Question

Photoelectric emission is observed from a metallic surface for frequencies $v_1$ and $v_2$ and of the incident light rays $(v_1>v_2)$. If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio of 1 : k, then the threshold frequency of the metallic surface is

• A. $\frac{v_1-v_2}{k-1}$
• B. $\frac{k{v}_1-v_2}{k-1}$
• C. $\frac{k{v}_2-v_1}{k-1}$
• D. $\frac{v_2-v_1}{k}$

Answer 1

The correct option is B

$\frac{k{v}_1-v_2}{k-1}$

By using $hv-h{v}_0=K_{mass}$
$\Rightarrow h(v_1-v_0)=K_1\dots \dots \left(i\right)$
And $h(v_2-v_0)=K_2\dots \dots \left(ii\right)$
$\Rightarrow \frac{v_1-v_0}{v_2-v_0}=\frac{K_1}{K_2}=\frac{1}{K}$,
Hence $v_0=\frac{K{v}_1-v_2}{K-1}$