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Question

A photoelectric surface is illuminated successively by monochromatic light of wavelength λ. If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case, the work function of the surface of the material is (h = Planck's constant, c = speed of light)

A
2hcλ
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B
hc3λ
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C
hc2λ
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D
hcλ
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Solution

The correct option is C hc2λ
Let ϕ0 be the work function of the surface of material. Then,
Kmax1=hcλϕ0in first caseand Kmax2=hcλ/2ϕ0=2hcλϕ0in second caseBut, Kmax2=3Kmax1 (given)2hcλϕ0=3(hcλϕ0)3ϕ0ϕ0=3hcλ2hcλ2ϕ0=hcλϕ0=hc2λ

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