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When the light of frequency $2\nu_0$ (where $\nu_0$ 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 $5\nu_0$, the maximum velocity of electrons emitted from the same plate is $v_2$. The ratio of $v_1$ to $v_2$ is
Explanation
KE ratio is $h\nu_0 : 4h\nu_0$, so the speed ratio is 1 : 2.
Detailed Solution
$E = W_0 + \frac{1}{2}mv^2$
$h(2\nu_0) = h\nu_0 + \frac{1}{2}mv_1^2 \Rightarrow h\nu_0 = \frac{1}{2}mv_1^2$ ...(i)
$h(5\nu_0) = h\nu_0 + \frac{1}{2}mv_2^2 \Rightarrow 4h\nu_0 = \frac{1}{2}mv_2^2$ ...(ii)
Divide (i) by (ii): $\frac{1}{4} = \frac{v_1^2}{v_2^2}$
$\frac{v_1}{v_2} = \frac{1}{2}$, i.e. 1 : 2