Question

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 then doubled
• D. Less than doubled

Answer 1

The correct option is C More then doubled

From Einstein's equation for photoelectric effect , the maximum kinetic energy , $K=h\nu -W$

where $\nu =$ frequency of incident light , $h=$ Planck's constant and $W=$ work function of metal.

When $\nu$ is made double, $K^{\prime }=2h\nu -W=2(h\nu -W)+W=2K+W$

Thus, kinetic energy of photoelectrons will be more than doubled when incident frequency is doubled.