When a light of wavelength $300 \mathrm{~nm}$ fall on a photoelectric emitter, photo electrons are emitted.…
When a light of wavelength $300 \mathrm{~nm}$ fall on a photoelectric emitter, photo electrons are emitted. For another emitter light of wavelength $600 \mathrm{~nm}$ is just sufficient for liberating photoelectrons. The ratio of the work function of the two emitters is
$1: 2$
$2: 1$
$4: 1$
$1: 4$
Solution
Work function $\phi_0=\frac{\mathrm{hc}}{\lambda_0}$
$\begin{aligned}
\therefore \quad \phi_0 & \propto \frac{1}{\lambda_0} \\
\frac{\phi_{0_1}}{\phi_{0_2}} & =\frac{\lambda_{0_2}}{\lambda_{0_1}}=\frac{600}{300}=\frac{2}{1}
\end{aligned}$
.