The surface of a metal is illuminated alternately

Solution:

$\mathrm{E}_{1}=\phi+\mathrm{K}_{1} \ldots(1)$

$\mathrm{E}_{2}=\phi+\mathrm{K}_{2}$.......(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)

$2.5=\phi+0.5 \mathrm{eV}$

$\phi=2 \mathrm{ev}$