Dual Nature of Matter

JEE Main level, numerical answer, medium

150 of 200

Question 150 of 200

When light of frequency twice the threshold frequency is incident on the metal plate, the maximum velocity of emitted electron is v1{v}_{1}. When the frequency of incident radiation is increased to five times the threshold value, the maximum velocity of emitted electron becomes v2{v}_{2}. If v2=xv1{v}_{2}=x{v}_{1}, the value of xx will be _____ .

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Answer: 2

Maximum kinetic energy is expressed as Kmax⁡=hf−ϕ{K}_{\max }=hf-\phi, here, ϕ\phi is work function of metal.
For first case, 12mv12=h[(2f0)−f0]\dfrac{1}{2}{m{v}_{1}}^{2}=h\left[\left(2{f}_{0}\right)-{f}_{0}\right]
For second case, 12mv22=h[(5f0)−f0]\dfrac{1}{2}{m{v}_{2}}^{2}=h\left[\left(5{f}_{0}\right)-{f}_{0}\right]
⇒v22v12=41⇒v2v1=2\Rightarrow \dfrac{{{v}_{2}}^{2}}{{{v}_{1}}^{2}}=\dfrac{4}{1}\Rightarrow \dfrac{{v}_{2}}{{v}_{1}}=2

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