Masterclass: 25 Solved JEE Advanced Numericals on Structure of Atom
Unravel the mysteries of quantum mechanics, Bohr's orbits, hydrogen spectra, and the photoelectric effect. Click "View Solution" to reveal the step-by-step mathematical breakdown.
Chapter 2 of Class 11 Chemistry is the gateway to modern physics and chemistry. Before attempting these rigorous problems, ensure your grasp of Planck's quantum theory, the Rydberg equation, and the four quantum numbers is absolute!
View Solution
Step 1: Use the General Bohr Energy Formula
The energy of an electron in the $n$-th orbit of a hydrogen-like species is given by:
$E_n = -13.6 \times \frac{Z^2}{n^2} \text{ eV}$
Step 2: Identify Variables
For ${He^+}$, the atomic number $Z = 2$.
The orbit number is $n = 3$.
Step 3: Calculate Energy ($E_3$)
$E_3 = -13.6 \times \frac{2^2}{3^2} = -13.6 \times \frac{4}{9} = -6.044 \text{ eV}$.
Step 4: Calculate Ionization Energy ($I.E.$)
Ionization energy is the energy required to move the electron from its current state to infinity ($E_{\infty} = 0$).
$I.E. = E_{\infty} - E_3 = 0 - (-6.044) = +6.044 \text{ eV}$.
View Solution
Step 1: Calculate Energy of Incident Photon ($E$)
Using the practical shortcut formula $E \text{ (in eV)} = \frac{12400}{\lambda \text{ (in \AA)}}$:
$E = \frac{12400}{4000} = 3.1 \text{ eV}$.
Step 2: Apply Einstein's Photoelectric Equation
$E = W + K.E._{\text{max}}$
Where Work Function ($W$) $= 2.0 \text{ eV}$.
$3.1 = 2.0 + K.E._{\text{max}} \implies K.E._{\text{max}} = 1.1 \text{ eV}$.
Step 3: Calculate Stopping Potential ($V_0$)
Kinetic energy is related to stopping potential by $K.E._{\text{max}} = eV_0$.
$1.1 \text{ eV} = e \times V_0 \implies V_0 = 1.1 \text{ V}$.
View Solution
Step 1: Bohr Radius Formula
The radius of the $n$-th orbit is given by $r_n = 0.529 \times \frac{n^2}{Z} \text{ \AA}$.
Therefore, $r \propto \frac{n^2}{Z}$.
Step 2: Setup for ${Li^{2+}}$ (2nd orbit)
For ${Li^{2+}}$, $Z = 3$, $n = 2$.
$r_{Li^{2+}} \propto \frac{2^2}{3} = \frac{4}{3}$.
Step 3: Setup for ${He^+}$ (3rd orbit)
For ${He^+}$, $Z = 2$, $n = 3$.
$r_{He^+} \propto \frac{3^2}{2} = \frac{9}{2}$.
Step 4: Calculate the Ratio
$\text{Ratio} = \frac{r_{Li^{2+}}}{r_{He^+}} = \frac{4/3}{9/2} = \frac{4}{3} \times \frac{2}{9} = \frac{8}{27}$.
View Solution
Step 1: Shortest Wavelength of Lyman Series
Lyman series: transition to ${n_1} = 1$.
Shortest wavelength corresponds to highest energy, i.e., transition from ${n_2} = \infty$.
$\frac{1}{\lambda_L} = R_H (1^2) \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R_H \times 1 = R_H$
$\lambda_L = \frac{1}{R_H}$.
Step 2: Longest Wavelength of Balmer Series
Balmer series: transition to ${n_1} = 2$.
Longest wavelength corresponds to lowest energy, i.e., transition from ${n_2} = 3$.
$\frac{1}{\lambda_B} = R_H (1^2) \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R_H \left( \frac{1}{4} - \frac{1}{9} \right) = R_H \left( \frac{5}{36} \right)$
$\lambda_B = \frac{36}{5 R_H}$.
Step 3: Calculate the Ratio
$\text{Ratio} = \frac{\lambda_L}{\lambda_B} = \frac{1/R_H}{36/(5 R_H)} = \frac{5}{36}$.
View Solution
Step 1: Formula for De Broglie Wavelength
The general formula is $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$.
When accelerated by potential $V$, kinetic energy $K = eV$.
$\lambda = \frac{h}{\sqrt{2meV}}$.
Step 2: Practical Shortcut for Electron
Substituting the constants $h, m_e, \text{ and } e$ into the formula gives a highly useful shortcut for electrons:
$\lambda \text{ (\AA)} \approx \sqrt{\frac{150}{V (\text{in Volts})}} \approx \frac{12.27}{\sqrt{V}}$.
Step 3: Calculate
$\lambda = \sqrt{\frac{150}{100}} = \sqrt{1.5} \approx 1.22 \text{ \AA}$.
Alternatively, $\frac{12.27}{\sqrt{100}} = \frac{12.27}{10} = 1.227 \text{ \AA}$.
View Solution
Step 1: Determine Uncertainty in Velocity ($\Delta v$)
Velocity $v = 45 \text{ m/s}$.
Accuracy is $2\%$, so uncertainty $\Delta v = 2\% \text{ of } 45 = \frac{2}{100} \times 45 = 0.9 \text{ m/s}$.
Step 2: Apply Heisenberg's Equation
$\Delta x \cdot \Delta p \ge \frac{h}{4\pi} \implies \Delta x \cdot (m \Delta v) \ge \frac{h}{4\pi}$
$\Delta x \ge \frac{h}{4\pi m \Delta v}$.
Step 3: Substitute Values (Use SI Units)
Mass $m = 40 \text{ g} = 0.040 \text{ kg}$.
$\Delta x = \frac{6.626 \times 10^{-34}}{4 \times 3.1416 \times 0.040 \times 0.9}$
$\Delta x = \frac{6.626 \times 10^{-34}}{0.4524} \approx 1.46 \times 10^{-33} \text{ m}$.
Note: This incomprehensibly tiny uncertainty proves that macroscopic objects do not display observable quantum mechanical effects.
View Solution
Step 1: Identify Quantum Numbers
For a $5d$ orbital, the principal quantum number $n = 5$.
The azimuthal quantum number $l$ for a $d$-orbital is $2$.
Step 2: Calculate Angular Nodes
Angular nodes are determined purely by $l$.
Angular nodes $= l = 2$.
Step 3: Calculate Radial Nodes
Radial nodes (spherical nodes) formula: $n - l - 1$.
Radial nodes $= 5 - 2 - 1 = 2$.
Step 4: Calculate Total Nodes
Total nodes $= n - 1 = 5 - 1 = 4$. (Which matches Radial + Angular = $2 + 2 = 4$).
View Solution
Step 1: Use the Spectral Lines Formula
When a sample of hydrogen atoms drops from an excited state $n_2$ to a lower state $n_1$, the maximum number of spectral lines produced is given by:
$\text{Number of lines} = \frac{(n_2 - n_1)(n_2 - n_1 + 1)}{2}$
Step 2: Substitute Values
Here, $n_2 = 5$ and $n_1 = 1$.
$\text{Number of lines} = \frac{(5 - 1)(5 - 1 + 1)}{2} = \frac{4 \times 5}{2} = \frac{20}{2} = 10$.
Step 3: Breakdown (Optional Verification)
Lyman series (drops to $n=1$): 5→1, 4→1, 3→1, 2→1 (4 lines).
Balmer series (drops to $n=2$): 5→2, 4→2, 3→2 (3 lines).
Paschen series (drops to $n=3$): 5→3, 4→3 (2 lines).
Brackett series (drops to $n=4$): 5→4 (1 line).
Total $= 4+3+2+1 = 10$.
View Solution
Step 1: Set up the Rydberg Equations
For ${He^+}$ ($Z=2$):
$\frac{1}{\lambda} = R_H (2^2) \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = 4 R_H \left( \frac{1}{4} - \frac{1}{16} \right)$
For Hydrogen ($Z=1$) making a transition from $n_2$ to $n_1$:
$\frac{1}{\lambda} = R_H (1^2) \left( \frac{1}{{n_1}^2} - \frac{1}{{n_2}^2} \right)$
Step 2: Equate and Solve
Since wavelengths are identical, equate the two expressions:
$4 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = \left( \frac{1}{{n_1}^2} - \frac{1}{{n_2}^2} \right)$
Step 3: Distribute the 4
$\left( \frac{4}{2^2} - \frac{4}{4^2} \right) = \left( \frac{1}{{n_1}^2} - \frac{1}{{n_2}^2} \right)$
$\left( \frac{4}{4} - \frac{4}{16} \right) = \left( \frac{1}{{n_1}^2} - \frac{1}{{n_2}^2} \right)$
$\left( 1 - \frac{1}{4} \right) = \left( \frac{1}{{n_1}^2} - \frac{1}{{n_2}^2} \right)$
$\left( \frac{1}{1^2} - \frac{1}{2^2} \right) = \left( \frac{1}{{n_1}^2} - \frac{1}{{n_2}^2} \right)$
Step 4: Conclusion
Comparing both sides, $n_1 = 1$ and $n_2 = 2$.
View Solution
Step 1: Identify Quantum Number
The orbital angular momentum depends entirely on the azimuthal quantum number ($l$), not the principal quantum number ($n$).
For any $p$-orbital, $l = 1$.
Step 2: Use the Formula
$\text{Orbital Angular Momentum} (\mu_l) = \sqrt{l(l+1)} \frac{h}{2\pi}$
Step 3: Substitute Values
$\mu_l = \sqrt{1(1+1)} \frac{h}{2\pi} = \sqrt{2} \frac{h}{2\pi}$.
(Sometimes $\frac{h}{2\pi}$ is written as $\hbar$, so it becomes $\sqrt{2}\hbar$).
Set A: $(3, 2, -2, +1/2)$
Set B: $(4, 0, 0, -1/2)$
Set C: $(2, 2, 1, +1/2)$
Set D: $(5, 3, -4, -1/2)$
View Solution
Step 1: Check Rules for Quantum Numbers
$n$ must be a positive integer ($1, 2, 3...$).
$l$ must range from $0$ to $(n-1)$.
$m_l$ must range from $-l$ to $+l$, including $0$.
$m_s$ must be exactly $+1/2$ or $-1/2$.
Step 2: Analyze Each Set
Set A $(3, 2, -2, +1/2)$: $n=3$, $l=2$ (allowed since $2 < 3$). $m_l=-2$ (allowed since $-2$ is between $-2$ and $+2$). Valid (a $3d$ electron).
Set B $(4, 0, 0, -1/2)$: $n=4$, $l=0$ (allowed). $m_l=0$ (allowed). Valid (a $4s$ electron).
Set C $(2, 2, 1, +1/2)$: $n=2$. Thus, $l$ can only be $0$ or $1$. Here $l=2$, which violates the rule $l < n$. A $2d$ orbital does not exist.
Set D $(5, 3, -4, -1/2)$: $n=5$, $l=3$ (allowed). But $m_l$ must be between $-3$ and $+3$. Here $m_l = -4$, which is out of bounds.
View Solution
Step 1: Wavelength of Photon ($\lambda_p$)
For a photon, energy $E = \frac{hc}{\lambda_p} \implies \lambda_p = \frac{hc}{E}$.
Step 2: De Broglie Wavelength of Electron ($\lambda_e$)
For a particle with mass $m$ and kinetic energy $E$, momentum $p = \sqrt{2mE}$.
$\lambda_e = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$.
Step 3: Calculate Ratio
$\text{Ratio} = \frac{\lambda_e}{\lambda_p} = \frac{h / \sqrt{2mE}}{hc / E}$
$\text{Ratio} = \frac{h}{\sqrt{2mE}} \times \frac{E}{hc} = \frac{E}{c\sqrt{2mE}} = \frac{1}{c} \sqrt{\frac{E^2}{2mE}} = \frac{1}{c} \sqrt{\frac{E}{2m}}$.
View Solution
Step 1: Calculate Velocity in $n=2$
Velocity in $n$-th orbit $v_n = v_1 \times \frac{Z}{n}$.
For H ($Z=1, n=2$), $v_2 = \frac{2.18 \times 10^6}{2} = 1.09 \times 10^6 \text{ m/s}$.
Step 2: Calculate Radius of $n=2$
Radius $r_n = r_1 \times \frac{n^2}{Z}$.
For H ($Z=1, n=2$), $r_2 = 0.529 \times 2^2 = 2.116 \text{ \AA} = 2.116 \times 10^{-10} \text{ m}$.
Step 3: Calculate Time Period ($T$)
Distance traveled in one revolution $= 2\pi r_2$.
$T = \frac{\text{Distance}}{\text{Velocity}} = \frac{2\pi r_2}{v_2}$
$T = \frac{2 \times 3.1416 \times 2.116 \times 10^{-10}}{1.09 \times 10^6}$
$T = \frac{13.29 \times 10^{-10}}{1.09 \times 10^6} \approx 1.22 \times 10^{-15} \text{ seconds}$.
View Solution
Step 1: Write Energy Expression
$E_n = -13.6 \frac{Z^2}{n^2} \text{ eV}$.
Step 2: Express the Difference ($\Delta E$)
$\Delta E = E_{n+1} - E_n = -13.6 Z^2 \left( \frac{1}{(n+1)^2} - \frac{1}{n^2} \right)$
$\Delta E = -13.6 Z^2 \left( \frac{n^2 - (n+1)^2}{n^2 (n+1)^2} \right)$
$\Delta E = -13.6 Z^2 \left( \frac{n^2 - (n^2 + 2n + 1)}{n^2 (n+1)^2} \right)$
$\Delta E = -13.6 Z^2 \left( \frac{-2n - 1}{n^2 (n+1)^2} \right) = 13.6 Z^2 \left( \frac{2n + 1}{n^2 (n+1)^2} \right)$
Step 3: Apply Limit as $n \rightarrow \infty$
As $n$ becomes very large, the highest power of $n$ in the numerator is $n^1$, and in the denominator, it is $n^4$.
$\lim_{n \rightarrow \infty} \frac{2n}{n^4} = \lim_{n \rightarrow \infty} \frac{2}{n^3} = 0$.
View Solution
Step 1: Write Electronic Configuration
Neutral Mn ($Z=25$): $[Ar] \ 4s^2 \ 3d^5$.
For ${Mn^{2+}}$, two electrons are removed. According to rules, electrons are removed from the outermost shell ($4s$) first.
${Mn^{2+}}$ configuration: $[Ar] \ 3d^5$.
Step 2: Determine Unpaired Electrons ($n$)
According to Hund's Rule of Maximum Multiplicity, the 5 electrons in the $d$-subshell will occupy 5 separate orbitals with parallel spins.
Number of unpaired electrons, $n = 5$.
Step 3: Calculate Spin-Only Magnetic Moment ($\mu$)
Formula: $\mu = \sqrt{n(n+2)} \text{ B.M.}$ (Bohr Magnetons)
$\mu = \sqrt{5(5+2)} = \sqrt{5 \times 7} = \sqrt{35}$.
Step 4: Approximate Square Root
Since $\sqrt{36} = 6$, $\sqrt{35}$ is slightly less, approximately $5.92$.
View Solution
Step 1: Set up Einstein Equations
Case 1: $h\nu = h\nu_0 + K \implies K = h\nu - h\nu_0$
Case 2: $h(2\nu) = h\nu_0 + 3K \implies 3K = 2h\nu - h\nu_0$
Step 2: Solve Simultaneous Equations
Substitute the expression for $K$ from Case 1 into Case 2:
$3(h\nu - h\nu_0) = 2h\nu - h\nu_0$
$3h\nu - 3h\nu_0 = 2h\nu - h\nu_0$
Step 3: Isolate terms
Bring $h\nu$ to one side and $h\nu_0$ to the other:
$3h\nu - 2h\nu_0 = 3h\nu_0 - h\nu_0$ (Wait, correct algebraic isolation):
$3h\nu - 2h\nu = 3h\nu_0 - h\nu_0$
$h\nu = 2h\nu_0$
Step 4: Cancel $h$
$\nu = 2\nu_0 \implies \nu_0 = \frac{\nu}{2}$.
View Solution
Step 1: Determine Total Electrons in a Principal Shell
The maximum number of electrons that can fit in any principal shell $n$ is given by $2n^2$.
For $n = 4$, Total electrons $= 2(4^2) = 2(16) = 32$ electrons.
Step 2: Apply Spin Restriction
According to Pauli's Exclusion Principle, exactly half of these electrons will have "spin up" ($m_s = +1/2$) and exactly half will have "spin down" ($m_s = -1/2$).
Step 3: Calculate
Electrons with $m_s = -1/2$ is $32 / 2 = 16$.
View Solution
Step 1: Bohr's Quantization Postulate
Bohr stated that angular momentum is quantized: $mvr = \frac{nh}{2\pi}$.
Step 2: Rearrange for Circumference
Circumference of a circle is $2\pi r$.
Rearrange Bohr's postulate: $2\pi r = \frac{nh}{mv}$.
Step 3: Apply De Broglie Relation
De Broglie stated that wavelength $\lambda = \frac{h}{p} = \frac{h}{mv}$.
Substitute this into our rearranged equation:
$2\pi r = n \left( \frac{h}{mv} \right) = n \lambda$.
View Solution
Step 1: Define "Series Limit"
The "series limit" corresponds to the shortest wavelength (highest energy) line in the series. This occurs when an electron transitions from $n_2 = \infty$ down to the base level of the series.
Step 2: Identify Base Level
For the Paschen series, the base level is $n_1 = 3$.
Step 3: Apply Rydberg Equation for Wave Number ($\bar{\nu}$)
$\bar{\nu} = \frac{1}{\lambda} = R_H (1^2) \left( \frac{1}{3^2} - \frac{1}{\infty^2} \right)$
$\bar{\nu} = R_H \left( \frac{1}{9} - 0 \right) = \frac{R_H}{9}$.
Step 4: Calculate
$\bar{\nu} = \frac{109677}{9} = 12186.33 \text{ cm}^{-1}$.
View Solution
Step 1: Define States
Ground state implies $n_1 = 1$.
First excited state implies $n_2 = 2$.
Step 2: Energy Formula for ${Li^{2+}}$
For ${Li^{2+}}$, Atomic number $Z = 3$.
$E_n = -13.6 \times \frac{Z^2}{n^2} = -13.6 \times \frac{9}{n^2} \text{ eV}$.
Step 3: Calculate Energies of Levels
$E_1 = -13.6 \times \frac{9}{1^2} = -122.4 \text{ eV}$.
$E_2 = -13.6 \times \frac{9}{2^2} = -13.6 \times \frac{9}{4} = -30.6 \text{ eV}$.
Step 4: Calculate Excitation Energy ($\Delta E$)
$\Delta E = E_2 - E_1 = -30.6 - (-122.4) = 122.4 - 30.6 = 91.8 \text{ eV}$.
View Solution
Step 1: Properties of Particles
Let proton mass = $m$ and charge = $+e$.
Alpha particle (${He^{2+}}$ nucleus) mass $\approx 4m$ and charge = $+2e$.
Step 2: Wavelength Formula
$\lambda = \frac{h}{\sqrt{2m q V}}$
Step 3: Set up Ratio
$\frac{\lambda_p}{\lambda_{\alpha}} = \frac{h / \sqrt{2 m_p q_p V}}{h / \sqrt{2 m_{\alpha} q_{\alpha} V}}$
$\frac{\lambda_p}{\lambda_{\alpha}} = \sqrt{\frac{m_{\alpha} q_{\alpha}}{m_p q_p}}$
Step 4: Substitute and Solve
$\frac{\lambda_p}{\lambda_{\alpha}} = \sqrt{\frac{(4m)(2e)}{(m)(e)}} = \sqrt{\frac{8me}{me}} = \sqrt{8} = 2\sqrt{2}$.
View Solution
Step 1: Apply Rydberg Formula
Transition to $n_1=3$ defines the Paschen series.
$\frac{1}{\lambda} = 109677 \left( \frac{1}{3^2} - \frac{1}{6^2} \right) \text{ cm}^{-1}$
$\frac{1}{\lambda} = 109677 \left( \frac{1}{9} - \frac{1}{36} \right) = 109677 \left( \frac{4-1}{36} \right) = 109677 \left( \frac{3}{36} \right) = 109677 \left( \frac{1}{12} \right)$
Step 2: Calculate Wavelength
$\lambda = \frac{12}{109677} \text{ cm} \approx 1.094 \times 10^{-4} \text{ cm}$.
Convert to nm: $1.094 \times 10^{-4} \text{ cm} = 1.094 \times 10^{-6} \text{ m} = 1094 \text{ nm}$.
Step 3: Analyze Electromagnetic Spectrum
Visible light is roughly $400 \text{ nm}$ to $700 \text{ nm}$. Wavelengths longer than $700 \text{ nm}$ fall into the Infrared (IR) region. $1094 \text{ nm}$ is clearly in the IR region.
View Solution
Step 1: Formula for Exchanges
Electrons with the same spin in degenerate orbitals can exchange positions. The number of exchanges is given by combination math: $\frac{n(n-1)}{2}$, where $n$ is the number of electrons with parallel spin.
Step 2: Calculate for $d^5$ (Half-filled)
In $d^5$, there are 5 electrons with parallel spin (all up).
Exchanges $= \frac{5(5-1)}{2} = \frac{5 \times 4}{2} = 10$ exchanges.
Step 3: Calculate for $d^4$
In $d^4$, there are 4 electrons with parallel spin.
Exchanges $= \frac{4(4-1)}{2} = \frac{4 \times 3}{2} = 6$ exchanges.
Step 4: Conclusion
Every exchange releases a tiny amount of energy (Exchange Energy). The leap from 6 to 10 exchanges provides a massive energetic stabilization, driving anomalous configurations like Chromium ($[Ar] 4s^1 3d^5$ instead of $4s^2 3d^4$).
View Solution
Step 1: Relate KE, PE, and TE
From Bohr's model mathematics:
$\text{Total Energy (TE)} = - \text{Kinetic Energy (KE)}$
$\text{Potential Energy (PE)} = 2 \times \text{Total Energy}$
$\text{PE} = -2 \times \text{KE}$
Step 2: Calculate Energies
$\text{TE} = -3.4 \text{ eV}$.
$\text{PE} = 2 \times (-3.4) = -6.8 \text{ eV}$.
Step 3: Calculate Quantum Number ($n$)
We know $E_n = \frac{-13.6}{n^2} \text{ eV}$ for Hydrogen.
$-3.4 = \frac{-13.6}{n^2}$
$n^2 = \frac{13.6}{3.4} = 4 \implies n = 2$.
View Solution
Step 1: Calculate Effective Power Output
Total Power $= 100 \text{ W} = 100 \text{ J/s}$.
Efficiency is $60\%$, so useful power emitting light $= 60\% \text{ of } 100 = 60 \text{ J/s}$.
Total Energy emitted per second ($E_{total}$) $= 60 \text{ J}$.
Step 2: Calculate Energy of a Single Photon ($E$)
$E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \text{ J s} \times 3 \times 10^8 \text{ m/s}}{589 \times 10^{-9} \text{ m}}$
$E = \frac{19.878 \times 10^{-26}}{589 \times 10^{-9}} \approx 3.37 \times 10^{-19} \text{ J}$.
Step 3: Calculate Number of Photons ($N$)
$E_{total} = N \times E \implies N = \frac{E_{total}}{E}$
$N = \frac{60}{3.37 \times 10^{-19}} \approx 17.8 \times 10^{19}$ photons.
Mastering the Quantum Realm
Congratulations on completing these 25 highly advanced numericals on the Structure of Atom. From navigating the subtle ratio traps of Bohr's radius equations to balancing the kinetic and potential energies of quantum orbits, you have built the ultimate foundation. Remember, in modern physics questions, managing exponents and units (especially Angstroms vs nanometers, and eV vs Joules) is just as important as the formulas themselves!
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