Q 12-11-102JEE MainJEE Main 2022 (24 Jun, Shift 1)Easy
When light of frequency twice the threshold frequency is incident on a metal plate, the maximum velocity of emitted electrons is $v_1$. When the frequency of incident radiation is increased to five times the threshold value, the maximum velocity of emitted electrons becomes $v_2$. If $v_2 = xv_1$, the value of $x$ will be ______.
Numerical value type. Enter your answer.
Answer: 2
$\dfrac12 mv^2 = h(\nu - \nu_0)$:
$$\frac12 mv_1^2 = h\nu_0,\qquad \frac12 mv_2^2 = 4h\nu_0$$
$$\frac{v_2}{v_1} = \sqrt4 = 2\ \Rightarrow\ x = 2$$
Solution by Sreeraj P, M.Sc Physics