Question

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

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

Answer 1

The correct option is C $912\times 4\stackrel{\circ }{A}$

$\frac{1}{\lambda }=R[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$
For limiting wavelength of Lyman series
$n_1=1,n_2=\infty$

$\frac{1}{{\lambda }_L}=R$
For limiting wavelength of Balmer series
$n_1=2,n_2=\infty$
$\frac{1}{{\lambda }_B}=R\left(\frac{1}{4}\right)\Rightarrow {\lambda }_B=\frac{4}{R}$
$\therefore {\lambda }_B=4{\lambda }_L=4\times 912\stackrel{\circ }{A}$.