The photoelectric threshold frequency of a material is υ.When light of frequency 4υ is incident on the metal, the maximum kinetic energy of the emitted photoelectron is :
A
4hυ
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B
3hυ
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C
5hυ
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D
5hυ2
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Solution
The correct option is C3hυ The maximum kinetic energy of the emitted electrons is given by Kmax=hυ−ϕ0=h(4υ)−h(υ)=3hυ