Question

If the threshold frequency of a metal for photoelectric effect is $v_0$, then which of the following will not happen?

• A. If the frequency of the incident radiation is $v_0$, the kinetic energy of the electrons ejected is zero.
• B. If the frequency of the incident radiation is $v$, the kinetic energy of the electrons ejected will be $hv$ - $h{v}_0$.
• C. If the frequency is kept same at $v$ but intensity is increased, the number of electrons ejected will increase.
• D. If the frequency of incident radiation is further increased, the number of photoelectrons ejected will increase.

Answer 1

The correct option is D

If the frequency of incident radiation is further increased, the number of photoelectrons ejected will increase.

We will take all the options one by one for option (a), We know

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

If $h\nu$ = $h{\nu }_0$ [or $v$ = $v_0$]

KE = 0

Option (a) is correct.

For option (b), we know from Einstein's equation,

$KE=h\nu -workfunction$

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

It is also correct.

$\Rightarrow$ We know that, The number of electrons ejected is proportional to the intensity or brightness of light.

So option (c) is also correct.

$\Rightarrow$ The number of photoelectrons ejected only depends on intensity not on frequency. If the frequency is increased, Kinetic energy of photoelectrons ejected will increase.

Therefore option (d) is not correct.