The smallest wavelength of Lyman series is 91 nm. The difference between the largest wavelengths of Paschen and Balmer series is nearly ____ nm.

### 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 E_n = fracE_0n^2$$E_n = \frac{E_0}{n^2}$$ L = fracnh2pi$$L = \frac{nh}{2\pi}$$ ### Core Logic From Bohr's theory, the energy of an electron in the n$n$-th orbit is inversely proportional to n^2$n^2$: E = -fracE_0n^2$$E = -\frac{E_0}{n^2}$$ Equating this to the given energy: -fracE_0n^2 = -0.04 E_0$$-\frac{E_0}{n^2} = -0.04 E_0$$ frac1n^2 = 0.04 = frac125$$\frac{1}{n^2} = 0.04 = \frac{1}{25}$$ n^2 = 25 implies n = 5$$n^2 = 25 \implies n = 5$$ ### Step 1: Final Conclusion Bohr's quantization condition for angular momentum: L = fracnh2pi$$L = \frac{nh}{2\pi}$$ Rearranging for the required term: frac2pi Lh = n$$\frac{2\pi L}{h} = n$$ Since n=5$n=5$, the value is 5$5$. ### Pattern Recognition The expression 2pi L / h$2\pi L / h$ is a direct request for the principal quantum number n$n$. E_n = E_1 / n^2$E_n = E_1 / n^2$ allows finding n$n$ instantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Atoms

### 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