Question

4eV is the energy of incident photon and the work function is 2eV. The stopping potential will be(a) 2V(b) 4V(c ) 6 V(d) 2√2V

Answer 1

option A should be the answer

Answer 2

Answer:

A) 2V

Explanation:

Energy of the incident of the photon = 4eV (Given)

Work function of the photon = 2eV (Given)

When stopping potential is applied no electron will reach the cathode and the current will becomes zero. This will be described by the equation as -

wo = 1/2mv² = eVs

h=4ev

wo =2eV

hu - work function

Thus,  4eV-2eV = 2eV

2ev = eVs

Therefore Vs = 2V

Thus, the stopping potential will be 2V