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

Answer 1

The correct option is D $912\times 4\mathring{A}$

Series Limit of Balmer series
$n=1$
$\frac{1}{\lambda }=R\left(\frac{1}{n^2}\right)=R$

Given: $\frac{1}{R}=912A$
Series Limit of Balmer series
$n=2$
$\frac{1}{\lambda }=R\left(\frac{1}{n^2}\right)=\frac{R}{4}$
$\lambda =\frac{4}{R}=4\times 912\mathring{A}=3648\mathring{A}$