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The work function of a metal is 6.4x10^-19J. If the metal is irradiated with radiation of wavelength 200nm,then the maximum K.E of the photoelectrons would be about

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6.4×10-19J

3.5×10-19J

1.0×10-18J

2.0×10-19J

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detailed solution

Correct option is B

Given, λ=200nm =200×10-9m E=hCλ=6.6×10-34×3×1082×10-7=9.9×10-19J Work function,W=6.4×10-19J According to photoelectric effect, E=W+K.E K.E=9.9×10-19-6.4×10-19=3.5×10-19J

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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 $ν_{1}, ν_{2}$ are incident frequencies in experiment one and two respectively and $ν_{0}$ is the threshold frequency of the metal ‘X’)

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