An electron in the hydrogen atom jumps from excited state $n$ to the ground state. The wavelength so emitted illuminates a photo-sensitive material having work function $2.75 \mathrm{eV}$. If the stopping potential of the photo-electron is $10 \mathrm{~V}$, the value of $n$ is

• Correct answer
  A

  3
• B

  4
• C

  5
• D

  2

Solution

$$ \begin{aligned} E & =\mathrm{KE}_{\max }+W \\ & =e V_0+W \\ & =10+2.75 \\ E & =12.75 \mathrm{eV} \end{aligned} $$
Difference of 4 and 1 energy level is $12.75 \mathrm{eV}$. So, higher energy level is 4 to ground and excited state is $n=3$.

Hence, the correct answer is option: 'A'.