Consider a hydrogen atom with v_k , r_k , and K_k denoting the...

From Bohr's quantization condition, the angular momentum of an electron in the $k^{\text{th}}$ orbit is given by:

$m v_k r_k = \dfrac{k h}{2\pi}$

The kinetic energy of the electron in the $k^{\text{th}}$ orbit is:

$K_k = \dfrac{1}{2} m v_k^2 = \dfrac{1}{2} (m v_k r_k) \dfrac{v_k}{r_k}$

Substituting the value of angular momentum:

$K_k = \dfrac{1}{2} \left( \dfrac{k h}{2\pi} \right) \dfrac{v_k}{r_k} = \dfrac{k h v_k}{4\pi r_k}$

The magnitude of change in kinetic energy for a transition from the $n^{\text{th}}$ orbit to the $1^{\text{st}}$ orbit is:

$|\Delta K| = |K_n - K_1| = \dfrac{h}{4\pi} \left| \dfrac{n v_n}{r_n} - \dfrac{v_1}{r_1} \right|$

This makes statement (A) correct.

Since the total energy $E_k = -K_k$, the magnitude of change in total energy is equal to the magnitude of change in kinetic energy:

$|\Delta E| = |\Delta K| = \dfrac{h}{4\pi} \left| \dfrac{n v_n}{r_n} - \dfrac{v_1}{r_1} \right|$

This makes statement (D) incorrect.

The electrostatic force provides the necessary centripetal force:

$\dfrac{m v_k^2}{r_k} = \dfrac{e^2}{4\pi\epsilon_0 r_k^2} \Rightarrow K_k = \dfrac{1}{2} m v_k^2 = \dfrac{e^2}{8\pi\epsilon_0 r_k}$

The total energy is $E_k = -K_k = -\dfrac{e^2}{8\pi\epsilon_0 r_k}$. The energy of the emitted photon during the transition is:

$\Delta E = E_n - E_1 = -\dfrac{e^2}{8\pi\epsilon_0 r_n} - \left( -\dfrac{e^2}{8\pi\epsilon_0 r_1} \right) = \dfrac{e^2}{8\pi\epsilon_0} \left( \dfrac{1}{r_1} - \dfrac{1}{r_n} \right)$

The frequency of the emitted radiation is $\nu = \dfrac{\Delta E}{h}$:

$\nu = \dfrac{e^2}{8\pi\epsilon_0 h} \left( \dfrac{1}{r_1} - \dfrac{1}{r_n} \right)$

This makes statement (C) correct.

The de Broglie wavelength of the electron is $\lambda_k = \dfrac{h}{m v_k}$. Using Bohr's quantization $m v_k = \dfrac{k h}{2\pi r_k}$, we get:

$\lambda_k = \dfrac{2\pi r_k}{k}$

From the kinetic energy relation $r_k = \dfrac{e^2}{8\pi\epsilon_0 K_k}$, substituting $r_k$ gives:

$\lambda_k = \dfrac{2\pi}{k} \left( \dfrac{e^2}{8\pi\epsilon_0 K_k} \right) = \dfrac{e^2}{4\epsilon_0 k K_k}$

The magnitude of change in de Broglie wavelength is:

$|\Delta \lambda| = |\lambda_n - \lambda_1| = \dfrac{e^2}{4\epsilon_0} \left| \dfrac{1}{n K_n} - \dfrac{1}{K_1} \right|$

This makes statement (B) incorrect.

Answer: Magnitude of change in kinetic energy of electron can be expressed as $\dfrac{h}{4\pi}\left|\dfrac{n v_n}{r_n} - \dfrac{v_1}{r_1}\right|$.; Frequency of the radiation emitted can be expressed as $\dfrac{e^2}{8\pi\epsilon_0 h}\left(\dfrac{1}{r_1} - \dfrac{1}{r_n}\right)$.