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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 hc2λLet ϕ0 be the work function of the surface of material. Then, Kmax1=hcλ−ϕ0in first caseand Kmax2=hcλ/2−ϕ0=2hcλ−ϕ0in second caseBut, Kmax2=3Kmax1 (given)∴2hcλ−ϕ0=3(hcλ−ϕ0)⇒3ϕ0−ϕ0=3hcλ−2hcλ⇒2ϕ0=hcλ⇒ϕ0=hc2λ

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