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Question: Answered & Verified by Expert
In a photoelectric experiment, the relation between the applied potential difference V and the photoelectric current I was found to be as shown in the graph below. If the work function of the cathode plate is 28.8 eV  and h=6.6×10-34 J s, the frequency of incident radiation would be nearly (in s-1)
PhysicsDual Nature of MatterJEE Main
Options:
  • A 0.436×1018 Hz
  • B 0.436×1017 Hz
  • C 0.775×1016 Hz
  • D 0.775×1015 Hz
Solution:
2990 Upvotes Verified Answer
The correct answer is: 0.775×1016 Hz
For the photoelectric effect,

hf=ϕ+KEmax

From the graph KEmax=3.2 eV

hf=(28.8+3.2) eV

f=32h eV=  32×1.6×10196.6×1034 Hz

=0.775×1016 Hz

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