Solution (a)

(i) Spectral Series Identification:

• Step 1 (Governing Law): [0.5 Mark] The spectral series is determined by the final energy level ($n_f$) to which the electron jumps. For the Balmer series, $n_f = 2$.
• Step 2 (Analysis): [0.5 Mark] The second excited state is $n_i = 3$ and the first excited state is $n_f = 2$. Since the transition ends at $n_f = 2$, the radiation belongs to the Balmer series.

(ii) Ratio of Wavelengths:

• Step 1 (Formula): [0.5 Mark] Using the Rydberg formula: $$\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$$ For the most energetic line, $n_i = \infty$.
• Step 2 (Mathematical Working): [1.0 Mark]
  For Balmer series ($n_f = 2$): $\frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = \frac{R}{4} \implies \lambda_B = \frac{4}{R}$
  For Paschen series ($n_f = 3$): $\frac{1}{\lambda_P} = R \left( \frac{1}{3^2} - \frac{1}{\infty^2} \right) = \frac{R}{9} \implies \lambda_P = \frac{9}{R}$
• Step 3 (Final Answer): [0.5 Mark] $$\frac{\lambda_B}{\lambda_P} = \frac{4/R}{9/R} = \frac{4}{9}$$

Solution (b)

(i) Distance of Closest Approach:

• Step 1 (Governing Law): [0.5 Mark] At the distance of closest approach ($d$), the entire initial kinetic energy ($K$) of the $\alpha$-particle is converted into electrostatic potential energy. $$K = \frac{1}{4\pi\epsilon_0} \cdot \frac{(Ze)(2e)}{d}$$
• Step 2 (Mathematical Working): [0.5 Mark] Rearranging for $d$: $$d = \frac{1}{4\pi\epsilon_0} \cdot \frac{2Ze^2}{K}$$

(ii) Ratio of Distances:

• Step 1 (Formula & Given Data): [0.5 Mark] Since $v_p = v_\alpha = v$, the kinetic energies are $K_p = \frac{1}{2}m_p v^2$ and $K_\alpha = \frac{1}{2}m_\alpha v^2$. Given $m_\alpha \approx 4m_p$ and charge $q_\alpha = 2e, q_p = e$.
• Step 2 (Mathematical Working): [1.0 Mark] $$d \propto \frac{q_1 q_2}{K} \implies d \propto \frac{Z \cdot q_{particle}}{\frac{1}{2}m v^2}$$ $$\frac{d_p}{d_\alpha} = \frac{q_p / m_p}{q_\alpha / m_\alpha} = \frac{e /