The surface of a metal is illuminated alternately with photons of energies $\mathrm{E}_{1}=4 \mathrm{eV}$ and $\mathrm{E}_{2}=2.5 \mathrm{eV}$ respectively. The ratio of maximum speeds of the photoelectrons emitted in the two cases is 2 . The work function of the metal in $(\mathrm{eV})$ is_________.
$\mathrm{E}_{1}=\phi+\mathrm{K}_{1} \ldots(1)$
$\mathrm{E}_{2}=\phi+\mathrm{K}_{2}$.......(2)
$E_{1}-E_{2}=K_{1}-K_{2}$
Now $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=2$
$\frac{\mathrm{K}_{1}}{\mathrm{~K}_{2}}=4$
$\mathrm{K}_{1}=4 \mathrm{~K}_{2}$
Now from equation (2)
$\Rightarrow 4-2.5=4 \mathrm{~K}_{2}-\mathrm{K}_{2}$
$1.5=3 \mathrm{~K}_{2}$
$1.5=3 \mathrm{~K}_{2}$
$\mathrm{~K}_{2}=0.5 \mathrm{eV}$
Now putting This
Value in equation (2)
$2.5=\phi+0.5 \mathrm{eV}$
$\phi=2 \mathrm{ev}$