Question

If frequency $(\nu >{\nu }_0)$ of incident light becomes n times the initial frequency ($\nu$), then KE of the emitted photoelectrons becomes $({\nu }_0$ threshold frequency)

• A. n times of the initial kinetic energy
• B. more than n times of the initial kinetic energy
• C. less than n times of the initial kinetic energy
• D. kinetic energy of the emitted photoelectrons remains unchanged

Answer 1

Answer: (B)

more than n times of the initial kinetic energy

Kinetic energy is given by:

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

So when $\nu$ becomes $n\nu$, Kinetic energy changes as

$K^{{}^{\prime }}=nh\nu -h{\nu }_0$

$K^{{}^{\prime }}-nK=(n-1)h{\nu }_0$

Thus, $K^{{}^{\prime }}>nK$