Question

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

Answer 1

Answer: (B)

$\frac{k\nu_{1} - \nu_{2}}{k - 1}$

Answer 2

By using $h\nu - h\nu_{0} = k_{\max}$ ⇒ $h(\nu_{1} - \nu_{0}) = k_{1}$ and

$h(\nu_{1} - \nu_{0}) = k_{2}$

Hence $\frac{\nu_{1} - \nu_{0}}{\nu_{2} - \nu_{0}} = \frac{k_{1}}{k_{2}} = \frac{1}{k}$ ⇒ $\nu_{0} = \frac{k\nu_{1} - \nu_{2}}{k - 1}$