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A metal was irradiated by light of frequency 3.25×1015 s1. 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 s1. What is its work function
[Given: NA=6×1023, h=6.6×1034 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/mol
Using 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×2×1015ϕ

putting value of K.E2 in equation (1) :
h×3.25×1015=ϕ+2(h×2×1015ϕ)

h×3.25×1015=ϕ+h×4×10152ϕ
ϕ=h×0.75×1015
ϕ=6.6×1034×0.75×1015 J/atom
ϕ=4.95×1019 J/atom
ϕ=4.95×1019×6×1023103 kJ/mol
=297 kJ/mol

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