Question

The frequency and the intensity of a beam of light falling on the surface of photoelectric material are increased by a factor of two. Treating efficiency of photoelectron generation as constant, this will :

• A. increase the photoelectric current by a factor of two and the maximum energy of the photoelectrons by a factor greater than two.
• B. increase the maximum kinetic energy of the photoelectrons and would increase the photoelectric current by a factor of two.
• C. increase the maximum kinetic energy of the photoelectrons by a factor of greater than two and will have no effect on the magnitude of the photoelectric current produced.
• D. not produce any effect on the kinetic energy of the emitted electrons but will increase the photoelectric current by a factor of two.

Answer 1

The correct option is A increase the photoelectric current by a factor of two and the maximum energy of the photoelectrons by a factor greater than two.

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

$hv=h{v}_0+\frac{m{v}_e^2}{2}$

$\Rightarrow \frac{m{v}_e^2}{2}$=hv-hv_0$

As the frequency is doubled

$2hv=h{v}_0+\frac{m(v_e{)}^2}{2}$

$\Rightarrow \frac{m(v^{1}e{)}^2}{2}=2hv-h{v}_0$

As, $2hv-h{v}_0>2hv-2h{v}_0$

$\Rightarrow \frac{m(v{e}^1{)}^2}{2}>2\left[\frac{m(ve{)}^2}{2}\right]$

So, the maximum energy of photoelectrons increases by a factor greater than two.