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Question

If the maximum kinetic energy of the photoelectrons is E when a radiation of frequency ν is incident on a photosensitive metal, then the maximum kinetic energy of the photoelectrons when the frequency of the incident radiation is doubled is:

A
2E
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B
E2
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C
E+hν
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D
Ehν
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Solution

The correct option is C E+hν
We know,
Energy of incident radiation = work function+KE
Given, maximum kinetic energy = E
So,
hν=hνo+E ...(1)
Here, ν and νo are the incident and the threshold frequencies respectively.

Let's assume that when the frequency of the incident radiation is doubled the new maximum kinetic energy = E.

Therefore,
2hν=hνo+E
2hν=(hνE)+E [From equation 1]
E=E+hν

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