Answer:2.2 MeV
Explanation:
Given
Energy of incoming Photo [tex]E=3.24 MeV[/tex]
and Rest mass of electron and Proton has 0.51 Mev which can be derived by
mass of Electron [tex]m=9.11\times 10^{-31}\ kg\approx \frac{0.51}{c^2}\ MeV[/tex]
Where [tex]c=velocity\ of\ light[/tex]
Energy associated [tex]E=\frac{0.51}{c^2}\times c^2=0.51 MeV[/tex]
This energy can be treated as work function i.e.
Work function [tex]W=2\times 0.51=1.02\ MeV[/tex]
Applying Einstein Equation we get
[tex]E=W+K.E.[/tex]
[tex]K.E.=3.24-1.02=2.2 MeV[/tex]
