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 = xv1, the value of x will be ______.

hv=hvth+12mv2 V=2Vth 2hvth=hvth+12mv12……(1) V=5Vth 5hvth=hvth+12mv22….(2) 12mv1212mv22=hvth4hhvth (v1v2)2=14⇒v2=2v1

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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 = xv₁, the value of x will be ______.

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answer is 2.

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Detailed Solution

$\begin{aligned} hv = {hv}_{th} + \frac{1}{2} {mv}^{2} \\ V = 2 V_{th} \\ 2 h v_{th} = {hv}_{th} + \frac{1}{2} {mv}_{1}^{2} \dots \dots (1) \\ V = 5 V_{th} \\ 5 {hv}_{th} = {hv}_{th} + \frac{1}{2} {mv}_{2}^{2} \dots . (2) \\ \frac{\frac{1}{2} {mv}_{1}^{2}}{\frac{1}{2} {mv}_{2}^{2}} = \frac{h v_{th}}{4 h {hv}_{th}} \\ {(\frac{v_{1}}{v_{2}})}^{2} = \frac{1}{4} \Rightarrow v_{2} = 2 v_{1} \end{aligned}$

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