When the energy of the incident radiation is increased by 20%, the kinetic energy of the photoelectrons emitted from a metal surface increased from 0.5 eV to 0.8 eV. The work function of the metal is 
 

According to Einstein's photoelectric equation, 

The kinetic energy of emitted photoelectrons is 

$K=hv-{\varphi }_0$

where h$\nu$ is the energy of incident radiation and ${\varphi }_0$ is work function of the metal. As per question, 

As per question,

$\begin{array}{l}0.5\mathrm{eV}=h\mathrm{v}-{\varphi }_0 ......................\left(i\right)\\ 0.8\mathrm{eV}=1.2h\mathrm{v}-{\varphi }_0 .................\left(ii\right)\end{array}$           

$\text{ On solving eqns. (i) and (ii), we get }$
${\varphi }_0=1.0\mathrm{eV}$