Question

Photoelectric emission is observed from a metallic surface for frequencies $v_1$ and $v_2$ of the incident light rays $\left(v_1\right)$ > $\left(v_2\right)$. 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. $\frac{k{v}_2-v_1}{k-1}$

Answer 1

The correct option is B $\frac{k{v}_1-v_2}{k-1}$

By using $hv-h{v}_0=k_{max}\Rightarrow h(v_1-v_0)=k_1$ and $h(v_1-v_0)=k_2$
Hence, $\frac{v_1-v_0}{v_2-v_0}=\frac{k_1}{k_2}=\frac{1}{k}\Rightarrow v_0=\frac{k{v}_1-v_2}{k-1}$