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Question: Answered & Verified by Expert
At an incident radiation frequency of v1, which is greater than the threshold frequency, the stopping potential for a certain metal is V1. At frequency 2v1 the stopping potential is 3V1. If the stopping potential at frequency 4v1 is nV1, then n is
PhysicsDual Nature of MatterJEE Main
Options:
  • A 2
  • B 3
  • C 6
  • D 7
Solution:
1515 Upvotes Verified Answer
The correct answer is: 7

We know,

hv=ϕ+K.E

From the given data,

hv1=ϕ+eV1

h2v1=ϕ+3eV1

On dividing 1 by 2, 12=ϕ+eV1ϕ+3eV1

 ϕ+3eV1=2ϕ+2ev1

  ϕ=eV1

From 1

hv1=eV1+eV1

 hv1=2eV1

Given, h4v1=ϕ+nV1

 42eV1=eV1+nV1

7eV1=nV1

 n=7

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