Question

When the light of frequency $2v_0$ (where $v_0$ is threshold frequency), is incident on a metal plate, the maximum velocity of electrons emitted is $v_1$. When the frequency of the incident radiation is increased to $5v_0$, the maximum velocity of electrons emitted from the same plate is $v_2$. The ratio of $v_1$ to $v_2$ is

• A. 2:01
• B. 1:02
• C. 4:01
• D. 1:04

Answer 1

The Correct Option is B

$E = W_{0} + \frac{1}{2} mv^{2}$
$ h\left(2v_{0}\right) = hv_{0} + \frac{1}{2} mv_{1}^{2}$
$ hv_{0 } = \frac{1}{2} mv_{1}^{2}$ .....(i)
$ h\left(5v_{0}\right) = hv_{0} + \frac{1}{2}mv_{2}^{2}$
$ 4 hv_{0} = \frac{1}{2} mv_{2}^{2} $ .....(ii)
Divide (i) by (ii),
$ \frac{1}{4} = \frac{v_{1}^{2}}{v_{2}^{2}} $
$ \frac{v_{1}}{v_{2}} = \frac{1}{2}$