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

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

Answer: (C)

$\frac{ hc }{\lambda}= eV _{0}+\phi_{0}=10 eV +2.75 eV =12.75 eV$
But $\frac{ e }{\lambda}=13.6\left[\frac{1}{12}-\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$