Question

A radiation of energy $E(E>E_0)$ falls on a metal whose threshold frequency is $E_0$. The Kinetic energy of the emitted electrons is given by:

• A. $\frac{E-E_o}{2}$
• B. $E-E_o$
• C. $E_o-E$
• D. $\frac{E}{E_o}$

Answer 1

The correct option is B $E-E_o$

If the energy of the incident light is greater than the threshold energy, the excess energy is imparted to the electron as kinetic energy.

$E=E_o+\text{K.E.}$
$\Rightarrow \text{K.E.}=E-E_o$