### Related Formula Bohr radius formula for hydrogen-like species: r_n = a_0 fracn^2Z$r_n = a_0 \frac{n^2}{Z}$ ### Core Logic For Lithium ion mathrmLi^++$\mathrm{Li}^{++}$: - Atomic number Z = 3$Z = 3$ - For ground state, the principal quantum number n = 1$n = 1$ Substitute Z = 3$Z = 3$ and n = 1$n = 1$ into Bohr's radius formula: r_1 = a_0 frac1^23 = fraca_03$r_1 = a_0 \frac{1^2}{3} = \frac{a_0}{3}$ Comparing with the given expression frac1X a_0$\frac{1}{X} a_0$: frac1X a_0 = fraca_03 implies X = 3$\frac{1}{X} a_0 = \frac{a_0}{3} \implies X = 3$ ### Pattern Recognition Sees: Ground state radius of hydrogen-like species. Trap: Confusing the atomic number Z$Z$ of Lithium with Helium (Z=2$Z=2$) or Beryllium (Z=4$Z=4$). Shortcut: Ground state radius of hydrogenic species is simply a_0 / Z$a_0 / Z$. Since Lithium has Z=3$Z=3$, the ground state radius must be a_0/3$a_0/3$ directly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Atoms

### Related Formula Bohr's electrostatic centripetal balance equation: fracm v^2r = frac14pivarepsilon_0 fracZ e^2r^2$\frac{m v^2}{r} = \frac{1}{4\pi\varepsilon_0} \frac{Z e^2}{r^2}$ This basic equation matches ONLY a single electron orbiting a nucleus of charge +Ze$+Ze$. ### Core Logic Assertion Analysis: - Bohr's model matches single-electron species (such as mathrmH$\mathrm{H}$, mathrmHe^+$\mathrm{He}^+$, mathrmLi^2+$\mathrm{Li}^{2+}$, etc.). Hence, Assertion A is true. Reason Analysis: - The model is restricted because it models ONLY the attractive force between the positive nucleus and one orbiting electron. It cannot handle multi-electron systems due to the presence of inter-electron repulsive forces, which are not integrated into Bohr's simple formulation. Hence, Reason R is true. Connection Check: - The lack of repulsive forces is exactly why the model fails for multi-electron species and remains applicable only to single-electron (hydrogen-like) species. Thus, R is the correct explanation of A. ### Pattern Recognition For single vs. multi-electron systems in atomic physics: Bohr model = single-electron ONLY. Quantum mechanics (Schrodinger) is required for multi-electron systems due to electron-electron interactions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Atoms

### Related Formula The energy of an electron in the n$n$-th shell of a hydrogen atom is: E_n = -frac13.6n^2mathrm~eV$E_n = -\frac{13.6}{n^2}\mathrm{~eV}$ The emitted photon energy during transition n_i rightarrow n_f$n_i \rightarrow n_f$ is: Delta E = E_n_i - E_n_f$\Delta E = E_{n_i} - E_{n_f}$ ### Core Logic Given state: - Initial state: "fourth excited state" $ ### Core Logic Given state: - Initial state: "fourth excited state" $\Rightarrow n_i = 5 - Emitted photon energy: $ - Emitted photon energy: $\Delta E = 2.86\mathrm{~eV} ### Step 1: Calculate energy of initial state ($ ### Step 1: Calculate energy of initial state ($E_5) $) $E_5 = -frac13.65^2 = -frac13.625 = -0.544mathrm~eV$E_5 = -\frac{13.6}{5^2} = -\frac{13.6}{25} = -0.544\mathrm{~eV}$ ### Step 2: Calculate final state energy ($E_n$) $ ### Step 2: Calculate final state energy ($E_n$) $2.86 = E_5 - E_n = -0.544 - E_n $ $E_n = -0.544 - 2.86 = -3.404\mathrm{~eV}$ ### Step 3: Determine the integer shell index ($n$) $$ ### Step 3: Determine the integer shell index ($n$) $-3.4 = -\frac{13.6}{n^2} $ $n^2 = \frac{13.6}{3.4} = 4 \Rightarrow n = 2$ ### Pattern Recognition Memorizing the first few energy levels of the hydrogen atom saves calculation time: - $ ### Pattern Recognition Memorizing the first few energy levels of the hydrogen atom saves calculation time: - $E_1 = -13.6\mathrm{~eV} - $ - $E_2 = -3.4\mathrm{~eV} - $ - $E_3 = -1.51\mathrm{~eV} - $ - $E_4 = -0.85\mathrm{~eV} - $ - $E_5 = -0.54\mathrm{~eV} Recognizing that a transition from $ Recognizing that a transition from $-0.54\mathrm{~eV} by emitting $ by emitting $2.86\mathrm{~eV} lands exactly at $ lands exactly at $-3.4\mathrm{~eV} maps directly to $ maps directly to $n=2$ instantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Atoms

### Related Formula For a hydrogen-like ion of atomic number Z$Z$, the transition energy between states n_2$n_2$ and n_1$n_1$ is: Delta E = 13.6 Z^2 left( frac1n_1^2 - frac1n_2^2 right) mathrm~eV$\Delta E = 13.6 Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \mathrm{~eV}$ ### Core Logic Identify the states: - Ground state: $ ### Core Logic Identify the states: - Ground state: $n_1 = 1 - $ - $2^{\mathrm{nd}} excitation energy state: $ excitation energy state: $n_2 = 3 Substitute values into the energy equation: $ Substitute values into the energy equation: $108.8 = 13.6 Z^2 left( frac11^2 - frac13^2 right)$108.8 = 13.6 Z^2 \left( \frac{1}{1^2} - \frac{1}{3^2} \right)$ $ $108.8 = 13.6 Z^2 left( 1 - frac19 right) = 13.6 Z^2 times frac89$108.8 = 13.6 Z^2 \left( 1 - \frac{1}{9} \right) = 13.6 Z^2 \times \frac{8}{9}$ ### Step 1: Solve for Z Rearrange the equation to isolate $ ### Step 1: Solve for Z Rearrange the equation to isolate $Z^2: $: $Z^2 = frac108.8 times 913.6 times 8$Z^2 = \frac{108.8 \times 9}{13.6 \times 8}$ Calculate the numerical value: $ Calculate the numerical value: $frac108.813.6 = 8$\frac{108.8}{13.6} = 8$ $ $Z^2 = frac8 times 98 = 9 implies Z = 3$Z^2 = \frac{8 \times 9}{8} = 9 \implies Z = 3$ ### Pattern Recognition Sees: "$ ### Pattern Recognition Sees: "$2^{\mathrm{nd}} excitation" $ excitation" $\implies n = 3. Ground state $. Ground state $\implies n = 1. Shortcut: Energy transition ratio for $. Shortcut: Energy transition ratio for $1 \to 3 is always $ is always $\frac{8}{9} of $ of $13.6 Z^2 \approx 12.1 Z^2. Since $. Since $108.8 / 12.1 \approx 9, $, $Z^2 = 9 \implies Z = 3$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Atoms

### Related Formula The wavelength lambda$\lambda$ for a transition in a hydrogen atom is given by the Rydberg formula: frac1lambda = R left( frac1n_1^2 - frac1n_2^2 right)$\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$ To find the largest wavelength (minimum energy transition), select the adjacent higher shell n_2 = n_1 + 1$n_2 = n_1 + 1$. ### Core Logic - **Lyman Series largest wavelength (lambda_L$\lambda_L$)**: Transition from n = 2 to 1$n = 2 \to 1$ frac1lambda_L = R left( frac11^2 - frac12^2 right) = frac3R4 implies lambda_L = frac43R$\frac{1}{\lambda_L} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = \frac{3R}{4} \implies \lambda_L = \frac{4}{3R}$ - **Balmer Series largest wavelength (lambda_B$\lambda_B$)**: Transition from n = 3 to 2$n = 3 \to 2$ frac1lambda_B = R left( frac12^2 - frac13^2 right) = R left( frac14 - frac19 right) = frac5R36 implies lambda_B = frac365R$\frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = \frac{5R}{36} \implies \lambda_B = \frac{36}{5R}$ ### Step 1: Ratio Calculation Now, compute the ratio of the wavelengths: fraclambda_Llambda_B = fracfrac43Rfrac365R = frac43 times frac536 = frac527$\frac{\lambda_L}{\lambda_B} = \frac{\frac{4}{3R}}{\frac{36}{5R}} = \frac{4}{3} \times \frac{5}{36} = \frac{5}{27}$ ### Pattern Recognition Sees: Ratio of largest wavelengths of series. Shortcut: The largest wavelength in a series starting at ground level $ ### Pattern Recognition Sees: Ratio of largest wavelengths of series. Shortcut: The largest wavelength in a series starting at ground level $n_1 is $ is $\lambda \propto \frac{n_1^2 (n_1+1)^2}{2n_1 + 1}. For Lyman ($. For Lyman ($n_1=1): $): $\lambda_L \propto \frac{4}{3}. For Balmer ($. For Balmer ($n_1=2): $): $\lambda_B \propto \frac{36}{5}. Ratio: $. Ratio: $\frac{4/3}{36/5} = \frac{5}{27}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Atoms