Question

A 5W source emits monochromatic light of wavelength \[5000\mathop A\limits^ \circ \]. When placed 0.5 m away, it liberates photoelectrons from a photosensitive metallic surface. When the source is moved to a distance of 1.0 m the number of photoelectrons liberated will be reduced by a factor of
A) Be reduced by a factor of 2
B) Be reduced by a factor of 4
C) Be reduced by a factor of 8
D) Be reduced by a factor of 16

Answer 1

Hint: As upon the striking of light, the electrons are emitted so this is the process of photoelectric effect. Then using the proportionalities of intensity with number of electrons emitted and the distance between the source and the surface.

Formula Used:
$I \propto N$ where N is number of electrons emitted per unit time.
$I \propto \dfrac{1}{{{d^2}}}$ where d is the distance between the source and the surface.

Complete step by step answer:
It is given that power of the source is 5 W and the light it emits has wavelength equal to \[5000\mathop A\limits^ \circ \].
The original distance of source from the metallic surface $\left( {{d_1}} \right)$ = 0.5 m
The new distance of source from the metallic surface $\left( {{d_2}} \right)$ = 1.0 m
Here, the metals are releasing electrons when the light strikes them (with an appropriate frequency), this process is called photoelectric effect and the electrons emitted are known as photoelectrons.
The number of photoelectrons liberated depends upon the intensity of the light striking the surface.
Intensity is directly proportional to the number of electrons emitted per unit time.
$I \propto N$ where I is intensity and $N = \dfrac{n}{t}$, number of electrons emitted per unit time.
Also, intensity is inversely proportional to the square of the distance between the source and surface.
$I \propto \dfrac{1}{{{d^2}}}$
$ \Rightarrow N \propto \dfrac{1}{{{d^2}}}$ [as both are equal to I]
For two cases this can be written as:
$\dfrac{{{N_1}}}{{{N_2}}} = {\left( {\dfrac{{{d_2}}}{{{d_1}}}} \right)^2}$
Substituting the values, we get:
$
  \dfrac{{{N_1}}}{{{N_2}}} = {\left( {\dfrac{{1.0m}}{{0.5m}}} \right)^2} \\
  \Rightarrow \dfrac{{{N_1}}}{{{N_2}}} = {\left( 2 \right)^2} \\
  \Rightarrow \dfrac{{{N_1}}}{{{N_2}}} = {\left( {\dfrac{{1.0m}}{{0.5m}}} \right)^2} \\
  \Rightarrow \dfrac{{{N_1}}}{{{N_2}}} = 4 \\
$
According to the question, we need to find the number of electrons emitted in the second case, so:
$ \Rightarrow {N_2} = \dfrac{{{N_1}}}{4}$

Therefore, it can be seen that the number of photoelectrons liberated are reduced by a factor of 4 and thus the correct option is (B).

Note:We were supposed to find the relationship between the number of electrons emitted (N) and the distance (d) from the source, but they did not have any direct relationship, so we found a physical quantity intensity which had relationships with both N and d which in turn gave us the required relationship among the two.
We generally use fractions for the comparison between two quantities, so whenever there is comparison, we try to keep those quantities in the terms of fraction only.