Question

When the frequency of light incident on a metallic plate is doubled, the maximum kinetic energy of emitted photoelectrons will be:

• A. doubled
• B. halved
• C. increases but more than the double of previous kinetic energy
• D. unchanged

Answer 1

The correct option is C increases but more than the double of previous kinetic energy

We know,
$K{E}_1=h{\nu }_1-h{\nu }_0$ and
$\Rightarrow h{\nu }_1=K{E}_1+h{\nu }_0$
Again,when the frequency is doubled
$K{E}_2=2h{\nu }_1-h{\nu }_0$
$\Rightarrow K{E}_2=2K{E}_1+2h{\nu }_0$ - $h{\nu }_0$
$\Rightarrow K{E}_2=2K{E}_1+h{\nu }_0$
$K{E}_2$ is more than the twice the $K{E}_1$