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 $E_1=4eV$ and $E_2=2.5eV$ respectively. If the work function of the metal is $2eV$, 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,
$K_{\max }=\frac{1}{2}m{v}_{\max }^2=hf-{\phi }_0$
For photons of energies $4eV$ and $2.5eV$, we have
(given, work function $=2eV$ )
$\frac{1}{2}m{v}_1^2=4-2=2\dots$(i)
and $\frac{1}{2}m{v}_2^2=2.5-2=0.5\dots$ (ii)
Dividing Eq. (i) by Eq. (ii), we get
$\frac{v_1^2}{v_2^2}=\frac{2}{0.5}=4$
$\Rightarrow \frac{v_1}{v_2}=2$