Energy of the incident photon on the metal surface is ' $3 \mathrm{~W}$ ' and then ' $5 \mathrm{~W}$ ',…
Energy of the incident photon on the metal surface is ' $3 \mathrm{~W}$ ' and then ' $5 \mathrm{~W}$ ', where 'W' is the work function for that metal. The ratio of velocities of emitted photoelectrons is
$1: \sqrt{2}$
$1: 1$
$1: 2$
$1: 4$
Solution
$\begin{aligned}
& \frac{1}{2} m v_{1}^{2}=3 W-W=2 W \\
& \frac{1}{2} m v_{2}^{2}=5 W-W=4 W \\
& \frac{v_{1}^{2}}{v_{2}^{2}}=\frac{1}{2} \\
\therefore & \frac{v_{1}}{v_{2}}=\frac{1}{\sqrt{2}}
\end{aligned}$