The ratio of energies of photons produced due to transition of electron of hydrogen atom from its (i) third to $2^{\text {nd }}$ energy level and (ii) highest energy level to $3^{\text {rd }}$ level is

The photon energy for an electron transition from $n_i$ to $n_f$ in hydrogen is given by $E = R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$.

For the $n_i = 3 \rightarrow n_f = 2$ transition:

$E_1 = R_H \left( \frac{1}{4} - \frac{1}{9} \right) = R_H \left( \frac{9 - 4}{36} \right) = R_H \cdot \frac{5}{36}$

For the $n_i = \infty \rightarrow n_f = 3$ transition:

$E_2 = R_H \left( \frac{1}{9} - 0 \right) = R_H \cdot \frac{1}{9}$

The energy ratio is:

$\frac{E_1}{E_2} = \frac{R_H \cdot \frac{5}{36}}{R_H \cdot \frac{1}{9}} = \frac{5}{36} \cdot \frac{9}{1} = \frac{45}{36} = \frac{5}{4}$

The ratio of photon energies is therefore $5:4$.

Answer: B