Assuming the validity of Bohr's atomic model for hydrogen like ions the radius of mathrmLi^++$\mathrm{Li}^{++}$ ion in its ground state is given by frac1mathrmXmathrma_0$\frac{1}{\mathrm{X}}\mathrm{a}_0$ , where mathrmX = \_ \_ \_ .$\mathrm{X} = \_ \_ \_ .$ (Where mathbfa_0$\mathbf{a}_0$ is the first Bohr's radius.)

### Related Formula K_textinitial = U_textclosest approach$$K_{\text{initial}} = U_{\text{closest approach}}$$ K = frac14pi epsilon_0 frac(2e)(Ze)r_0$$K = \frac{1}{4\pi \epsilon_{0}} \frac{(2e)(Ze)}{r_0}$$ ### Core Logic By energy conservation, the entire kinetic energy of the alpha particle gets converted to electrostatic potential energy at the distance of closest approach (r_0$r_0$). K_i + U_i = K_f + U_f$K_i + U_i = K_f + U_f$ K_i + 0 = 0 + frac14piepsilon_0 frac(2e)(79e)r_0$$K_i + 0 = 0 + \frac{1}{4\pi\epsilon_0} \frac{(2e)(79e)}{r_0}$$ ### Step 1: Convert Energy and Solve Initial kinetic energy K_i = 7.7text MeV = 7.7 times 10^6 times 1.6 times 10^-19text J$K_i = 7.7\text{ MeV} = 7.7 \times 10^6 \times 1.6 \times 10^{-19}\text{ J}$. 7.7 times 10^6 times 1.6 times 10^-19 = frac9 times 10^9 times (2 times 1.6 times 10^-19) times (79 times 1.6 times 10^-19)r_0$$7.7 \times 10^6 \times 1.6 \times 10^{-19} = \frac{9 \times 10^9 \times (2 \times 1.6 \times 10^{-19}) \times (79 \times 1.6 \times 10^{-19})}{r_0}$$ r_0 = frac9 times 10^9 times 2 times 79 times (1.6 times 10^-19)^27.7 times 10^6 times 1.6 times 10^-19$$r_0 = \frac{9 \times 10^9 \times 2 \times 79 \times (1.6 \times 10^{-19})^2}{7.7 \times 10^6 \times 1.6 \times 10^{-19}}$$ r_0 = frac9 times 10^9 times 158 times 1.6 times 10^-197.7 times 10^6$$r_0 = \frac{9 \times 10^9 \times 158 \times 1.6 \times 10^{-19}}{7.7 \times 10^6}$$ r_0 = frac2275.2 times 10^-107.7 times 10^6 = 295.48 times 10^-16text m approx 2.95 times 10^-14text m$$r_0 = \frac{2275.2 \times 10^{-10}}{7.7 \times 10^6} = 295.48 \times 10^{-16}\text{ m} \approx 2.95 \times 10^{-14}\text{ m}$$ ### Pattern Recognition Distance of closest approach problem: Simply equate initial Kinetic Energy (in Joules) to Potential Energy k(Z_1e)(Z_2e)/r_0$k(Z_1e)(Z_2e)/r_0$. Alpha particle has charge 2e$2e$. ### 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