Q: The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 A° . The wavelength of the second spectral line in the Balmer series of singly ionized helium atom is

(A) 1215 A°

(B) 1640 A°

(C) 2430 A°

(D) 4687 A°

Ans: (A)

Sol: $\large \frac{\lambda_1}{\lambda_2} = \frac{Z_1^2 (\frac{1}{n_f^2} – \frac{1}{n_i^2})_1}{Z_2^2 (\frac{1}{n_f^2} – \frac{1}{n_i^2})_2} $

$\large \lambda_2 = \frac{\lambda_1 Z_1^2 (\frac{1}{n_f^2} – \frac{1}{n_i^2})_1}{Z_2^2 (\frac{1}{n_f^2} – \frac{1}{n_i^2})_2} $

On Putting the given values ,

$\large \lambda_2 = \frac{(6561 A^o) ( 1)^2 (\frac{1}{2^2} – \frac{1}{3^2})_1}{(2)^2 (\frac{1}{2^2} – \frac{1}{4^2})_2} $

= 1215 A°