Question

Radiation of wavelength $\lambda$, is incident on a photocell. The fastest emitted electron has speed v. If the wavelength is changed to $\frac{3\lambda }{4}$, the speed of the fastest emitted electron will be:

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

$\frac{hc}{\lambda }-\varphi =\frac{1}{2}m{v}^2...\left(i\right)4\frac{hc}{3\lambda }-\varphi =\frac{1}{2}m{v}_1^2...\left(ii\right)\left(ii\right)-\frac{4}{3}\times \text{(i) gives,}\frac{\varphi }{3}=\frac{1}{2}m{v}^2-\frac{1}{2}m{v}_1^2\times \frac{4}{3}\Rightarrow V_1^2>V^2\times \frac{4}{3}\Rightarrow V_1>{\left(\frac{4}{3}\right)}^{\frac{1}{2}}V$