400nm wavelength falls on the photosensitive materials of work function 2.3eV. Compute the maximum energy of photoelectron?

Solution:

Energy of incident photon is (E) =$ \frac{{1240}}{\lambda }eV = \frac{{1240}}{{400}}eV = 3.1eV$

And work function is W=2.5eV

So maximum energy of ejected electron will be, KE=E−W

=3.1eV−2.3eV

=0.8eV

=0.8×1.6×10⁻¹⁹Joule

=12.8×10^-20J

Thus, Maximum Kinetic Energy is 12.8×10^-20J.