When light of frequency twice the threshold frequency is incident on the metal plate, the maximum velocity of emitted election is \(v _{1}\). When the frequency of incident radiation is increased to five times the threshold value, the maximum velocity of emitted electron becomes \(v _{2}\). If \(v _{2}= x\) \(v _{1}\), the value of \(x\) will be...........

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Correct Answer

(C) \(2\)

Detailed explanation

\(hv =h v _{\text {th }}+\frac{1}{2} mv ^{2}\)

\(v =2 v _{\text {th }}\)

\(2 hv _{\text {th }}=h v _{\text {th }}+\frac{1}{2} mv _{1}^{2} \ldots \ldots(1)\)

\(v =5 v _{\text {th }}\)

\(5 hv _{\text {th }}=h v _{\text {th }}+\frac{1}{2} mv _{2}^{2} \ldots(2)\)

\(\frac{\frac{1}{2} m v_{1}^{2}}{\frac{1}{2} m v_{2}^{2}}=\frac{h v_{t h}}{4 h v_{\text {th }}}\)

\(\left(\frac{ v _{1}}{ v _{2}}\right)^{2}=\frac{1}{4} \Rightarrow v _{2}=2 v _{1}\)

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