Question

The wavelength of a certain line in Balmer series is observed to be $4341\mathring{A}$. To what value of '$n$' does this correspond? $(R_H=109678{cm}^{-1})$.

Answer 1

$\frac{1}{\lambda }=R_H[\frac{1}{2^2}-\frac{1}{n^2}]$
$\frac{1}{n^2}=\frac{1}{4}-\frac{1}{\lambda \times R_H}$
$=\frac{1}{4}-\frac{1}{4341\times {10}^{-8}\times 109678}$
$=0.04$
$n^2=\frac{1}{0.04}=25$
or $n=5$.