Question

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

• A. $4h\nu$
• B. $3h\nu$
• C. $5h\nu$
• D. $\frac{5}{2}h\nu$

Answer 1

Answer: (B)

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