The threshold frequency of a photoelectric metal is $v_0$. If light of frequency $2 v_0$ is incident on the surface of that metal, maximum velocity of emitted electrons is v. What will be the maximum velocity of emitted electrons if light of frequency $4 v_0$ is incident on the same metal surface.

$\sqrt{3} v$

The correct answer is Option (2) → $\sqrt{3} v$

By Einstein photoelectric equation

$\phi=\phi_0+K . E_{\max }$

$h 2 v_0=h v_0+\frac{1}{2} m v^2$

$\frac{1}{2} m v^2=h v_0$             .....(i)

$4 h v_0=h v_0+\frac{1}{2} m v_1^2$

$\frac{1}{2} m v_1^2=3 h v_0$          .....(ii)

On putting the value of $h v_0$ from equation (i)

$\frac{1}{2} m v_1^2=3 \frac{1}{2} m v^2$

$v_1=\sqrt{3} v$