Two photons of energy $2.5 \mathrm{eV}$ and $3.5 \mathrm{eV}$ fall on a metal surface of work function $1.5 \mathrm{eV}$. The ratio of the maximum velocities of the photoelectrons emitted from the metal surface is

• A

  1 : 4
• B

  2 : 1
• C

  1 : 2
• Correct answer
  D

  $1: \sqrt{2}$

Solution

Using, the equation
Dividing Eq. (i) by Eq. (ii) we get $\begin{aligned} & \frac{\frac{1}{2} m v_1^2}{\frac{1}{2} m v_2^2}=\frac{h v_{0_1}-\phi_0}{h v_{0_2}-\phi_0} \\ & \frac{v_1^2}{v_2^2}=\frac{(2.5-1.5)}{(3.5-1.5)}=\frac{1}{2} \\ & \frac{v_1^2}{v_2^2}=\frac{1}{2} \text { or } \frac{v_1}{v_2}=\frac{1}{\sqrt{2}} \end{aligned}$

Hence, the correct answer is option: 'D'.