The energy of first (lowest) Balmer line of H atom is x J. The energy (in J) of second Balmer line of H atom is :

We need to find the ratio of energies of the second and first Balmer lines of hydrogen.

Balmer series: Transitions to $$n = 2$$.

Energy of a transition: $$E = 13.6 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$ eV.

First Balmer line ($$n = 3 \to 2$$):

$$E_1 = 13.6\left(\frac{1}{4} - \frac{1}{9}\right) = 13.6 \times \frac{9-4}{36} = 13.6 \times \frac{5}{36} = x$$

Second Balmer line ($$n = 4 \to 2$$):

$$E_2 = 13.6\left(\frac{1}{4} - \frac{1}{16}\right) = 13.6 \times \frac{4-1}{16} = 13.6 \times \frac{3}{16}$$

Finding the ratio:

$$\frac{E_2}{E_1} = \frac{3/16}{5/36} = \frac{3}{16} \times \frac{36}{5} = \frac{108}{80} = \frac{27}{20} = 1.35$$

Therefore $$E_2 = 1.35x$$.

The correct answer is Option 2: 1.35x.