The hydrogen spectnun consists of several spectral lines in Lyman series ($$L_{1},L_{2},L_{3}....;L_{1}$$ has lowest energy among Lyman series). Similarly it consists of several spectral lines in Balmer series($$B_{1},B_{2},B_{3}....;B_{1}$$ has lowest energy among Balmer lines). The energy of $$L_{1}$$ is x times the energy of $$B_{1}$$. The value of x is_____ $$\times 10^{-1}$$. (Nearest integer)

We need to find the ratio of energy of $$L_1$$ (first Lyman line) to $$B_1$$ (first Balmer line).

In hydrogen, the energy of spectral lines is given by $$E = 13.6 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) \text{ eV}$$.

For $$L_1$$ in the Lyman series, the transition from n = 2 to n = 1 yields $$E_{L_1} = 13.6\left(\frac{1}{1^2} - \frac{1}{2^2}\right) = 13.6\left(1 - \frac{1}{4}\right) = 13.6 \times \frac{3}{4} = 10.2 \text{ eV}$$.

For $$B_1$$ in the Balmer series, the transition from n = 3 to n = 2 gives $$E_{B_1} = 13.6\left(\frac{1}{2^2} - \frac{1}{3^2}\right) = 13.6\left(\frac{1}{4} - \frac{1}{9}\right) = 13.6 \times \frac{5}{36} = \frac{68}{36} = \frac{17}{9} \text{ eV}$$.

The ratio is then $$x = \frac{E_{L_1}}{E_{B_1}} = \frac{13.6 \times 3/4}{13.6 \times 5/36} = \frac{3/4}{5/36} = \frac{3}{4} \times \frac{36}{5} = \frac{108}{20} = \frac{27}{5} = 5.4$$.

Since $$x = 5.4 = 54 \times 10^{-1}$$, the value is 54.