Q. An electron in the hydrogen atom jumps 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 $10\ eV,$ then the value of $n$ is
A) 5
B) 2
C) 3
D) 4
Explanation
(d) $KE_{\max} = 10eV$
$\phi = 2.75eV$
$E = \phi + KE_{\max} = 12.75eV$
= Energy difference between $n = 4$ and $n = 1$
$\Rightarrow$ value of $n = 4$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.