Question

which of the following is correct expression for kinetic energy of photo electron​

Answer 1

To find:

The expression for Kinetic Energy of electron emitted in Photoelectric Effect.

Derivation:

When photon particles (with adequate energy) are incident on a metal plate a certain amount of energy is used up to overcome the threshold energy.

And the rest of the the energy is transmitted to the electron as its kinetic energy with which it is emitted.

So, energy of photon be E and kinetic energy of electron be KE;

$\therefore \: E= w_{0} + KE$

$= > \: h \nu= h \nu_{0} + KE$

$= > \: KE = h \nu - h \nu_{0}$

$= > \: KE = h (\nu - \nu_{0} )$

So, final expression is:

$\boxed{ \bf{\: KE = h (\nu - \nu_{0} )}}$