
A photocell is illuminated by a small bright source placed one metre away. When the same source of light is placed two metres away, the electrons emitted by the photo cathode
A. each carry one-quarter of their previous energy
B. each carry one-quarter of their previous momentum
C. are half as numerous
D. are one-quarter as numerous
Answer
162.9k+ views
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 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 the energy when a photon incident on 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 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 metre and finally it got shifted to the distance of 2 metres.
\[\dfrac{{{I_1}}}{{{I_2}}} = {\left( {\dfrac{{2m}}{{1m}}} \right)^2}\]
\[\Rightarrow \dfrac{{{I_1}}}{{{I_2}}} = 4\]
\[\therefore {I_2} = \dfrac{{{I_1}}}{4}\]
As the number of ejected intensity is proportional to the intensity of the photon, so the number of the electrons crossing the photocell is one-quarter of the initial.
Therefore, the correct option is D.
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 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 the energy when a photon incident on 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 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 metre and finally it got shifted to the distance of 2 metres.
\[\dfrac{{{I_1}}}{{{I_2}}} = {\left( {\dfrac{{2m}}{{1m}}} \right)^2}\]
\[\Rightarrow \dfrac{{{I_1}}}{{{I_2}}} = 4\]
\[\therefore {I_2} = \dfrac{{{I_1}}}{4}\]
As the number of ejected intensity is proportional to the intensity of the photon, so the number of the electrons crossing the photocell is one-quarter of the initial.
Therefore, the correct option is D.
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).
Recently Updated Pages
JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Charging and Discharging of Capacitor

Wheatstone Bridge for JEE Main Physics 2025

Instantaneous Velocity - Formula based Examples for JEE

Formula for number of images formed by two plane mirrors class 12 physics JEE_Main
