Question

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

• A. $912 \, Å$
• B. $1824 \, Å$
• C. $3648 \, Å$
• D. $456 \, Å$

Answer 1

Answer: (C)

For series limit of Balmer series,
$n_{2}=2, \, n_{1}=\infty$
$\frac{1}{\lambda }=R\left[\frac{1}{n_{2}^{2}} - \frac{1}{n_{1}^{2}}\right]$
$=\left[\frac{1}{(2)^2}-\frac{1}{(\infty)^2}\right]=\frac{R}{4}$
$\therefore \lambda =\frac{4}{R}=\frac{4}{10967800}m$
$=4\times 912\times 10^{- 10 }m$
$=3648 \, Å$