Light of energy 5eV falls on a metal of work function 3eV and electrons are liberated. The stopping potential is?

A. 1.7V
B. 2.0V
C. 8.0V
D. 15.0V

Correct Answer:

Option B – 2.0V

Explanation

For a photoemitted electron,
eV = K.E._max, where V = stopping potential and K.E. = maximum kinetic energy of the photo emitted electron given by K.E. = Energy of incident rad – Work – function
= 5ev – 3ev = 2ev
∴ eV = 2eV
V = 2eV/e = 2V
∴ Stopping Potential = 2.0V