answer:** 18 permissible sets: $(0, 0, \pm\frac{1}{2})$, $(1, \{-1, 0, 1\}, \pm\frac{1}{2})$, and $(2, \{-2, -1, 0, 1, 2\}, \pm\frac{1}{2})$ Using $0 \leq l \leq n-1$ $l \in \{0, 1, 2\}$ Using $-l \leq m_{l} \leq +l$: For $l=0, m_{l}=0$; For $l=1, m_{l} \in \{-1, 0, 1\}$; For $l=2, m_{l} \in \{-2, -1, 0, 1, 2\}... | Chemistry 10th JEE |
Identify the quantum number that primarily determines the size of the orbit and the main energy level of an electron in a hydrogen-like atom. Explain how this quantum number relates to the distance of the electron from the nucleus. | **Final Answer:** Principal Quantum Number ($n$) Principal Quantum Number ($n$) $E_{n} = -13.6 \frac{Z^{2}}{n^{2}} \text{ eV}$ $r_{n} = 0.529 \frac{n^{2}}{Z} \text{ \AA}$ $r_{n} \propto n^{2} \implies \text{Radius increases with } n$ | Chemistry 10th JEE |
Determine the value of the azimuthal quantum number ($l$) for the differentiating (last) electron added to an Aluminum atom ($Z=13$) in its ground state. Show the electronic configuration to justify your answer. | **Final Answer:** $l = 1$ $Z = 13$ $\text{Configuration} = 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{1}$ $\text{Last Orbital} = 3p$ $l = 1$ | Chemistry 10th JEE |
State the general mathematical expression used to calculate the maximum number of electrons that can be contained within the $n^{th}$ energy level (shell) of an atom. | **Final Answer:** Maximum electrons = $2n^{2}$ $\text{Total number of orbitals in shell } n = n^{2}$ $\text{Maximum electrons} = 2 \times (\text{Number of orbitals})$ $\text{Maximum electrons} = 2 \times n^{2} = 2n^{2}$ | Chemistry 10th JEE |
Calculate the total number of electrons that can have a spin quantum number of $s = -\frac{1}{2}$ in the fourth energy level ($n=4$). | **Final Answer:** 16 Given: $n = 4$ Total orbitals $N = n^{2} = 4^{2} = 16$ $$N_{e} = 16 \times 1 = 16$$ | Chemistry 10th JEE |
Identify which of the following sets of quantum numbers $(n, l, m, s)$ is physically impossible for an electron in an atom and provide the specific reason for its impossibility:
1) $n=3, l=1, m=0, s=+\frac{1}{2}$
2) $n=3, l=2, m=-1, s=+\frac{1}{2}$
3) $n=2, l=2, m=1, s=-\frac{1}{2}$
4) $n=4, l=1, m=1, s=-\frac{1}{2}$ | **Final Answer:** Option (3): $n=2, l=2, m=1, s=-\frac{1}{2}$ is impossible because $l$ must be strictly less than $n$. Constraint for azimuthal quantum number: $0 \le l \le n-1$ For Set 3: $n=2, l=2$ $l_{max} = 2 - 1 = 1$ $$2 \not\le 1 \implies \text{Impossible}$$ | Chemistry 10th JEE |
Which rule of electronic configuration specifies that for a given electron configuration, the lowest energy term is the one with the greatest value of spin multiplicity, meaning degenerate orbitals are each occupied by a single electron before any pairing occurs? | **Final Answer:** Hund's Rule of Maximum Multiplicity Spin Multiplicity $M = 2S + 1$ $S = \sum s_{i} = n \times \frac{1}{2}$ (for $n$ parallel spins) $$ M = n + 1 $$ | Chemistry 10th JEE |
Provide a comprehensive analysis of the properties of anode rays (canal rays). In your answer, specifically address their origin in a discharge tube, the dependence of their $e/m$ ratio on the identity of the gas used, and their direction of deflection in an electric field. | **Final Answer:** Anode rays are positive ions originating from gas ionization with a gas-dependent $e/m$ ratio and deflect toward the negative plate. $X(\text{gas}) + e^{-}(\text{cathode ray}) \to X^{+} + 2e^{-}$ $\frac{e}{m} = \frac{z \cdot e}{m_{\text{ion}}}$. Since $m_{\text{ion}}$ depends on the atomic/molar mass ... | Chemistry 10th JEE |
Explain the primary theoretical limitation of Rutherford's atomic model in the context of classical physics. Discuss why a charged particle like an electron, while undergoing continuous acceleration in a circular orbit, was expected to radiate energy and what the ultimate impact of this energy loss would be on the atom... | **Final Answer:** Classical physics predicts Rutherford's model is unstable because electrons lose energy via radiation and spiral into the nucleus within $10^{-10} \text{ s}$. Centripetal acceleration for circular motion: $a = \frac{v^{2}}{r}$ Power radiated by accelerating charge: $P = \frac{1}{4\pi \epsilon_{0}} \fr... | Chemistry 10th JEE |
Illustrate the three-dimensional geometric shapes for $s$, $p$, and $d$ atomic orbitals. For each type, describe the specific symmetry (spherical or directional) associated with the distribution of electron density. | **Final Answer:** $s$: Spherical (Non-directional); $p$: Dumbbell (Directional); $d$: Double-dumbbell (Directional) For $s$ orbitals ($l=0$): $\psi$ depends only on $r$. Shape: Spherical. Symmetry of $s$: Spherical symmetry; non-directional distribution of electron density. For $p$ orbitals ($l=1$): Orientations $m_{l}... | Chemistry 10th JEE |
Consider the two sets of orbital filling diagrams, labeled (A) and (B), representing the $3s$ and $3p$ subshells of a particular atom. Analyze the distribution of electrons in each case and determine which diagram represents a violation of **Pauli’s Exclusion Principle**. In your answer, clearly identify the specific o... | **Final Answer:** Diagram (B) violates Pauli's Exclusion Principle in the $3s$ orbital because it contains two electrons with the same spin quantum number ($m_{s} = +\frac{1}{2}$). Identify spins in $3s$ orbital: Diagram (A) has $\uparrow \downarrow$ (opposite); Diagram (B) has $\uparrow \uparrow$ (parallel). Pauli Exc... | Chemistry 10th JEE |
A local radio station, "Star FM," broadcasts its signal at a frequency of $98.5\text{ MHz}$. Determine the wavelength of this electromagnetic radiation in meters. (Use the speed of light, $c = 3 \times 10^8\text{ m/s}$) | **Final Answer:** $\lambda \approx 3.05 \text{ m}$ Given: $f = 98.5 \text{ MHz} = 98.5 \times 10^{6} \text{ Hz}$ and $c = 3 \times 10^{8} \text{ m/s}$ Using the wave equation: $\lambda = \frac{c}{f}$ Substitution: $\lambda = \frac{3 \times 10^{8}}{98.5 \times 10^{6}} = \frac{300}{98.5}$ Result: $\lambda \approx 3.05 \t... | Chemistry 10th JEE |
Define the specific wave parameter that represents the linear distance measured between two consecutive crests or two consecutive troughs in a periodic wave. | **Final Answer:** Wavelength ($\lambda$) Definition of wavelength ($\lambda$) as the spatial period: $d(\text{crest}_{n}, \text{crest}_{n+1}) = \lambda$ Relation to wave speed $v$ and frequency $f$: $\lambda = \frac{v}{f}$ | Chemistry 10th JEE |
Identify the wave characteristic that is mathematically defined as the reciprocal of the wavelength ($\frac{1}{\lambda}$) and state its common units. | **Final Answer:** Wave number; $\text{m}^{-1}$ or $\text{cm}^{-1}$ $\text{Definition: Wave Number } (\bar{\nu}) = \frac{1}{\lambda}$ $\text{SI unit substitution: } \frac{1}{\text{m}} = \text{m}^{-1}$ $\text{CGS unit substitution: } \frac{1}{\text{cm}} = \text{cm}^{-1}$ | Chemistry 10th JEE |
While mechanical waves require a medium to propagate, electromagnetic waves possess a unique ability to travel through a specific environment devoid of all matter. Name this environment. | **Final Answer:** Vacuum (or Free Space) Mechanical waves require a material medium (solid, liquid, or gas) to transmit energy via particle oscillations. Electromagnetic waves consist of coupled, self-sustaining oscillating electric $\vec{E}$ and magnetic $\vec{B}$ fields. In regions devoid of matter, the propagation s... | Chemistry 10th JEE |
The electromagnetic spectrum consists of various regions. Specify the name of the narrow band of radiation, centered around a frequency of $10^{15} \text{ Hz}$, which is detectable by the human eye. | **Final Answer:** Visible Light $\nu = 10^{15} \text{ Hz}$ $\lambda = \frac{c}{\nu} = \frac{3 \times 10^{8} \text{ m/s}}{10^{15} \text{ Hz}} = 3 \times 10^{-7} \text{ m}$ $\lambda = 300 \text{ nm}$ (Near visible-ultraviolet boundary) $\text{Radiation detectable by the human eye} = \text{Visible Light}$ | Chemistry 10th JEE |
Electromagnetic radiation in the radiofrequency region, typically characterized by frequencies near $10^6 \text{ Hz}$, is primarily utilized for which major technological application? | **Final Answer:** AM Radio Broadcasting Given: Frequency $f \approx 10^{6} \text{ Hz}$ Identification: This frequency falls within the Radio Wave spectrum ($3 \text{ kHz}$ to $300 \text{ GHz}$) Specific Application: Frequencies around $1 \text{ MHz}$ ($10^{6} \text{ Hz}$) are used for AM (Amplitude Modulation) radio br... | Chemistry 10th JEE |
Express the mathematical formula for wavenumber ($\bar{\nu}$) in terms of wavelength ($\lambda$). State the symbol used to denote wavenumber and calculate its value for a radiation with a wavelength of $2 \times 10^{-5} \text{ cm}$. | **Final Answer:** $\bar{\nu} = 5 \times 10^{4} \text{ cm}^{-1}$ Symbol: $\bar{\nu}$ $\bar{\nu} = \frac{1}{\lambda}$ Given $\lambda = 2 \times 10^{-5} \text{ cm}$ $\bar{\nu} = \frac{1}{2 \times 10^{-5}} = 0.5 \times 10^{5}$ $\bar{\nu} = 5 \times 10^{4} \text{ cm}^{-1}$ | Chemistry 10th JEE |
Given that photons in the X-ray region are more energetic than those in the visible region, compare the energy of visible light photons with those in the Infrared (IR) and Ultraviolet (UV) regions. Determine which region (IR or UV) contains photons that are less energetic than visible light. | **Final Answer:** Infrared (IR) region $$E = h\nu \implies E \propto \nu$$ $$\nu_{\text{UV}} > \nu_{\text{visible}} > \nu_{\text{IR}}$$ $$E_{\text{UV}} > E_{\text{visible}} > E_{\text{IR}}$$ | Chemistry 10th JEE |
Arrange the following types of electromagnetic radiations in increasing order of their wavelengths: Ultraviolet rays, Radio waves, X-rays, and Infrared rays. From this list, identify which radiation possesses the maximum wavelength. | **Final Answer:** Increasing order: X-rays < Ultraviolet rays < Infrared rays < Radio waves; Maximum wavelength: Radio waves $\lambda_{\text{X-rays}} \approx 10^{-10} \text{ m}, \lambda_{\text{UV}} \approx 10^{-8} \text{ m}, \lambda_{\text{IR}} \approx 10^{-5} \text{ m}, \lambda_{\text{Radio}} \approx 10^{2} \text{ m}$... | Chemistry 10th JEE |
Calculate the ratio of the energy of a photon with a wavelength of $2500 \text{ \AA}$ to the energy of a photon with a wavelength of $5000 \text{ \AA}$. Show the relationship used for this evaluation. | **Final Answer:** $E_{1} : E_{2} = 2 : 1$ Given: $\lambda_{1} = 2500 \text{ \AA}$, $\lambda_{2} = 5000 \text{ \AA}$ Using $E = \frac{hc}{\lambda} \implies E \propto \frac{1}{\lambda}$ $$\frac{E_{1}}{E_{2}} = \frac{\lambda_{2}}{\lambda_{1}} = \frac{5000}{2500} = 2$$ | Chemistry 10th JEE |
List four different units commonly used to measure the wavelength of electromagnetic radiation, ranging from the macroscopic scale to the atomic scale (e.g., Meter to Angstrom). | **Final Answer:** The four units are Meter ($m$), Micrometer ($\mu m$), Nanometer ($nm$), and Angstrom ($\text{\AA}$). Unit 1 (Macroscopic): $1 \text{ m} = 10^{0} \text{ m}$ Unit 2 (Micro-scale): $1 \mu\text{m} = 10^{-6} \text{ m}$ Unit 3 (Nano-scale): $1 \text{ nm} = 10^{-9} \text{ m}$ Unit 4 (Atomic-scale): $1 \text{... | Chemistry 10th JEE |
Identify and name two distinct types of sources—one natural and one artificial—that are capable of emitting radiations in the visible part of the electromagnetic spectrum. | **Final Answer:** Natural Source: The Sun; Artificial Source: Incandescent Bulb/Electric Lamp Visible spectrum definition: $\lambda \in [380\text{ nm}, 750\text{ nm}]$ Natural source: The Sun ($T \approx 5800\text{ K}$, $\lambda_{max} \approx 500\text{ nm}$) Artificial source: Incandescent Lamp ($T \approx 2500\text{ K... | Chemistry 10th JEE |
Determine the standard notation used for representing the wavelength of a wave and define what distance this parameter represents physically in a wave train. | **Final Answer:** Notation: $\lambda$; Physical distance: The distance between any two consecutive points having the same phase, such as two consecutive crests or troughs. $$\text{Standard Notation} = \lambda$$ $$k = \frac{2\pi}{\lambda}$$ $$\Delta \phi = k \Delta x \implies 2\pi = \left( \frac{2\pi}{\lambda} \right) \... | Chemistry 10th JEE |
Identify and describe a primary practical application for electromagnetic radiation belonging to the microwave region, specifically characterized by a frequency of approximately $10^{10} \text{ Hz}$. | **Final Answer:** RADAR systems for aircraft navigation (Microwaves) Given: Frequency $f \approx 10^{10} \text{ Hz}$ and speed of light $c \approx 3 \times 10^{8} \text{ m/s}$ Using $\lambda = \frac{c}{f} = \frac{3 \times 10^{8}}{10^{10}}$ $$\lambda = 3 \times 10^{-2} \text{ m} = 3 \text{ cm}$$ As $\lambda \approx 3 \t... | Chemistry 10th JEE |
Define the term wavenumber ($\bar{\nu}$) and determine its standard units in the CGS system. Explain how this unit is derived from its relationship with wavelength ($\lambda$). | **Final Answer:** $\bar{\nu} = \text{cm}^{-1}$ $\bar{\nu} = \frac{1}{\lambda}$ $\text{Unit of } \bar{\nu} = \frac{1}{\text{Unit of } \lambda}$ $\text{In CGS system, Unit of } \lambda = \text{cm}$ $\text{Unit of } \bar{\nu} = \frac{1}{\text{cm}} = \text{cm}^{-1}$ | Chemistry 10th JEE |
Define the relationships between the fundamental properties of electromagnetic radiation by providing the mathematical expressions for each of the following:
(a) The velocity of light ($c$) in terms of frequency ($\nu$) and wavelength ($\lambda$).
(b) The wavenumber ($\bar{\nu}$) in terms of wavelength ($\lambda$).
(c)... | **Final Answer:** (a) $c = \nu \lambda$, (b) $\bar{\nu} = \frac{1}{\lambda}$, (c) $\nu = \frac{c}{\lambda}$, (d) $\lambda = \frac{c}{\nu}$ $c = \nu \lambda$ $\bar{\nu} = \frac{1}{\lambda}$ $\nu = \frac{c}{\lambda}$ $\lambda = \frac{c}{\nu}$ | Chemistry 10th JEE |
Analyze the following propositions regarding the energy ($E$) of electromagnetic radiation and determine which dependencies are physically correct according to Planck's Quantum Theory:
(A) The energy of the radiation is directly proportional to its frequency ($\nu$).
(B) The energy of the radiation is directly proporti... | **Final Answer:** All propositions (A), (B), and (C) are correct. Generalized equation: $E = h\nu = \frac{hc}{\lambda} = hc\bar{\nu}$ Using Planck's equation: $E = h\nu$ Substituting $\nu = \frac{c}{\lambda} \implies E = \frac{hc}{\lambda}$ Substituting $\frac{1}{\lambda} = \bar{\nu} \implies E = hc\bar{\nu}$ Concludin... | Chemistry 10th JEE |
Identify the wave property of electromagnetic radiation that is defined as the reciprocal of the wavenumber ($\bar{\nu}$). | **Final Answer:** Wavelength ($\lambda$) $\bar{\nu} = \frac{1}{\lambda}$ $\text{Reciprocal} = \frac{1}{\bar{\nu}} = \lambda$ | Chemistry 10th JEE |
For a single photon of electromagnetic radiation, evaluate the ratio of its energy ($E$) to its frequency ($\nu$). Identify the fundamental physical constant that this ratio represents. | **Final Answer:** The ratio is $h$ (Planck's constant) Using Planck-Einstein relation: $E = h\nu$ Ratio calculation: $\frac{E}{\nu} = h$ | Chemistry 10th JEE |
Based on the fundamental principles of wave motion, derive and state the mathematical relationships between the following parameters:
(i) Velocity of light ($c$), frequency ($\nu$), and wavelength ($\lambda$).
(ii) Wavenumber ($\bar{\nu}$) and wavelength ($\lambda$).
(iii) Wavelength ($\lambda$) expressed in terms of v... | **Final Answer:** (i) $c = \nu \lambda$, (ii) $\bar{\nu} = \frac{1}{\lambda}$, (iii) $\lambda = \frac{c}{\nu}$ $c = \frac{\lambda}{T} = \nu \lambda$ $\bar{\nu} = \frac{1}{\lambda}$ $c = \nu \lambda \Rightarrow \lambda = \frac{c}{\nu}$ | Chemistry 10th JEE |
For a beam of visible light propagating through a vacuum, determine the mathematical expression for its wavelength ($\lambda$) as a function of its frequency ($\nu$) and the speed of light ($c$). Explain whether the wavelength is directly or inversely proportional to the frequency. | **Final Answer:** $\lambda = \frac{c}{\nu}$; Wavelength is inversely proportional to frequency. Using the fundamental relation for electromagnetic waves in vacuum: $c = \nu \lambda$ Rearranging to solve for wavelength: $\lambda = \frac{c}{\nu}$ As $c$ is a constant, $\lambda \propto \frac{1}{\nu}$ | Chemistry 10th JEE |
If the wavenumber ($\bar{\nu}$) of a specific radiation is given as $2.5 \times 10^4 \text{ cm}^{-1}$, calculate its frequency ($\nu$) in units of $\text{s}^{-1}$. Take the velocity of light ($c$) as $3 \times 10^{10} \text{ cm/s}$. | **Final Answer:** $\nu = 7.5 \times 10^{14} \text{ s}^{-1}$ Given: $\bar{\nu} = 2.5 \times 10^{4} \text{ cm}^{-1}$, $c = 3 \times 10^{10} \text{ cm/s}$ Using relation $\nu = c \bar{\nu}$ $$\nu = (3 \times 10^{10}) \times (2.5 \times 10^{4}) = 7.5 \times 10^{14} \text{ s}^{-1}$$ | Chemistry 10th JEE |
According to Planck's quantum theory, the energy ($E$) of a photon is related to its frequency ($\nu$). State this relationship clearly and use it to derive an expression for frequency ($\nu$) in terms of energy and Planck's constant ($h$). Furthermore, show how energy ($E$) is related to wavelength ($\lambda$) and the... | **Final Answer:** $\nu = \frac{E}{h}$ and $E = \frac{hc}{\lambda}$ Using Planck's equation: $E = h \nu$ Rearranging for frequency: $$\nu = \frac{E}{h}$$ Using relation $c = \nu \lambda \implies \nu = \frac{c}{\lambda}$ Substituting $\nu$ into energy equation: $$E = h \left( \frac{c}{\lambda} \right) = \frac{hc}{\lambda... | Chemistry 10th JEE |
For the wave parameters listed below, identify and state the most appropriate standard SI units or commonly used spectroscopic units:
(a) Wavelength ($\lambda$)
(b) Frequency ($\nu$)
(c) Wavenumber ($\bar{\nu}$)
(d) Velocity of propagation ($c$) | **Final Answer:** (a) $m, \text{nm}, \text{\AA}$; (b) $\text{Hz}, \text{s}^{-1}$; (c) $m^{-1}, \text{cm}^{-1}$; (d) $m \cdot \text{s}^{-1}$ $$\text{Wavelength } (\lambda): \text{SI Unit} = \text{m}, \; \text{Spectroscopic Units} = \text{nm} \text{ or } \text{\AA}$$ $$\text{Frequency } (\nu): \text{SI Unit} = \text{s}^{... | Chemistry 10th JEE |
Determine the number of distinct colors that are observed when a narrow beam of white light undergoes dispersion by passing through a glass prism. Briefly explain the physical reason why white light separates into these different components. | **Final Answer:** 7 distinct colors; Reason: Wavelength-dependent refractive index $\mu(\lambda)$ leads to different angles of deviation. $N_{\text{colors}} = 7 \text{ (Violet, Indigo, Blue, Green, Yellow, Orange, Red)}$ Using Cauchy's formula: $\mu(\lambda) = A + \frac{B}{\lambda^{2}}$ Using deviation formula: $\delta... | Chemistry 10th JEE |
Identify the classification of a spectrum that is characterized by the absence of specific wavelengths, appearing as dark lines or gaps in an otherwise bright field. Determine whether this specific type of spectrum is categorized as continuous or discontinuous and define the condition that causes these wavelengths to b... | **Final Answer:** The spectrum is an Absorption Spectrum, classified as a Discontinuous spectrum, caused by the absorption of photons where $h\nu = \Delta E$. Identification: Dark lines on a bright background $\rightarrow$ Absorption Spectrum Classification: Presence of discrete gaps $\rightarrow$ Discontinuous spectru... | Chemistry 10th JEE |
Explain how Niels Bohr utilized the concepts of stationary orbits and the quantization of angular momentum to resolve the problem of atomic stability inherent in classical physics models. | **Final Answer:** Bohr resolved atomic instability by postulating that electrons occupy discrete stationary orbits where they do not radiate energy, with a minimum allowed radius $r_{1}$ that prevents the electron from collapsing into the nucleus. Postulating angular momentum quantization: $mvr = \frac{nh}{2\pi}$ Equat... | Chemistry 10th JEE |
Bohr's model relies on specific constraints regarding the movement and energy of electrons. Evaluate whether the statement "the transition of an electron between different energy levels is a continuous process" is consistent with Bohr’s postulates, and explain the correct nature of energy changes during such transition... | **Final Answer:** The statement is incorrect; transitions in Bohr's model are discrete quantum jumps where electrons move instantaneously between orbits without existing in an intermediate state. Quantized energy levels: $E_{n} = -\frac{13.6 Z^{2}}{n^{2}} \text{ eV}$ Energy change during transition: $\Delta E = E_{fina... | Chemistry 10th JEE |
Identify the specific physical state in which an element must be present in order to produce its characteristic emission spectrum when excited by thermal or electrical energy. | **Final Answer:** Gaseous state Using $\Delta E = E_{2} - E_{1} = h\nu$ Characteristic spectra require $\text{isolated atoms}$ to maintain quantized energy levels. $\text{State with minimal interaction} \implies \text{Gaseous state}$ | Chemistry 10th JEE |
Describe the characteristics of a spectrum where the various wavelengths of light overlap and transition into one another without any observable dark gaps or interruptions, and provide the technical name for such a spectrum. | **Final Answer:** Continuous Spectrum $$\text{Wavelength range: } \lambda \in [\lambda_{min}, \lambda_{max}]$$ $$I(\lambda) > 0 \, \forall \lambda \in [\lambda_{min}, \lambda_{max}]$$ $$\text{Technical Name: Continuous Spectrum}$$ | Chemistry 10th JEE |
Determine the name of the spectrum that is generated when a substance is excited to a higher energy state and subsequently releases that energy in the form of electromagnetic radiation while returning to a ground state. | **Final Answer:** Emission Spectrum Excitation process: An electron in ground state $E_{1}$ absorbs energy to reach an excited state $E_{n}$. De-excitation process: The electron returns to a lower state, releasing energy as a photon: $\Delta E = E_{n} - E_{1}$. Photon energy equation: $h\nu = \frac{hc}{\lambda} = \Delt... | Chemistry 10th JEE |
According to the postulates of Niels Bohr, describe the geometry of the orbits in which electrons move and explain the energy state of an electron while it remains within one of these specified orbits. | **Final Answer:** Geometry: Circular orbits; Energy State: Constant (Stationary state) Geometry of orbits $\implies$ Circular: $\frac{mv^{2}}{r} = \frac{kZe^{2}}{r^{2}}$ Angular momentum quantization $\implies mvr = n\frac{h}{2\pi}, n \in \{1, 2, 3, \dots\}$ Energy state $\implies E = \text{constant}$ (Stationary state... | Chemistry 10th JEE |
Provide the mathematical expression used to calculate the radius $r_n$ of the $n^{th}$ orbit of a hydrogen-like atom, expressing the result in terms of the principal quantum number $n$, Planck’s constant $h$, the mass of the electron $m$, the electronic charge $e$, and the atomic number $Z$. | **Final Answer:** $$r_{n} = \frac{n^{2}h^{2}}{4\pi^{2}mZe^{2}}$$ $$mvr = \frac{nh}{2\pi} \implies v = \frac{nh}{2\pi mr}$$ $$\frac{mv^{2}}{r} = \frac{Ze^{2}}{r^{2}} \implies mv^{2} = \frac{Ze^{2}}{r}$$ $$m \left( \frac{nh}{2\pi mr} \right)^{2} = \frac{Ze^{2}}{r}$$ $$\frac{n^{2}h^{2}}{4\pi^{2}mr^{2}} = \frac{Ze^{2}}{r}$... | Chemistry 10th JEE |
For an electron revolving in a stable Bohr orbit of a hydrogen-like species, determine the numerical ratio of its Kinetic Energy ($KE$) to its Total Energy ($TE$). | **Final Answer:** $-1$ $PE = -\frac{kZe^{2}}{r}$ Using $\frac{mv^{2}}{r} = \frac{kZe^{2}}{r^{2}} \implies KE = \frac{1}{2}mv^{2} = \frac{kZe^{2}}{2r}$ $TE = KE + PE = \frac{kZe^{2}}{2r} - \frac{kZe^{2}}{r} = -\frac{kZe^{2}}{2r}$ $\text{Ratio} = \frac{KE}{TE} = \frac{\frac{kZe^{2}}{2r}}{-\frac{kZe^{2}}{2r}} = -1$ | Chemistry 10th JEE |
Calculate or determine the following values for a Hydrogen atom ($Z=1$):
a) The radius (in $\text{\AA}$) of the ground state orbit ($n=1$).
b) The orbital angular momentum of an electron in the second excited state ($n=3$).
c) The energy (in $eV/atom$) of an electron when it is in the second energy level ($n=2$).
d) Th... | **Final Answer:** a) $0.529 \text{ \AA}$, b) $1.5 \frac{h}{\pi}$, c) $-3.4 \text{ eV}$, d) $8$ Using $r_{n} = 0.529 \frac{n^{2}}{Z} \text{ \AA}$ $\rightarrow$ $r_{1} = 0.529 \frac{1^{2}}{1} = 0.529 \text{ \AA}$ Using $L = \frac{nh}{2\pi}$ for $n=3$ $\rightarrow$ $L = \frac{3h}{2\pi} = 1.5 \frac{h}{\pi}$ Using $E_{n} = ... | Chemistry 10th JEE |
Identify the specific scientist who proposed the postulate that the angular momentum of a revolving electron is quantized and can only be an integral multiple of $\frac{h}{2\pi}$. | **Final Answer:** Niels Bohr $L = n\frac{h}{2\pi}$ $n \in \{1, 2, 3, \dots\}$ $\text{Scientist} = \text{Niels Bohr}$ | Chemistry 10th JEE |
State the precise numerical value of Planck's constant ($h$) in both SI units ($J \cdot s$) and CGS units ($erg \cdot s$). | **Final Answer:** $h = 6.62607015 \times 10^{-34} \text{ J} \cdot \text{s}$ (SI) and $6.62607015 \times 10^{-27} \text{ erg} \cdot \text{s}$ (CGS) SI value: $h = 6.62607015 \times 10^{-34} \text{ J} \cdot \text{s}$ Conversion factor: $1 \text{ J} = 10^{7} \text{ erg}$ CGS value: $h = 6.62607015 \times 10^{-34} \times 1... | Chemistry 10th JEE |
Discuss the validity of the Bohr model by stating whether it successfully explains the following phenomena: the existence of stationary energy states, the origin of the hydrogen line spectrum, and the mathematical condition for the quantization of angular momentum ($mvr = \frac{nh}{2\pi}$). | **Final Answer:** The Bohr model successfully explains the existence of stationary energy states, the origin of the hydrogen line spectrum, and the mathematical condition for the quantization of angular momentum. Bohr's first postulate: $$E_{n} = -\frac{me^{4}}{8\epsilon_{0}^{2}n^{2}h^{2}} = \text{constant}$$ Bohr's th... | Chemistry 10th JEE |
Calculate the magnitude of the angular momentum for an electron moving in the third stationary orbit ($n=3$) of a hydrogen atom, expressing your answer in terms of Planck's constant $h$ and $\pi$. | **Final Answer:** $L = \frac{3h}{2\pi}$ Given: $n = 3$ Using Bohr's quantization condition $L = \frac{nh}{2\pi}$ $$L = \frac{3 \times h}{2\pi} = \frac{3h}{2\pi}$$ | Chemistry 10th JEE |
In the Bohr model of the hydrogen atom, the radius of the ground state (first shell) is denoted as $a_0$. Calculate the radius of the fourth shell ($n=4$) in terms of $a_0$. | **Final Answer:** $16 a_{0}$ Using $r_{n} = \frac{n^{2}}{Z} a_{0}$ $r_{4} = \frac{4^{2}}{1} a_{0}$ $$r_{4} = 16 a_{0}$$ | Chemistry 10th JEE |
Consider the following two propositions regarding atomic structure:
1. Show whether the radius of the second orbit of a $Be^{3+}$ ion is equal to, greater than, or less than the radius of the first orbit of a hydrogen atom.
2. State the mathematical relationship that describes how the radius of an orbit in a hydrogen-l... | **Final Answer:** The radius of the second orbit of $Be^{3+}$ is equal to the radius of the first orbit of H-atom; the relationship is $r \propto \frac{n^{2}}{Z}$. Using Bohr's radius formula $r_{n} = a_{0} \frac{n^{2}}{Z}$ For $Be^{3+}$: $n = 2$, $Z = 4$, so $r_{Be^{3+}} = a_{0} \frac{2^{2}}{4} = a_{0}$ For $H$: $n = ... | Chemistry 10th JEE |
Identify the scientist who proposed the azimuthal quantum number to account for the fine structure observed in atomic spectra, and briefly explain how this model modified Bohr's original circular orbit theory. | **Final Answer:** Arnold Sommerfeld; he modified Bohr's theory by introducing elliptical orbits where the ratio of the semi-minor axis $b$ to the semi-major axis $a$ is given by $\frac{b}{a} = \frac{k}{n}$. $L = mvr = \frac{nh}{2\pi}$ $\oint p_{\theta} d\theta = kh$ $\frac{b}{a} = \frac{k}{n}$ $k \in \{1, 2, \dots, n\}... | Chemistry 10th JEE |
Determine the simplified ratio of the radii of the first, second, and fourth Bohr orbits ($r_1 : r_2 : r_4$) for a hydrogen-like atom. | **Final Answer:** $1 : 4 : 16$ $r_{n} = \frac{n^{2} h^{2} \epsilon_{0}}{\pi m e^{2} Z}$ $r_{n} \propto n^{2}$ $r_{1} : r_{2} : r_{4} = 1^{2} : 2^{2} : 4^{2}$ $r_{1} : r_{2} : r_{4} = 1 : 4 : 16$ | Chemistry 10th JEE |
Based on the energy expression $E_n = -\frac{2\pi^2 Z^2 m e^4}{n^2 h^2}$, arrange the principal electronic shells $K, L, M,$ and $N$ in increasing order of their energy levels. Justify your answer by considering the relationship between the energy and the principal quantum number $n$. | **Final Answer:** $K < L < M < N$ $E_{n} = -\frac{2\pi^{2} Z^{2} m e^{4}}{n^{2} h^{2}} \implies E_{n} = -\frac{k}{n^{2}}$ where $k = \frac{2\pi^{2} Z^{2} m e^{4}}{h^{2}} > 0$ $n_{K} = 1, n_{L} = 2, n_{M} = 3, n_{N} = 4$ $E_{K} = -k, E_{L} = -\frac{k}{4}, E_{M} = -\frac{k}{9}, E_{N} = -\frac{k}{16}$ Comparing values: $-... | Chemistry 10th JEE |
Determine the numerical value of the radius of the first Bohr orbit for a Hydrogen atom ($Z=1$) in units of meters ($m$), and express this value in scientific notation. | **Final Answer:** $r_{1} = 5.29 \times 10^{-11} \text{ m}$ Using $r_{n} = \frac{n^{2} h^{2} \epsilon_{0}}{\pi m e^{2} Z}$ $r_{1} = \frac{1^{2} \times (6.626 \times 10^{-34})^{2} \times 8.854 \times 10^{-12}}{\pi \times 9.109 \times 10^{-31} \times (1.602 \times 10^{-19})^{2} \times 1}$ $$r_{1} = 5.29 \times 10^{-11} \t... | Chemistry 10th JEE |
Calculate the radius of the first Bohr orbit for the $Li^{2+}$ ion in Angstrom ($\text{\AA}$) units. | **Final Answer:** $r_{1} = 0.1763 \text{ \AA}$ Given: $n = 1$ and $Z = 3$ for $Li^{2+}$ ion Using $r_{n} = 0.529 \frac{n^{2}}{Z} \text{ \AA}$ $$r_{1} = 0.529 \times \frac{1^{2}}{3} = 0.1763 \text{ \AA}$$ | Chemistry 10th JEE |
If an electron in a hydrogen atom is found to have a total energy of $-34.84 \text{ kcal mol}^{-1}$, determine the principal quantum number ($n$) of the stationary orbit in which the electron is revolving. | **Final Answer:** $n = 3$ Given: $E_{n} = -34.84 \text{ kcal mol}^{-1}$ Using $E_{n} = \frac{E_{1}}{n^{2}}$, where $E_{1} \approx -313.6 \text{ kcal mol}^{-1}$ $-34.84 = \frac{-313.6}{n^{2}}$ $n^{2} = \frac{313.6}{34.84} \approx 9$ $n = \sqrt{9} = 3$ | Chemistry 10th JEE |
Analyze the properties of Bohr's orbits for a hydrogen atom and answer the following:
(i) Describe how the distance between two successive orbits ($r_{n+1} - r_n$) changes as the value of the principal quantum number $n$ increases.
(ii) Describe how the magnitude of the energy difference between two successive orbits (... | **Final Answer:** (i) Increases, (ii) Decreases, (iii) $0.529 \text{ \AA}$ $r_{n} = a_{0}n^{2} \implies \Delta r = r_{n+1} - r_{n} = a_{0}((n+1)^{2} - n^{2})$ $\Delta r = a_{0}(2n + 1) \implies \text{As } n \uparrow, \Delta r \uparrow$ $E_{n} = -13.6 \frac{1}{n^{2}} \text{ eV} \implies |\Delta E| = 13.6 \left( \frac{1}... | Chemistry 10th JEE |
Determine the numerical values for the following properties related to the Bohr model of the atom:
a) The radius of the third orbit ($n=3$) of a hydrogen atom in $\text{\AA}$.
b) The angular momentum of an electron moving in the $6^{th}$ orbit in terms of $\frac{h}{\pi}$.
c) The energy of the electron in the second orb... | **Final Answer:** (a) $4.761 \text{ \AA}$, (b) $\frac{3h}{\pi}$, (c) $-3.4 \text{ eV/atom}$, (d) $50$ Using $r_{n} = 0.529 \times \frac{n^{2}}{Z} \text{ \AA}$ for $n=3, Z=1$: $r_{3} = 0.529 \times \frac{3^{2}}{1} = 4.761 \text{ \AA}$ Using $L = \frac{nh}{2\pi}$ for $n=6$: $L = \frac{6h}{2\pi} = \frac{3h}{\pi}$ Using $E... | Chemistry 10th JEE |
Calculate the maximum electron capacity of the 'N' shell by applying the $2n^2$ rule, where $n$ represents the principal quantum number. | **Final Answer:** 32 Given: $n = 4$ (for 'N' shell) Maximum number of electrons $= 2n^{2}$ $$ \text{Maximum number of electrons} = 2 \times (4)^{2} = 32 $$ | Chemistry 10th JEE |
The three-dimensional geometry or "shape" of an atomic orbital is defined by a specific quantum number. Determine which quantum number this is and state the shapes associated with its values 0, 1, and 2. | **Final Answer:** Azimuthal Quantum Number ($l$); Shapes: $0$: Spherical, $1$: Dumbbell, $2$: Double-dumbbell $\text{Orbital Shape} \implies \text{Azimuthal Quantum Number } (l)$ $l = 0 \implies s\text{-orbital} \implies \text{Spherical}$ $l = 1 \implies p\text{-orbital} \implies \text{Dumbbell}$ $l = 2 \implies d\text... | Chemistry 10th JEE |
If a subshell is characterized by the azimuthal quantum number $l=2$ (the d-subshell), calculate the total number of electrons that can be accommodated in this subshell using the formula $2(2l+1)$. | **Final Answer:** 10 Given: $l=2$ Using formula $N = 2(2l+1)$ $$N = 2(2(2)+1) = 2(5) = 10$$ | Chemistry 10th JEE |
For a subshell where the azimuthal quantum number $l=2$, determine the total number of possible values for the magnetic quantum number ($m_l$). List these values to show the total count of orbitals in that subshell. | **Final Answer:** 5 Given: $l = 2$ $m_{l} \in \{-l, -(l-1), \dots, 0, \dots, (l-1), l\}$ $m_{l} = \{-2, -1, 0, +1, +2\}$ $N = 2l + 1 = 2(2) + 1 = 5$ | Chemistry 10th JEE |
For the fourth principal energy level ($n=4$), calculate the total number of orbitals (which corresponds to the total number of possible 'm' values) available in the entire shell. | **Final Answer:** 16 Given: $n = 4$ Formula for total orbitals in shell $n$: $N = n^{2}$ $$N = 4^{2} = 16$$ | Chemistry 10th JEE |
Determine the total number of orientations (values of the magnetic quantum number $m_l$) possible for an electron residing in any $d$ subshell. | **Final Answer:** 5 For any $d$ subshell, the azimuthal quantum number is $l = 2$ The number of possible orientations is given by the formula $2l + 1$ Substituting $l = 2$ into the formula: $2(2) + 1 = 5$ The allowed values for $m_{l}$ are $-2, -1, 0, +1, +2$ | Chemistry 10th JEE |
The orbital angular momentum of an electron is given by the expression $L = \sqrt{l(l+1)} \frac{h}{2\pi}$. Based on this formula, identify which specific quantum number determines the magnitude of the orbital angular momentum. | **Final Answer:** Azimuthal Quantum Number ($l$) Given: $L = \sqrt{l(l+1)} \frac{h}{2\pi}$ Constants: $h$ and $\pi$ are Planck's constant and mathematical constant respectively. Variable: $l$ is the Azimuthal (or Orbital) Quantum Number. | Chemistry 10th JEE |
In the context of the spin quantum number ($s$), the two possible orientations of an electron are clockwise and anti-clockwise. State the numerical value conventionally assigned to the spin orientation representing the clockwise direction. | **Final Answer:** $+\frac{1}{2}$ $m_{s} \in \{+\frac{1}{2}, -\frac{1}{2}\}$ $\text{Clockwise orientation} \implies m_{s} = +\frac{1}{2}$ | Chemistry 10th JEE |
Identify the specific subshell whose orbitals are characterized by a "double dumb-bell" geometric shape and state its corresponding azimuthal quantum number ($l$). | **Final Answer:** $d \text{ subshell, } l = 2$ $\text{Orbital Shape} = \text{double dumb-bell}$ $l = 2$ $\text{Subshell} = d$ | Chemistry 10th JEE |
Evaluate and state the maximum number of electrons allowed in the $p$, $d$, and $f$ subshells respectively. Furthermore, state the name and definition of the principle which dictates that no two electrons in an atom can possess an identical set of four quantum numbers. | **Final Answer:** $p = 6, d = 10, f = 14$; Pauli Exclusion Principle Formula for maximum electrons: $N = 2(2l + 1)$ $p$ subshell ($l=1$): $N = 2(2 \times 1 + 1) = 6$ $d$ subshell ($l=2$): $N = 2(2 \times 2 + 1) = 10$ $f$ subshell ($l=3$): $N = 2(2 \times 3 + 1) = 14$ Principle: Pauli Exclusion Principle (No identical $... | Chemistry 10th JEE |
For the $M$-shell of an atom, determine the corresponding principal quantum number ($n$), the total number of subshells present, the names of these subshells, and the maximum electron capacity of the entire shell. | **Final Answer:** $n = 3$, subshells: $3s, 3p, 3d$, maximum capacity: $18$ electrons For the $M$-shell, the principal quantum number is $n = 3$ Number of subshells in a shell is equal to $n$, so total subshells $= 3$ The subshells correspond to $l = 0, 1, 2$, which are named $3s, 3p, \text{ and } 3d$ Maximum electron c... | Chemistry 10th JEE |
Describe the properties of the magnetic quantum number ($m_l$) by stating what physical characteristic of the orbital it represents, its relationship to the azimuthal quantum number ($l$), and its role in explaining the splitting of spectral lines in a magnetic field. | **Final Answer:** The magnetic quantum number $m_{l}$ represents the orbital orientation, takes $2l+1$ values ranging from $-l$ to $+l$, and explains the Zeeman effect. Physical Characteristic: $m_{l}$ describes the spatial orientation of the orbital in three-dimensional space relative to a defined coordinate system. R... | Chemistry 10th JEE |
Using the general formula $2(2l+1)$ for the maximum number of electrons in a sub-energy level, calculate the maximum occupancy for a subshell where the azimuthal quantum number $l = 3$. | **Final Answer:** $14$ $N_{max} = 2(2l + 1)$ $N_{max} = 2(2(3) + 1)$ $N_{max} = 2(7) = 14$ | Chemistry 10th JEE |
Determine the total number of individual orbitals (sub-orbitals) contained within a $p$-subshell. | **Final Answer:** 3 For a $p$-subshell, the azimuthal quantum number $l = 1$ The number of orbitals in a subshell is given by the formula $2l + 1$ $$ 2(1) + 1 = 3 $$ | Chemistry 10th JEE |
Calculate the maximum number of electrons that can be accommodated in a subshell characterized by an azimuthal quantum number $l=2$. Show the formula used for your calculation. | **Final Answer:** 10 Formula for number of orbitals in a subshell: $n_{orb} = 2l + 1$ Substitute $l = 2$: $n_{orb} = 2(2) + 1 = 5$ Formula for max electrons: $N_{e} = 2 \times n_{orb}$ $N_{e} = 2 \times 5 = 10$ | Chemistry 10th JEE |
Consider the following three statements regarding the azimuthal quantum number ($l$):
(A) It describes the three-dimensional geometric shape of the orbital.
(B) For a given principal quantum number $n$, the possible values of $l$ range from $0$ to $n-1$.
(C) An azimuthal quantum number of $l=1$ corresponds to the 'p' s... | **Final Answer:** All three statements (A), (B), and (C) are correct. Statement (A): $\text{Orbital Shape} = f(l) \implies \text{True}$ Statement (B): $l \in \{0, 1, 2, \dots, n-1\} \implies \text{True}$ Statement (C): $l=1 \rightarrow \text{p subshell} \implies \text{True}$}], | Chemistry 10th JEE |
Based on the laws of quantum mechanics and Pauli’s exclusion principle, determine the maximum number of electrons that can reside in any single d-orbital (e.g., $d_{xy}$) and state the required relationship between their spin quantum numbers. | **Final Answer:** The maximum number of electrons is 2, and they must have opposite spins ($m_{s} = \pm \frac{1}{2}$) $$ \text{For a single orbital (e.g., } d_{xy} \text{): } n, l, m_{l} = \text{constant} $$ $$ \text{Pauli Exclusion Principle: Unique } (n, l, m_{l}, m_{s}) $$ $$ m_{s} \in \{+\frac{1}{2}, -\frac{1}{2}\}... | Chemistry 10th JEE |
Determine the standard spectroscopic designation (e.g., $1s$, $2p_x$, etc.) for an atomic orbital defined by the quantum numbers $n=3$, $l=2$, and $m=0$. | **Final Answer:** $3d_{z^{2}}$ $$n = 3 \implies \text{3rd Principal Shell}$$ $$l = 2 \implies d \text{ subshell}$$ $$m_{l} = 0 \text{ for } l = 2 \implies d_{z^{2}} \text{ orbital orientation}$$ $$\text{Designation} = 3d_{z^{2}}$$ | Chemistry 10th JEE |
Name the principle which states that in the ground state of an atom or ion, electrons fill atomic orbitals of the lowest available energy levels before occupying higher energy levels. | **Final Answer:** Aufbau Principle $$\text{Ground state electron filling} \rightarrow \text{Lowest Energy Orbitals first}$$ $$E_{1s} < E_{2s} < E_{2p} < E_{3s} < E_{3p} < E_{4s} < E_{3d}$$ $$\text{Identification} = \text{Aufbau Principle}$$ | Chemistry 10th JEE |
Identify the rule or principle which asserts that no two electrons in the same orbital can have the same set of four quantum numbers, effectively meaning they must have opposite spins. | **Final Answer:** Pauli Exclusion Principle Quantum state of an electron: $S = \{n, l, m_{l}, m_{s}\}$ Pauli Exclusion Principle condition: $S_{1} \neq S_{2}$ for any two electrons For same orbital, $(n, l, m_{l})$ are identical $\implies m_{s1} \neq m_{s2} \implies m_{s} \in \{+1/2, -1/2\}$ | Chemistry 10th JEE |
List all possible sets of the four quantum numbers $(n, l, m, s)$ for the electrons residing in the $3p$ subshell of a ground-state Phosphorus atom ($Z = 15$). | **Final Answer:** $(3, 1, -1, +\frac{1}{2}), (3, 1, 0, +\frac{1}{2}), (3, 1, 1, +\frac{1}{2})$ Ground state electronic configuration of $P (Z = 15)$: $1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{3}$ For $3p$ subshell: $n = 3$ and $l = 1$ According to Hund's Rule, the 3 electrons in $3p^{3}$ occupy separate orbitals with parallel s... | Chemistry 10th JEE |
A neutral Phosphorus atom ($Z = 15$) is found to have three unpaired electrons in its outermost $p$-subshell. Identify and explain the specific rule of electronic configuration that accounts for this observation rather than having the electrons pair up in a single orbital. | **Final Answer:** Hund's Rule of Maximum Multiplicity $Z = 15$ $1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{3}$ $\text{For } 3p^{3}: 3p_{x}^{1} 3p_{y}^{1} 3p_{z}^{1}$ $S = \sum m_{s} = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = \frac{3}{2}$ $\text{Multiplicity } = 2S + 1 = 2\left(\frac{3}{2}\right) + 1 = 4$ | Chemistry 10th JEE |
Evaluate the $(n+l)$ values for the following set of atomic orbitals: $4f, 5d, 6p,$ and $5p$. Based on your calculations, determine which of these orbitals possess identical $(n+l)$ sums. | **Final Answer:** The orbitals $4f, 5d,$ and $6p$ have identical $(n+l)$ sums of $7$. For $4f$: $n=4, l=3 \implies n+l = 4+3 = 7$ For $5d$: $n=5, l=2 \implies n+l = 5+2 = 7$ For $6p$: $n=6, l=1 \implies n+l = 6+1 = 7$ For $5p$: $n=5, l=1 \implies n+l = 5+1 = 6$ | Chemistry 10th JEE |
For each of the four quantum numbers—Principal ($n$), Azimuthal ($l$), Magnetic ($m$), and Spin ($s$)—describe the specific physical property of the electron or the orbital it defines, such as size, shape, spatial orientation, or the nature of electron rotation. | **Final Answer:** $n$: Size/Energy; $l$: Shape; $m$: Orientation; $s$: Spin direction Principal ($n$): Defines the main shell, size, and energy of the orbital. $E_{n} = -13.6 \frac{Z^{2}}{n^{2}} \text{ eV}$. Azimuthal ($l$): Defines the shape of the orbital and subshell. $L = \sqrt{l(l+1)} \frac{h}{2\pi}$. Magnetic ($m... | Chemistry 10th JEE |
Describe the fundamental characteristics of cathode rays and explain how their behavior in the presence of external electric and magnetic fields helps determine their nature. | **Final Answer:** Cathode rays are negatively charged particles (electrons) with a constant specific charge $\frac{e}{m} \approx 1.76 \times 10^{11} \text{ C/kg}$. Force in electric field: $F_{e} = eE$ Force in magnetic field: $F_{m} = evB$ Velocity selector for undeflected rays: $eE = evB \implies v = \frac{E}{B}$ Rad... | Chemistry 10th JEE |
Evaluate the various practical applications of X-rays, Infrared ($\text{IR}$) radiation, and Microwaves in the fields of medicine, diagnostic imaging, and telecommunications. | **Final Answer:** X-rays: Diagnostic Radiography; IR: Thermal physiotherapy and remotes; Microwaves: Satellite and mobile telecommunications. X-rays: $\nu \text{ range } 10^{16} - 10^{19} \, \text{Hz}$. Using $E = h\nu$, high energy allows penetration of soft tissues but absorption by dense materials. Result: Diagnosti... | Chemistry 10th JEE |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.