Find the ratio of energies of photons produced due to transition of an electron of hydrogen atom from its (i) second permitted energy level to the first level, and (ii) the highest permitted energy level to the first permitted level.

We need to find the ratio of energies of photons for two transitions in hydrogen: (i) from $$n = 2$$ to $$n = 1$$, and (ii) from $$n = \infty$$ to $$n = 1$$.

The energy of a photon emitted during a transition from level $$n_i$$ to $$n_f$$ is given by $$E = 13.6\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$$ eV.

Case (i): Transition from $$n = 2$$ to $$n = 1$$: $$E_1 = 13.6\left(\frac{1}{1^2} - \frac{1}{2^2}\right) = 13.6\left(1 - \frac{1}{4}\right) = 13.6 \times \frac{3}{4}$$.

Case (ii): Transition from $$n = \infty$$ to $$n = 1$$ (highest permitted level to the ground state): $$E_2 = 13.6\left(\frac{1}{1^2} - \frac{1}{\infty^2}\right) = 13.6\left(1 - 0\right) = 13.6 \times 1$$.

The ratio is $$\frac{E_1}{E_2} = \frac{13.6 \times \frac{3}{4}}{13.6 \times 1} = \frac{3}{4}$$, which gives us $$E_1 : E_2 = 3 : 4$$.

Hence, the correct answer is Option A.