Question

The photoelectric threshold frequency of a metal is $v$. When light of frequency $ 4v $ is incident on the metal. The maximum kinetic energy of the emitted photoelectrons is

• A. $ 4\,\,hv $
• B. $ 3\,\,hv $
• C. $ 5\,\,hv $
• D. $ \frac{5}{2}\,\,hv $

Answer 1

Answer: (B)

From Einstein's photoelectric equation the maximum kinetic energy of photoelectrons emitted from metal surface is given by
$E_{k}=h v_{1}-W$
where $W$ is work function of metal.
Given, $W=h v$ and $v_{1}=4 v$
$\therefore E_{k}=4 h v-h v=3 \,h v$