Ans. Let the hydrogen atoms be excited to $n^{\;\text{th}}$ energy level.$E\;=-13.6 \left(\;\frac{1}{n_f^2}-\;\frac{1}{n_i^2}\right) \;\; [ \because n_i=1 ]$$\Rightarrow 12.3\; = \left(\;\frac{1}{1^2}-\;\frac{1}{n^2}\right)$$\Rightarrow \;\; 12.3\; = 13.6 - \;\frac{13.6}{n^2}$$\Rightarrow \;\; \;\frac{13.6}{n^2}\; = 13.6-12.3 = 1.3$$\Rightarrow \;\; n^2\; = \;\frac{13.6}{1.3}$$\Rightarrow n \;\; \approx 3$ The formula for calculating the wavelength of Lyman series is given below: $\;\frac{1}{\lambda}=R \left(\;\frac{1}{1^2}-\;\frac{1}{n^2}\right)$ For second member of Lyman series, $n=3$ $\therefore \;\;\frac{1}{\lambda}\;=R \left(1-\;\frac{1}{3^2}\right)$ $\Rightarrow \;\frac{1}{\lambda}= (1.09737 \times 10^7) \left(\;\frac{8}{9}\right)$ $\Rightarrow \lambda = 1025.1\ \text{Å}$ The formula for calculating the wavelength of Balmer series is given below: $\therefore \;\; \;\frac{1}{\lambda}=R \left(\;\frac{1}{4}-\;\frac{1}{n^2}\right)$ For second member of Balmer series: $\;n\;=4$ $\;\;\frac{1}{\lambda}\;=R \left(\;\frac{1}{4}-\;\frac{1}{4^2}\right)$ $\Rightarrow\;\;\frac{1}{\lambda}= (1.09737 \times 10^7) \left(\;\frac{3}{16}\right)$ $\Rightarrow \lambda\; = 4861\ \text{Å}$