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Question

A certain metal when irradiated by a light of frequency ν=3.2×1016 Hz emits photoelectrons with twice the kinetic energy of photoelectrons that are emitted when the same metal is irradiated by a light of frequency ν=2.0×1016 Hz. Calculate the work function of the metal.

A
6.0×1015 Hz
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B
8.0×1017 Hz
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C
8.0×1015 Hz
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D
6.0×1017 Hz
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Solution

The correct option is C 8.0×1015 Hz
Applying photoelectric equation,
KE=hνhν0
Where ν0 = threshold frequency
ν = frequency of incident radiation
h = Planck's constant
K.E. = Planck's constant
(νν0)=KEh

Given: KE2=2KE1

(ν2ν0)=KE2h ...(i)

and (ν1ν0)=KE1h ...(ii)

Dividing equation (i) by (ii),
ν2ν0ν1ν0=KE2KE1=2KE1KE1=2

ν2ν0=2ν12ν0
ν0=2ν1ν2=2(2×1016)(3.2×1016)
=8×1015 Hz

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