Question

An electron in the hydrogen atom jumps from excited state $n$ to the ground state. The wavelength so emitted illuminates a photosensitive material having work function $2.75eV$. If the stopping potential of the photoelectron is $10V$, then the value of $n$ is

Answer 1

Answer: 4

$\frac{hc}{\lambda }=e{V}_0+{\varphi }_0$
$=10eV+2.75eV=12.75eV$
But $\frac{hc}{\lambda }=13.6\left[\frac{1}{1^2}-\frac{1}{n^2}\right]eV$
$\Rightarrow 1-\frac{1}{n^2}=\frac{12.75}{13.6}$
$\Rightarrow \frac{1}{n^2}=0.0625$
$\Rightarrow n^2=\frac{10000}{625}=16$
$\Rightarrow n=4$