JEE MAIN - Physics (2022 - 24th June Morning Shift - No. 23)
When light of frequency twice the threshold frequency is incident on the metal plate, the maximum velocity of emitted electron is v1. When the frequency of incident radiation is increased to five times the threshold value, the maximum velocity of emitted electron becomes v2. If v2 = x v1, the value of x will be __________.
Odpowiedź
2
Wyjaśnienie
Let us say that work function is $$\phi$$
$$ \Rightarrow 2\phi = \phi + {1 \over 2}mv_1^2$$ ...... (1)
and $$5\phi = \phi + {1 \over 2}mv_2^2$$ ..... (2)
From (1) and (2)
$${{v_2^2} \over {v_1^2}} = {4 \over 1}$$ or $${{{v_2}} \over {{v_1}}} = 2$$