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QJEEM22Phae15: JEE Main 2022 Physics - Class 12 JEEM22Phae15 Solution: When light of frequency twice the threshold frequency is…

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 __________.

Step-by-Step Verified Solution

Let us say that work function is ϕ\phi

⇒2ϕ=ϕ+12mv12\Rightarrow 2\phi = \phi + {1 \over 2}mv_1^2 ...... (1)

and 5ϕ=ϕ+12mv225\phi = \phi + {1 \over 2}mv_2^2 ..... (2)

From (1) and (2)

v22v12=41{{v_2^2} \over {v_1^2}} = {4 \over 1} or v2v1=2{{{v_2}} \over {{v_1}}} = 2

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