Answer
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Hint: The intensity is the energy per unit area per unit time. The energy of the beam of electrons per unit time will be the total sum of the energy of the individual electrons incident per unit time. From the source, the intensity is inversely proportional to the square distance of the point.
Formula used:
\[I = \dfrac{P}{{4\pi {d^2}}}\]
Here I is the intensity of the beam of a source having power P at a distance of d from the source.
Complete step by step solution:
When a photocell is illuminated by small bright light then the photons carry energy. The electrons in the photocell gain energy when a photon encounters the surface of the metal and if the energy is sufficient to overcome the threshold energy of the metal then it gets ejected from the surface of the metal.
The number of electrons ejected from the surface of the metal is proportional to the number of photons incident on the surface of the metal per unit of time, i.e. the intensity of the photon. As the intensity of the photon is inversely proportional to the square of the distance from the source of light,
\[I \propto \dfrac{1}{{{d^2}}}\]
So, if the source is kept at two distances \[{d_1}\] and \[{d_2}\] then the intensities of the corresponding distances will be in the ratio as,
\[\dfrac{{{I_1}}}{{{I_2}}} = {\left( {\dfrac{{{d_2}}}{{{d_1}}}} \right)^2}\]
It is given that initially the light source is kept at a distance of 1 meter and finally it gets shifted to a distance of 2 meters.
\[\dfrac{{{I_1}}}{{{I_2}}} = {\left( {\dfrac{{2\,m}}{{1\,m}}} \right)^2} \\ \]
\[\Rightarrow \dfrac{{{I_1}}}{{{I_2}}} = 4 \\ \]
\[\therefore {I_2} = \dfrac{{{I_1}}}{4}\]
As the number of ejected intensities is proportional to the intensity of the photon, so the number of electrons emitted is one-quarter of the initial.
Therefore, the correct option is B.
Note: The number of electrons ejected from the surface of the metal is independent of the energy of the photons (given the photon contains sufficient energy to overcome the work function of the metal).
Formula used:
\[I = \dfrac{P}{{4\pi {d^2}}}\]
Here I is the intensity of the beam of a source having power P at a distance of d from the source.
Complete step by step solution:
When a photocell is illuminated by small bright light then the photons carry energy. The electrons in the photocell gain energy when a photon encounters the surface of the metal and if the energy is sufficient to overcome the threshold energy of the metal then it gets ejected from the surface of the metal.
The number of electrons ejected from the surface of the metal is proportional to the number of photons incident on the surface of the metal per unit of time, i.e. the intensity of the photon. As the intensity of the photon is inversely proportional to the square of the distance from the source of light,
\[I \propto \dfrac{1}{{{d^2}}}\]
So, if the source is kept at two distances \[{d_1}\] and \[{d_2}\] then the intensities of the corresponding distances will be in the ratio as,
\[\dfrac{{{I_1}}}{{{I_2}}} = {\left( {\dfrac{{{d_2}}}{{{d_1}}}} \right)^2}\]
It is given that initially the light source is kept at a distance of 1 meter and finally it gets shifted to a distance of 2 meters.
\[\dfrac{{{I_1}}}{{{I_2}}} = {\left( {\dfrac{{2\,m}}{{1\,m}}} \right)^2} \\ \]
\[\Rightarrow \dfrac{{{I_1}}}{{{I_2}}} = 4 \\ \]
\[\therefore {I_2} = \dfrac{{{I_1}}}{4}\]
As the number of ejected intensities is proportional to the intensity of the photon, so the number of electrons emitted is one-quarter of the initial.
Therefore, the correct option is B.
Note: The number of electrons ejected from the surface of the metal is independent of the energy of the photons (given the photon contains sufficient energy to overcome the work function of the metal).
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