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 $2.75\, eV$. If the stopping potential of the photoelectron is $10\, V$, then the value of $n$ is

Answer 1

$\frac{h c}{\lambda}=e V_{0}+\phi_{0}$
$=10 \,eV +2.75 \,eV =12.75 \,eV$
But $\frac{h c}{\lambda}=13.6\left[\frac{1}{1^{2}}-\frac{1}{n^{2}}\right] eV$
$ \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$