Question

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

$\frac{1}{\lambda }=R\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$
For limiting wavelength of Lyman series
$n_1=1,n_2=\infty$
$\frac{1}{{\lambda }_L}=R$
For limiting wavelength of Balmer series
$n_1=2,n_2=\infty$
$\frac{1}{{\lambda }_L}=R$
For limiting wavelength of Balmer series
$n_1=2,n_2=\infty$
$\frac{1}{{\lambda }_L}=R\left(\frac{1}{4}\right)$
$\Rightarrow {\lambda }_B=\frac{4}{R}$
$\therefore {\lambda }_B=4{\lambda }_L=4{\lambda }_L=4\times 912\mathring{A}$