Hydrogen Atom Transition Lyman Series Properties Analysis

To determine the correct statement(s), we analyze the Bohr model equations for a hydrogen atom ($Z = 1$).

1. Analysis of Option A:

According to Bohr's quantization condition for angular momentum in the $k^{\text{th}}$ orbit: $m v_k r_k = \frac{k h}{2\pi}$

Expressing the mass $m$ of the electron as: $m = \frac{k h}{2\pi v_k r_k}$

The kinetic energy $K_k$ of the electron in the $k^{\text{th}}$ orbit is: $K_k = \frac{1}{2} m v_k^2 = \frac{1}{2} \left( \frac{k h}{2\pi v_k r_k} \right) v_k^2 = \frac{h}{4\pi} \frac{k v_k}{r_k}$

For the $n^{\text{th}}$ orbit ($k = n$): $K_n = \frac{h}{4\pi} \frac{n v_n}{r_n}$

For the $1^{\text{st}}$ orbit ($k = 1$): $K_1 = \frac{h}{4\pi} \frac{v_1}{r_1}$

The magnitude of change in kinetic energy when the electron transitions between the $n^{\text{th}}$ and $1^{\text{st}}$ orbit is: $|\Delta K| = |K_n - K_1| = \frac{h}{4\pi}\left|\frac{n v_n}{r_n} - \frac{v_1}{r_1}\right|$

Thus, Option A is correct.

---

2. Analysis of Option B:

The de Broglie wavelength of the electron in the $k^{\text{th}}$ orbit is: $\lambda_k = \frac{h}{p_k} = \frac{h}{m v_k}$

Using $m v_k r_k = \frac{k h}{2\pi}$, we get: $\lambda_k = \frac{2\pi r_k}{k}$

The electrostatic force provides the necessary centripetal force: $\frac{1}{4\pi\epsilon_0} \frac{e^2}{r_k^2} = \frac{m v_k^2}{r_k} \implies K_k = \frac{e^2}{8\pi\epsilon_0 r_k} \implies r_k = \frac{e^2}{8\pi\epsilon_0 K_k}$

Substituting $r_k$ into the expression for $\lambda_k$: $\lambda_k = \frac{2\pi}{k} \left(\frac{e^2}{8\pi\epsilon_0 K_k}\right) = \frac{e^2}{4 k \epsilon_0 K_k}$

For $k = 1$: $\lambda_1 = \frac{e^2}{4\epsilon_0 K_1}$

For $k = n$: $\lambda_n = \frac{e^2}{4 n \epsilon_0 K_n}$

The magnitude of change in de Broglie wavelength is: $|\Delta \lambda| = |\lambda_n - \lambda_1| = \frac{e^2}{4\epsilon_0}\left|\frac{1}{n K_n} - \frac{1}{K_1}\right|$

Because of the missing factor of $n$ in the denominator of the first term, Option B is incorrect.

---

3. Analysis of Option C:

For the Lyman series, the electron undergoes a transition from $n^{\text{th}}$ orbit to the $1^{\text{st}}$ orbit ($n \to 1$). The energy of the emitted photon is equal to the energy difference between the initial and final states: $h \nu = E_n - E_1$

The total energy in the $k^{\text{th}}$ orbit is $E_k = -K_k = -\frac{e^2}{8\pi\epsilon_0 r_k}$. $h \nu = \left(-\frac{e^2}{8\pi\epsilon_0 r_n}\right) - \left(-\frac{e^2}{8\pi\epsilon_0 r_1}\right) = \frac{e^2}{8\pi\epsilon_0} \left(\frac{1}{r_1} - \frac{1}{r_n}\right)$

Therefore, the frequency of the emitted radiation is: $\nu = \frac{e^2}{8\pi\epsilon_0 h}\left(\frac{1}{r_1} - \frac{1}{r_n}\right)$

Thus, Option C is correct.

---

4. Analysis of Option D:

Since the magnitude of total energy in any orbit equals its kinetic energy ($|E_k| = K_k$), the magnitude of change in total energy is equal to the magnitude of change in kinetic energy: $|\Delta E| = |E_n - E_1| = |-K_n - (-K_1)| = |K_1 - K_n| = |\Delta K|$

Using the result from Option A: $|\Delta E| = \frac{h}{4\pi}\left|\frac{v_1}{r_1} - \frac{n v_n}{r_n}\right|$

Since Option D has a prefactor of $\frac{h}{2\pi}$ instead of $\frac{h}{4\pi}$, Option D is incorrect.

---

Conclusion:

The correct statements are A and C.