Question

Photoelectric emission is observed from a metallic surface for frequencies $n_{1}$ and $n_{2}$ of the incident light rays $\left(n_{1} > n_{2}\right)$ . 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. $\frac{n_{1} - n_{2}}{k - 1}$
• B. $\frac{k n_{1} - n_{2}}{k - 1}$
• C. $\frac{k n_{2} - n_{1}}{k - 1}$
• D. $\frac{n_{2} - n_{1}}{k}$

Answer 1

Answer: (B)

$\frac{K E_{1}}{K E_{2}}=\frac{h n_{1} - h n_{0}}{h n_{2} - h n_{0}}=\frac{1}{k}$
$n_{0}=\frac{k n_{1} - n_{2}}{k - 1}$