Question

The frequency and intensity of the incident beam of light falling on the surface of photoelectric material is increased by a factor of two. This will

• A. increase the maximum kinetic energy of the photoelectrons as well as photoelectric current by a factor of two
• B. increase the maximum kinetic energy of photoelectorns and would increase the photoelectric current by a factor of two
• C. increase the maximum kinetic energy of photoelectrons by a factor of two and will have no effect on photoelectric current
• D. increase the photoelectric current by a factor of two but will have no effect on kinetic energy of emitted electrons.

Answer 1

Answer: (D)

As the intensity of the beam is doubled, the no. of photons in the beam is doubled. As a consequence, the photoelectrons emitted will also be doubled. Thus, the photoelectric current increases by a factor of two.
To find the energy of photoelectrons, let initial frequency of beam be $v$ and threshold frequency be $v_0$.
Then, by Einsteins eqn
$\begin{array}{l}hv=h{v}_0+\frac{m{v}_e^2}{2}\\ \Rightarrow \frac{m{v}_e^2}{2}=hv-hv\_o\$\$\end{array}$
As the frequency is doubled
$\begin{array}{l}2hv=h{v}_0+\frac{m{\left(v_e\right)}^2}{2}\\ \Rightarrow \frac{m{\left(v^{1}e\right)}^2}{2}=2hv-h{v}_0\end{array}$
As, $2hv-h{v}_0>2hv-2h{v}_0$
$\Rightarrow \frac{m{\left(v{e}^1\right)}^2}{2}>2\left[\frac{m(ve{)}^2}{2}\right]$
So, the maximum energy of photoelectrons increases by a factor greater than two.