Question

A \[5{\text{ watt}}\] Source emits monochromatic light of wavelength \[5000{\text{ A}}\] when placed \[0.5{\text{ m}}\]away, it liberates photoelectron from a photosensitive metallic surface. When the source is moved to a distance of\[1.0{\text{ m}}\], the numbers of photoelectrons liberated will:
(A) be reduced by the factor of \[2\]
(B) be reduced by the factor of \[4\]
(C) be reduced by the factor of \[8\]
(D) be reduced by the factor of \[16\]

Answer 1

Hint:When a monochromatic light of certain intensity is incident on the metallic surface if the intensity of light is strong enough. It will proceed to eject photoelectrons from the metal surface atoms. And the number of photoelectrons ejected depends upon the intensity of the monochromatic light incident on the surface.
The intensity of light is inversely proportional to the square of distance.
\[I \propto \dfrac{1}{{{r^2}}}\]

Formula used:
Write the expression between the number of photoelectron\[N\] liberated and Intensity of Light\[I\]
\[\
I = \dfrac{{hN{\text{v}}}}{{At}} \\
I \propto N \\
\ \]
Write the relationship between intensity of light and distance
\[I \propto \dfrac{1}{{{r^2}}}\]

Complete step by step answer:
Understand that Intensity of light is directly proportional to the number of photoelectrons\[N\] liberated and it is inversely proportional to the square of distance\[r\].
Write the expression between the number of photoelectron\[N\] liberated and Intensity of Light \[I\]
\[\
I = \dfrac{{hN{\text{v}}}}{{At}} \\
\Rightarrow I \propto N \\
\ \] ….. (1)
Here,\[h\]is Planck’s constant, \[{\text{v}}\] is frequency of light, \[A\] is unit area and \[t\] is unit time.
Write the relationship between intensity of light and distance
\[I \propto \dfrac{1}{{{r^2}}}\] ….. (2)
Use equation \[(1)\] and \[(2)\]
\[N \propto \dfrac{1}{{{r^2}}}\]
Write the relationship between numbers of photoelectron\[{N_1}\] and distance\[{r_1}\], when monochromatic light is placed at\[0.5{\text{ m}}\]
\[{N_1} \propto \dfrac{1}{{r_{^1}^2}}\]
Substitute\[0.5{\text{ m}}\] for \[{r_1}\]
\[{N_1} \propto \dfrac{1}{{{{(0.5)}^2}}}\] …… (3)
Write the relationship between numbers of photoelectron\[{N_2}\] and distance\[{r_2}\] , when monochromatic light is placed at\[1.0{\text{ m}}\]
\[{N_2} \propto \dfrac{1}{{r_{^2}^2}}\]
Substitute\[1.0{\text{ m}}\] for\[{r_2}\]
\[{N_2} \propto \dfrac{1}{{{{(1)}^2}}}\] …… (4)
Divide equation (4) by (3)
\[\
\dfrac{{{N_2}}}{{{N_1}}} = \dfrac{{{{(0.5)}^2}}}{1} \\
\therefore{N_2} = \dfrac{{{N_1}}}{4} \\
\ \]
Therefore, Option B is the correct choice. The numbers of photoelectrons liberated will be reduced by 4.

Note:The relationship between the number of photoelectrons and distance is established using the expression of intensity in terms of number of photoelectrons and Inverse square law of Intensity and distance. The established relation shows that, number of photoelectrons are inversely proportional to the square of distance. The result is obtained by comparing the expressions for two different distances of monochromatic light given.