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Question

Photoelectric emission is observed from a metal surface when the metal is subjected to radiations of incident frequencies ν1 and ν2 (where ν1>ν2). If the kinetic energies of the photoelectrons emitted in the two cases are in the ratio 2:1, then the threshold frequency v0 of the metal is:

A
ν1ν2
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B
ν1ν2h
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C
2ν1ν2
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D
2ν2ν1
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Solution

The correct option is D 2ν2ν1
hν1=hν0+E ...(i)
where E is the kinetic energy of the electron.
hν2=hν0+12E or 2hν2=2hν0+E ...(ii)
Eqn. (i) - Eqn. (ii) gives:
hν12hν2=hν02hν0
(or)
ν12ν2=ν0
or
ν0=2ν2ν1

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