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# A metal was irradiated by light of frequency 3.25×1015 s−1. The photoelectron produced had its kinetic energy, 2 times the kinetic energy of the photoelectrons which was produced when the same metal was irradiated with a light of frequency 2×1015 s−1. What is its work function [Given: NA=6×1023, h=6.6×10−34 J s]

A
297 kJ/mol
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B
900 kJ/mol
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C
385 kJ/mol
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D
none of these
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Solution

## The correct option is A 297 kJ/molUsing photoelectric equation: E=ϕ+KE or hν=ϕ+KE where E= energy of incident radiation ϕ= work function of metal KE= kinetic energy of ejected electrons in first case we have : h×3.25×1015=ϕ+KE1 ……(1) and KE1=2KE2 in second case we have : h×2×1015=ϕ+KE2 ……(2) KE2=h×2×1015−ϕ putting value of K.E2 in equation (1) : h×3.25×1015=ϕ+2(h×2×1015−ϕ) h×3.25×1015=ϕ+h×4×1015−2ϕ ⇒ϕ=h×0.75×1015 ⇒ϕ=6.6×10−34×0.75×1015 J/atom ⇒ϕ=4.95×10−19 J/atom ⇒ϕ=4.95×10−19×6×1023103 kJ/mol =297 kJ/mol  Suggest Corrections  0      Similar questions  Explore more