Question 5

According to Bohr's second postulate, the angular momentum of an electron in the $k^{th}$ orbit is quantized: $m v_k r_k = \frac{kh}{2\pi} \implies v_k = \frac{kh}{2\pi m r_k}$

The kinetic energy of the electron is $K_k = \frac{1}{2} m v_k^2$. Substituting $v_k$, we get:

$K_k = \frac{1}{2} m v_k \left( \frac{kh}{2\pi m r_k} \right) = \frac{kh v_k}{4\pi r_k}$

For a transition from $n^{th}$ orbit to the Lyman series ($k=1$), the magnitude of change in kinetic energy is:

$|\Delta K| = |K_n - K_1| = \left| \frac{nh v_n}{4\pi r_n} - \frac{1 \cdot h v_1}{4\pi r_1} \right| = \frac{h}{4\pi} \left| \frac{n v_n}{r_n} - \frac{v_1}{r_1} \right|$

Thus, option (A) is correct. For option (C), the total energy in the $k^{th}$ orbit is $E_k = -K_k = -\frac{e^2}{8\pi\epsilon_0 r_k}$.

The energy of the emitted photon is $h\nu = |E_n - E_1| = K_1 - K_n = \frac{e^2}{8\pi\epsilon_0} \left( \frac{1}{r_1} - \frac{1}{r_n} \right)$.

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

Thus, option (C) is correct.

Option (D) is incorrect because the coefficient should be $\frac{h}{4\pi}$, same as the change in kinetic energy.

Option (B) is incorrect because the de Broglie wavelength $\lambda = \frac{h}{mv} = \frac{h}{\sqrt{2mK}}$ is not linearly proportional to $1/K$.