Question

A $5\,{\text{watt}}$ source emits a monochromatic light of wavelength $5000\,\mathop {\text{A}}\limits^{\text{o}}$. When placed $0.5\,{\text{m}}$ away, it liberates photoelectrons from a photosensitive metallic surface. When the source is moved to a distance of $1.0\,{\text{m}}$, the number of photoelectrons liberated will be reduced by a factor of:
A. 8
B. 16
C. 2
D. 4

Answer 1

Use the relation between the intensity of the light at a particular distance from the light source, the power of the source and the distance from the source. Also use the relation between the light intensity at a distance and the number of photoelectrons emitted when the metal surface is illuminated by light. Using these two relations, one will obtain the relation between the intensity of the light at a distance and the number of photoelectrons.

Formula used:
The intensity $I$ of light at a distance from the source is
$I = \dfrac{P}{{4\pi {r^2}}}$ …… (1)
Here, $P$ is the power of the light source and $r$ is the distance at which the intensity of light is $I$.

Complete step by step answer:
We have given that the power of the light source is $5\,{\text{W}}$ and the wavelength of the monochromatic light is $5000\,\mathop {\text{A}}\limits^{\text{o}}$.
$P = 5\,{\text{W}}$
$\lambda = 5000\,\mathop {\text{A}}\limits^{\text{o}}$

From equation (1), it can be concluded that the intensity $I$ of the light at a distance $r$ from the source of light is inversely proportional to the square of distance $r$ from the source of light.
$I \propto \dfrac{1}{{{r^2}}}$

We also know that the intensity $I$ of light at any distance $r$ is directly proportional to the number $N$ of photos in the light.
$I \propto N$

From these two relations of intensity, we can conclude that the number $N$ of photoelectrons emitted from the metallic surface is inversely proportional to the square of distance of that light from the light source.
$N \propto \dfrac{1}{{{r^2}}}$

We have given two distances $0.5\,{\text{m}}$ and $1.0\,{\text{m}}$ of metallic surfaces from the source of light.
${r_1} = 0.5\,{\text{m}}$
${r_2} = 1.0\,{\text{m}}$

For the two distances ${r_1}$ and ${r_2}$ of the light from the light source, write the relation between the numbers ${N_1}$ and ${N_2}$ of photoelectrons liberated and distance of metallic surface from the source.
$\dfrac{{{N_2}}}{{{N_1}}} = \dfrac{{r_1^2}}{{r_2^2}}$

Substitute $1.0\,{\text{m}}$ for ${r_2}$ and $0.5\,{\text{m}}$ for ${r_1}$ in the above equation.
$\dfrac{{{N_2}}}{{{N_1}}} = \dfrac{{{{\left( {0.5\,{\text{m}}} \right)}^2}}}{{{{\left( {1.0\,{\text{m}}} \right)}^2}}}$
$\Rightarrow \dfrac{{{N_2}}}{{{N_1}}} = \dfrac{{0.25}}{1}$
$\Rightarrow \dfrac{{{N_2}}}{{{N_1}}} = \dfrac{1}{4}$
$\Rightarrow {N_1} = 4{N_2}$

From the above equation, it can be concluded that the number of photoelectrons emitted will be reduced by a factor 4.

Hence, the correct option is D.

Note:
The students may get confused that the intensity of light is proportional to the number of photons in the light then how it is related with the number of photoelectrons emitted from the metallic surface. But one should keep in mind that each photon in the light liberates one photoelectron from the metallic surface. SO the number of photons will be equal to the number of photoelectrons liberated from the metallic surface.