Question

calculate kinetic energy of emitted photon when photon of wavelength 300 nanometer falls on a metal having work function 3 electron volt ​

Answer 1

Energy of incident radiation is given by:

E=

λ

hc

=

300∗10

−9

6.64∗10

−34

∗3∗10

8

=6.64∗10

−19

=4.05eV

The work function of the metal is equal to 2.54eV

Hence the maximum kinetic energy of an ejected photoelectron would be 1.59eV

And the stopping potential needed to stop this would be 1.59V.