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The frequency of incident light falling ...

The frequency of incident light falling on a photosensitive metal plate is doubled, the KE of the emitted photoelectrons is

A

double the earlier value

B

unchanged

C

more than doubled

D

less than doubled

Text Solution

Verified by Experts

The correct Answer is:
C
Le `hv_0-W_0=K`
If frequency is doubled, let kinetic energy of photoelectrons be `K_1`.
`2hv_0-W_0=K_1`
`implies2(hv-W_0)+W_0=K_1`
`implies2K+W_0=K_1`
i.e., kinetic energy is more than doubled.
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