Question

Let $V_{1}$ and $V_{2}$ be the maximum velocities of the emitted electrons when the surface of a metal is illuminated with light waves of energy $E_{1}=4 \,eV$ and $E_{2}=2.5 \,eV$ respectively. If the work function of the metal is $2\, eV$, then the ratio $\frac{V_{1}}{V_{2}}$ is

• A. 1.6
• B. 4
• C. 2
• D. 0.5

Answer 1

Answer: (C)

In case of photoemission,
$K_{\max }=\frac{1}{2} m v_{\max }^{2}=h f-\varphi_{0}$
For photons of energies $4 \,eV$ and $2.5\, eV$, we have
(given, work function $=2\, eV$ )
$\frac{1}{2} m v_{1}^{2}=4-2=2 \ldots $(i)
and $\frac{1}{2} m v_{2}^{2}=2.5-2=0.5 \ldots$ (ii)
Dividing Eq. (i) by Eq. (ii), we get
$\frac{v_{1}^{2}}{v_{2}^{2}}=\frac{2}{0.5}=4$
$ \Rightarrow \frac{v_{1}}{v_{2}}=2$