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The work function of a metal is 4.0ev. If the metal is irradiated with radiation of wavelength 300nm,then the maximum K.E  of the photoelectrons would be about

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a
6.4×10-19J
b
0.2×10-20J
c
2.0×10-20J
d
2.0×10-19J

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detailed solution

Correct option is C

Given, λ=300nm =300×10-9m           E=hCλ=6.6×10-34×3×1083×10-7=6.6×10-19J Work function,W=4ev=4×1.6×10-19=6.4×10-19J According to photoelectric effect, E=W+K.E   K.E=6.6×10-19-6.4×10-19=0.2×10-19J=2×10-20J

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In a Photo electric effect experiment, A metal ‘X’ is irradiated by two incident radiations in two different experiments. If the ratio of incident frequencies is 1:2 and the ratio of kinetic energy of photo electrons is 1:3. Then the correct relation is (where ν1, ν2 are incident frequencies in experiment one and two respectively and ν0 is the threshold frequency of the metal ‘X’)


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