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.
Answer & explanation
Correct answer: option 2
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)
$\frac{1}{2} m v_1^2=3 \frac{1}{2} m v^2$
$v_1=\sqrt{3} v$