Question

If the frequency of incident light falling on a photosensitive material is doubled, then the kinetic energy of the emitted photoelectron will be :

• A. Same as its initial value
• B. Two times its initial value
• C. More than two times its initial value
• D. Less than two times its initial value

Answer 1

The correct option is C More than two times its initial value

Let the initial kinetic energy of emitted photoelectron be $KE$.

Hence $KE=h\nu -h{\nu }_0$

When the frequency is doubled, let the kinetic energy of emitted photoelectron be $K{E}^{\prime }$
Hence $K{E}^{\prime }=2h\nu -h{\nu }_0$

$=KE+h\nu$

$=2KE+h{\nu }_0$
which is more than two times the initial kinetic energy.

Hence correct answer is option C.