Question

Maximum kinetic energy of photoelectron is $E$ and its wavelength of incident light is $\frac{\lambda }{2}$ . If energy becomes double when wavelength becomes $\frac{\lambda }{3}$ , then work function of metal is

• A. $\frac{3hc}{\lambda }$
• B. $\frac{hc}{\lambda }$
• C. $\frac{hc}{2\lambda }$
• D. $\frac{hc}{3\lambda }$

Answer 1

The correct option is B $\frac{hc}{\lambda }$

$E=\frac{2hc}{\lambda }\phi ...\left(\text{i}\right)$

$2E=\frac{3hc}{\lambda }\phi ...\left(\text{i}\right)$

Solve & then $\phi =\frac{hc}{\lambda }$