Dual Nature of Matter

JEE Main level, single correct, medium

132 of 200

Question 132 of 200

The threshold frequency of metal is f0{f}_{0}. When the light of frequency 2f02{f}_{0} is incident on the metal plate, the maximum velocity of photoelectron is v1{v}_{1}. When the frequency of incident radiation is increased to 5f05{f}_{0}. the maximum velocity of photoelectrons emitted is v2{v}_{2}. The ratio of v1{v}_{1} to v2{v}_{2} is:

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Answer: (A) v1v2=12\dfrac{{v}_{1}}{{v}_{2}}=\dfrac{1}{2}

Maximum kinetic energy is related to the incident frequency ff and the threshold frequency f0{f}_{0} as:
Kmax⁡=hf−hf0{K}_{\max }=hf-h{f}_{0}
When, f=2f0f=2{f}_{0}
12mv12=2hf0−hf0=hf0...(1)\dfrac{1}{2}m{v}_{1}^{2}=2h{f}_{0}-h{f}_{0}=h{f}_{0}...\left(1\right)
When, f=5f0f=5{f}_{0}
12mv22=5hf0−hf0=4hf0...(2)\dfrac{1}{2}m{v}_{2}^{2}=5h{f}_{0}-h{f}_{0}=4h{f}_{0}...\left(2\right)
Hence, the ratio, v1v2=12\dfrac{{v}_{1}}{{v}_{2}}=\dfrac{1}{2}

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