Atoms and Nuclei

Concept:Use Bohr’s quantization and energy relations to evaluate changes in kinetic energy, de Broglie wavelength, radiation frequency, and total energy of the electron.Explanation:For the electron in a hydrogen atom, electrostatic force provides centripetal acceleration:$$\frac{m v_k^2}{r_k} = \frac{e^2}{4\pi\varepsilon_0 r_k^2} \quad \Rightarrow \quad m v_k^2 = \frac{e^2}{4\pi\varepsilon_0 r_k}$$Thus kinetic energy $K_k = \frac12 m v_k^2 = \frac{e^2}{8\pi\varepsilon_0 r_k}$, total energy $E_k = -K_k$.Angular momentum is quantized: $m v_k r_k = \frac{k h}{2\pi}$.For a transition from $n$ to $1$ (Lyman series), change in kinetic energy:$$|\Delta K| = |K_n - K_1| = \left| \frac{n h v_n}{4\pi r_n} - \frac{h v_1}{4\pi r_1} \right| = \frac{h}{4\pi}\left| \frac{n v_n}{r_n} - \frac{v_1}{r_1} \right|$$Hence option (A) is correct.de Broglie wavelength: $\lambda_k = \frac{h}{p_k} = \frac{h}{\sqrt{2m K_k}}$. Using $K_k = \frac{e^2}{8\pi\varepsilon_0 r_k}$ and Bohr’s relation $2\pi r_k = k \lambda_k$, we get $\lambda_k = \frac{e^2}{4\varepsilon_0 k K_k}$.Change in wavelength: $|\Delta \lambda| = \frac{e^2}{4\varepsilon_0}\left| \frac{1}{n K_n} - \frac{1}{1 \cdot K_1} \right|$, not as in option (B). So (B) is incorrect.Frequency of emitted radiation: $h\nu = E_n - E_1 = -K_n - (-K_1) = K_1 - K_n$.Using $K_k = \frac{e^2}{8\pi\varepsilon_0 r_k}$, we obtain $h\nu = \frac{e^2}{8\pi\varepsilon_0}\left(\frac{1}{r_1} - \frac{1}{r_n}\right)$, so $\nu = \frac{e^2}{8\pi\varepsilon_0 h}\left(\frac{1}{r_1} - \frac{1}{r_n}\right)$. Option (C) is correct.Total energy change equals magnitude of kinetic energy change because $|E_k| = K_k$. Thus $|\Delta E| = \frac{h}{4\pi}\left| \frac{n v_n}{r_n} - \frac{v_1}{r_1} \right|$, not the expression in option (D). Therefore (D) is incorrect.Answer:Options (A) and (C) are correct.