When the energy of the incident radiation is increased by $20\%$, the kinetic energy of the photoelectrons emitted from a metal surface increased from $0.5 \, eV$ to $0.8 \, eV$. The work function of the metal is:

Question

Aicharya · Accepted Answer

We know that, K E m a x = h v - ψ Let initial frequency of the light be ν and final frequency be ν ' . According to question, final energy of light is 20 % more than initial energy. ∴ ν ' = 1 . 2 ν In the 1st case, 0.5 e V = h v - ψ ...(i) In the 2nd case, 0.8 e V = 1.2 h v - ψ ...(ii) Solving (i) and (ii) simultaneously, we get, ψ = 1 e V

When the energy of the incident radiation is increased by $20\%$, the kinetic energy of the photoelectrons…

Solution