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Question

When the frequency of light incident on a metallic plate is doubled, the maximum kinetic energy of emitted photoelectrons will be:
  1. halved
  2. doubled
  3. unchanged
  4. increases but more than the double of previous kinetic energy


Solution

The correct option is D increases but more than the double of previous kinetic energy
We know,
KE1=hν1hν0  and
hν1=KE1+hν0
Again,when the frequency is doubled
KE2=2hν1hν0
KE2=2KE1+2hν0 - hν0
KE2=2KE1+hν0
KE2 is more than the twice the KE1

 

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