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Question

When electromagnetic radiation of wavelength 396 A is incident on a metal sheet that has a work function of 15.6 eV, find the maximum kinetic energy of the photoelectrons emitted.
Take h=6.6×1034 Js

A
15.65 eV
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B
10.48 eV
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C
9.65 eV
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D
16.48 eV
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Solution

The correct option is A 15.65 eV
Given λ = 396 A=396×1010 m
From photoelectric effect,
hν=W+(KE)max
hcλ=W+(K.E)max
where λ = Wavelength of incident radiation
W = Work function of the metal sheet
(KE)max = Kinetic energy of the emitted photoelectron.
So,
6.6×1034×3×108396×1010=15.6×1.6×1019+(KE)max
K.Emax=(0.05×101624.96×1019) J
K.Emax=25.04×1019 J=15.65 eV

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