A photoelectric cell is illuminated by a point source of light 1m away. When the source is shifted to 2m then :
A. each emitted electron carries half the initial energy
B. Number of electrons emitted is a quarter of the initial number
C. each emitted electron carries one-quarter of the initial energy
D. number of electrons emitted is half the initial number
Answer
299.7k+ 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 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).
Recently Updated Pages
If the magnetizing field on a ferromagnetic material class 12 physics JEE_Main

A point charge is placed at the corner of a cube The class 12 physics JEE_Main

What changes occur if the monochromatic light used class 12 physics JEE_Main

The unit of specific conductance is A Ohm B Ohmmetre class 12 physics JEE_Main

Which lens is used in magnifying glass A Concave lens class 12 physics JEE_Main

Sir C V Raman won the Nobel Prize in which year A 1928 class 12 physics JEE_Main

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Understanding the Electric Field of a Uniformly Charged Ring

Electron Gain Enthalpy and Electron Affinity Explained

Derivation of Equation of Trajectory Explained for Students

Understanding Atomic Structure for Beginners

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
CBSE Class 12 Physics Question Paper 2026: Download SET-wise PDF with Answer Key & Analysis

JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

Effective Nuclear Charge for JEE

What Are Current and Potential Difference in Electricity?

Understanding Uniform Acceleration in Physics

Hybridisation in Chemistry – Concept, Types & Applications

