Question

If the series limit wavelength of the Lyman series for hydrogen atom is $ 912\,?$ then the series limit wavelength for the Balmer series for the hydrogen atom is

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

Answer 1

The Correct Option is 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 $
$=4 \times 912 \times 10^{-10}\, m$
$=4 \times 912\,?$