We know that, KEmax=hv− 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′=1.2ν
In the 1st case,
0.5eV=hv− ψ ...(i)
In the 2nd case,
0.8eV=1.2hv− ψ ...(ii)
Solving (i) and (ii) simultaneously, we get,
ψ =1eV