Question

Photoelectric emission is observed from a surface for frequencies $v_1$ and $v_2$ of the incident radiation ($v_1>v_2$). If the maximum kinetic energies of the photoelectrons in the two cases are in the ratio 1 : k then the threshold frequency $v_0$ is given by

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

$K.E.=h(v-v_0)$
For the two photon
$I=h(v_1-v_0)$ .................(i)
$k=h(v_2-v_0)$ .................(ii)
Taking a ratio, $\frac{1}{k}=\frac{v_1-v_0}{v_2-v_0}$
$k(v_1-v_0)=v_2-v_0$
$k{v}_1-k{v}_0=v_2-v_0$
$k{v}_1-v_2=k{v}_0-v_0$
$k{v}_1-v_2=v_0(k-1)$
$v_0=\frac{k{v}_1-v_2}{k-1}$