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

$\mathrm{E}=\mathrm{KE}+{\mathrm{W}}_0=0.5 \mathrm{eV}+{\mathrm{W}}_0 .................. \left(\mathrm{i}\right)$          

When the energy of the incident photon is increased by 20%, we have:

$E\text{'}=120\% \mathrm{of} E = E\left(1 +\frac{20}{100}\right) = \frac{6}{5}E$

$\Rightarrow \frac{6}{5}\mathrm{E}=\mathrm{KE}\text{'}+{\mathrm{W}}_0=0.8 \mathrm{eV}+{\mathrm{W}}_0 ...........\left(\mathrm{ii}\right)$

Subtracting (ii) from (i), we get,    ${\mathrm{W}}_0= 1 \mathrm{eV}$