Physics - Modern Physics Question with Solution | TestHub

Using Einstein's photoelectric equation, the maximum kinetic energy of the emitted photoelectrons is given by $K=E-W$, where $E$ is the energy of the incident photons and $W$ is the work function of the metal.

For the first case, the energy of the incident photons is $E_1=3W$.
The maximum kinetic energy is $K_1=3W-W=2W$.

For the second case, the energy of the incident photons is $E_2=5W$.
The maximum kinetic energy is $K_2=5W-W=4W$.

The kinetic energy is related to the velocity by $K=\frac{1}{2}m{v}^2$, which implies $v\propto \sqrt{K}$.

The ratio of the velocities of the emitted photoelectrons is:
$\frac{v_1}{v_2}=\sqrt{\frac{K_1}{K_2}}=\sqrt{\frac{2W}{4W}}=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}}$

The ratio is $1:\sqrt{2}$.