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Question

# When a light of frequency v1 is incident on a metal surface, the photoelectrons emitted have twice the kinetic energy as did the photoelectrons emitted when the same metal has irradiated with light of frequency v2. What will be the value of threshold frequency?

A
v0=v1v2
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B
v0=v12v2
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C
v0=2v2v1
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D
v0=v1+v2
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Solution

## The correct option is C v0=2v2−v1hv1=hv0+KE1.....(1)hv2=hv0+KE2....(2)But KE1=2KE2.....(3)Substitute equation (3) in equation (1) hv1=hv0+2KE2....(4)From equation (2)hv2−hv0=KE2....(5)Substitute equation (5) in equation (4)hv1=hv0+2(hv2−hv0)hv1=hv0+2hv2−2hv0 hv1=2hv2−hv0 hv0=2hv2−hv1 v0=2v2−v1 Hence, the value of threshold frequency will be v0=2v2−v1.

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