Question

If the series limit wavelength of the Lyman series for hydrogen atom is $912\mathring{A}$ then the series limit wavelength for the Balmer series for the hydrogen atom is

• A. $912\mathring{A}$
• B. $912\times 2\mathring{A}$
• C. $912\times 4\mathring{A}$
• D. $\frac{912}{2}\mathring{A}$

Answer 1

Answer: (C)

For series limit of Balmer series
$n_2=2,n_1=\infty$
$\frac{1}{\lambda }=R\left[\frac{1}{n_2^2}-\frac{1}{n_1^2}\right]$
$=\left[\frac{1}{(2{)}^2}-\frac{1}{(\infty {)}^2}\right]=\frac{R}{4}$
$\therefore \lambda =\frac{4}{R}$
$=\frac{4}{10967800}m$
$=4\times 912\times 10^{-10}m$
$=4\times 912\mathring{A}$