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 $
$=4 \times 912 \times 10^{-10}\, m$
$=4 \times 912\,\mathring{A}$