Question

The maximum kinetic energy of a photoelectron is $E$ when wavelength of incident light is $\frac{\lambda }{2}$ . If maximum kinetic energy becomes $2E$ when wavelength is reduced to $\frac{\lambda }{3}$ . The work function of the metal $\text{(in eV)}$ is $(\text{Given}\frac{hc}{\lambda }=15)$

Answer 1

By Einstein’s photoelectric equation
$K{E}_{max}=\frac{hc}{\lambda }-\varphi$

Case I: When the incident wavelength is $\frac{\lambda }{2}$
$E=\frac{2hc}{\lambda }-\varphi ......\left(i\right)$

Case II : When the incident wavelength is $\frac{\lambda }{3}$
$2E=\frac{3hc}{\lambda }-\varphi .....\left(ii\right)$

From equation (i) and (ii)

$2=\frac{\frac{3hc}{\lambda }-\varphi }{\frac{2hc}{\lambda }-\varphi }$

$\Rightarrow \frac{4hc}{\lambda }-2\varphi =\frac{3hc}{\lambda }-\varphi$

$\therefore \varphi =\frac{hc}{\lambda }=\frac{15}{2}=7.5\text{eV}$

Correct answers: $7.5,7.5,7.500$