Photoelectric cell or photocell or photovoltaic cell is an electronic device which works on the principle of the photoelectric effect and converts light energy into electrical energy.

Construction:

Photocell consists of an evacuated glass tube containing two electrodes emitter (C) and Collector (A). The emitter is shaped in the form of a semi-hollow cylinder. It is always kept at a negative potential. The collector is in the form of a metal rod and fixed at the axis of the semi-cylindrical emitter. The collector is always kept at a positive potential. The glass tube is fitted on a non-metallic base and pins are provided at the base for external connection.

Working:

The emitter is connected to a negative terminal and the collector is connected to the positive terminal of a battery. Radiation of frequency more than the threshold frequency of material of emitter is made incident on the emitter. Photo-emission takes place. The photo-electrons are attracted to the collector which is positive w.r.t. the emitter. Thus current flows in the circuit. If the intensity of incident radiation is increased the photoelectric current increases.

Applications of the Photoelectric Cell:

• The photoelectric cell is used in the reproduction of sound which is recorded on a movie film.
• The photoelectric cell is used in exposure meter.  The exposure meter is used along with a camera to know the correct time of exposure for having a good photograph.
• The photoelectric cell is used in lux-meter. It is used to determine the intensity of light.
• The photoelectric cell is used in a burglar alarm.  This device is kept near a safe to be protected from a thief.

Use of Photoelectric Cell in Sound Reproduction from the Motion Picture:

The photoelectric cell is used in the reproduction of sound which is recorded on a movie film. In a movie, film sound is recorded on the film of actions in the form of a thin transparent strip. This thin transparent strip is called the soundtrack. The transparency of the soundtrack depends on the variation of the frequency of sound recorded. Using photocell sound is reproduced from the soundtrack.

When the film is run in a projector the light of the projector this soundtrack and falls on the photocell. Due to variation in the soundtrack, the variation of intensity of sound takes place and thus the photo-electric current varies. The current is amplified and is fed to speakers.

Use of Photoelectric Cell in Burglar Alarm:

The photovoltaic cell is used in a burglar alarm.  This device is kept near a safe to be protected from a thief. A burglar alarm is a device which is used for locating intruder, thief near precious, valuable things like safe.

The device consists of a photocell and an infrared source of light. The light from the infra-red source is made continuously incident on the photocell making photoelectric effect continuous. Thus the photoelectric current in the cell flows continuously. When the path of infra-red light is obstructed by the thief, the light falling on photocell is cut-off and photo-electric current in the cell stops and relay circuit is activated and a siren starts hooting.

Use of Photoelectric Cell in Exposure meter:

The photovoltaic cell is used in exposure meter.  The exposure meter is used along with a camera to know the correct time of exposure of film for having a good photograph. To have a good photograph if the intensity of light is more the exposure of film should be less. If the intensity of light is less the exposure of film should be more. The exposure meter is a device attached to the camera which decides the exposure of the film.

The exposure meter consists of a photo-electric cell with a sensitive milliammeter and battery connected in series to it. The photoelectric current produced in the cell is directly proportional to the intensity of light.

If deflection in the milliammeter is small the photoelectric current is small. It indicates that the intensity of light is small. Thus the exposure time should be more. If deflection in the milliammeter is large the photoelectric current is large. It indicates that the intensity of light is more. Thus the exposure time should be less.

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In this article. we are going to study to calculate the stopping potential and maximum kinetic energy of the photoelectron using Einstein’s photoelectric equation.

Example 01:

A metal whose work function is 4.2 eV is irradiated by radiation whose wavelength is 2000 A. Find the maximum kinetic energy emitted electron.

Given: Work function Φ = 4.2 eV = 4.2 x 1.6 x 10^-19 J = 6.72 x 10^-19 J, wavelength of radiation = l = 2000 A = 2000 x 10^-10 m = 2 x 10^-7 m, Planck’s constant = 6.63 x 10^-34 Js.

To find: maximum kinetic energy = K.E._max = ?

Solution:

By Einstein’s photoelectric equation

Ans: Thus maximum kinetic energy of emitted electron is 2.015 eV

Example 02:

If photoelectrons are to be emitted from potassium surface with speed of 6 x 10⁵ m/s, what frequency of radiation must be used? Threshold frequency for potassium is 4.22 x 10¹⁴ Hz.

Given: speed of electron = v = 6 x 10⁵ m/s, Threshold frequency = ν_o = 4.22 x 10¹⁴ Hz. Planck’s constant = h = 6.63 x 10^-34 Js.

To find: Frequency of incident radiation = n = ?

Ans: Frequency of incident radiation is 6.69 x 10¹⁴ Hz.

Example – 03:

When radiation of certain wavelength shines on the cathode of the photoelectric cell, the photocurrent produced can be reduced to zero by applying stopping potential of 2.63 V. If the work function of the photo emitter is 4 eV, find the wavelengths of radiation.

Given: Stopping potential = V_s = 2.63 V, work function = Φ = 4 eV = 4 x 1.6 x 10^-19 J = 6.4 x 10^-19 J, speed of light = c = 3 x 10⁸ m/s, Planck’s constant = h = 6.63 x 10^-34 Js, Charge on electron = e = 1.6 x 10^-19 C.

To Find: Wavelength of radiation = λ =?

Solution:

By Einstein’s photoelectric equation

K.E. _max = hν – Φ = hc/λ – Φ

But K.E. _max = eV_s

eV_s =  hc/λ – Φ

∴ (1.6 x 10^-19) x (2.63) = (6.63 x 10^-34) (3 x 10⁸)/λ – 6.4 x 10^-19

∴ 4.21 x 10^-19 + 6.4 x 10^-19 = (19.89 x 10^-26)/λ

∴ 10.61 x 10^-19 = (19.89 x 10^-26)/λ

∴ λ = (19.89 x 10^-26) / (10.61 x 10^-19)

∴ λ = 1.871 x 10^-7  = 1871x 10^-10 m = 1871 Å

Ans: The wavelength of radiation is 1871 Å

Example – 04:

Radiation of wavelength 3000 A falls on a photoelectric surface for which work function is 1.6 eV. What is the stopping potential for emitted electron?

Given:  Wavelength of radiation = λ = 3000 Å = 3000 x 10^-10 m, work function = Φ = 1.6 eV = 1.6 x 1.6 x 10^-19 J = 2.56 x 10^-19 J, speed of light = c = 3 x 10⁸ m/s, Planck’s constant = h = 6.63 x 10^-34 Js, Charge on electron = e = 1.6 x 10^-19 C.

To Find: Stopping potential = V_s =?

Solution:

By Einstein’s photoelectric equation

K.E. _max = hν – Φ = hc/λ – Φ

But K.E. _max = eV_s

eV_s =  hc/λ – Φ

∴ (1.6 x 10^-19) x V_s = (6.63 x 10^-34) (3 x 10⁸)/(3000 x 10^-10) – 2.56 x 10^-19

∴ (1.6 x 10^-19) x V_s = 6.63 x 10^-19 – 2.56 x 10^-19

∴ V_s = 4.07 x 10^-19

∴ V_s = (4.07 x 10^-19)/(1.6 x 10^-19)

∴ V_s = 2.54 V

Ans: The stopping potential is 2.54 V.

Example – 05:

When radiation of certain wavelength shines on the cathode of the photoelectric cell, the photocurrent produced can be reduced to zero by applying stopping potential of 3 V. If the work function of the photo emitter is 3.63 eV, find the frequency of radiation.

Given: Stopping potential = V_s = 3 V, work function = Φ = 3.63 eV = 3.63 x 1.6 x 10^-19 J, speed of light = c = 3 x 10⁸ m/s, Planck’s constant = h = 6.63 x 10^-34 Js, Charge on electron = e = 1.6 x 10^-19 C.

To Find: frequency of radiation = ν =?

Solution:

By Einstein’s photoelectric equation

K.E. _max = hν – Φ = hc/λ – Φ

But K.E. _max = eV_s

eV_s =  hν – Φ

∴ (1.6 x 10^-19) x (3) = (6.63 x 10^-34) ν – 3.63 x 1.6 x 10^-19

∴ 4.8 x 10^-19 = (6.63 x 10^-34) ν – 5.808 x 10^-19

∴ 4.8 x 10^-19 + 5.808 x 10^-19 = (6.63 x 10^-34) ν

∴ 10.608 x 10^-19 = (6.63 x 10^-34) ν

∴ ν = (10.61 x 10^-19)/(6.63 x 10^-34)

∴ ν = 1.6 x 10¹⁵ Hz

Ans: The frequency of radiation is 1.6 x 10¹⁵ Hz

Example – 06:

Photoelectrons emitted by a surface have maximum kinetic energy of 4 x 10^-19 J. What is the stopping potential for photo emission from the surface for the incident radiation?

Given: Maximum kinetic energy of photoelectron = K.E. _max = 4 x 10^-19 J, Charge on electron = e = 1.6 x 10^-19 C.

To Find: frequency of radiation = ν =?

Solution:

K.E. _max = eV_s

V_s = K.E. _max/e

∴   V_s = (4 x 10^-19)/(1.6 x 10^-19) = 2.5 V

Ans: The stopping potential = 2.5 V

Example – 07:

Light of wavelength 2000 Å is incident on the cathode of a photocell. The current in the photocell is reduced to zero by stopping potential of 2 V. Find the threshold wavelength of the material of cathode.

Given: Stopping potential = V_s = 2 V, wavelength of incident light = λ = 2000 Å = 2000 x 10^-10 m, speed of light = c = 3 x 10⁸ m/s, Planck’s constant = h = 6.63 x 10^-34 Js, Charge on electron = e = 1.6 x 10^-19 C.

To Find: frequency of radiation = ν =?

Solution:

By Einstein’s photoelectric equation

K.E. _max = hν – Φ = hc/λ – Φ

But K.E. _max = eV_s

eV_s =  hc/λ – Φ

∴ (1.6 x 10^-19) x 2 = (6.63 x 10^-34) (3 x 10⁸)/(2000 x 10^-10)- Φ

∴3.2 x 10^-19 = 9.945 x 10^-19 – Φ

∴ Φ = 9.945 x 10^-19 – 3.2 x 10^-19 = 6.745 x 10^-19 J

We have Φ = h ν_o = hc/λ_o

∴ λ_o = hc/Φ =(6.63 x 10^-34) x (3 x 10⁸) / ( 6.745 x 10^-19) = 2.949 x 10^-7 m

∴  λ_o= 2949 x 10^-10 m = 2949 Å

Ans: The threshold wavelength is 2949 Å

Example – 08:

Photoelectrons are ejected from metal surface when radiation of wavelength 160 nm is incident on the surface. Find stopping potential of emitted electrons if the limiting wavelength is 240 nm for photoelectric emission from the surface.

Given: Stopping potential = V_s = 2 V, wavelength of incident light = λ = 160 nm = 160 x 10^-9 m, Threshold wavelength = λ_o = 240 nm = 240 x 10^-9 m, speed of light = c = 3 x 10⁸ m/s, Planck’s constant = h = 6.63 x 10^-34 Js, Charge on electron = e = 1.6 x 10^-19 C.

To Find: Stopping potential = V_s =?

Solution:

By Einstein’s photoelectric equation

K.E. _max = hν – Φ = hc/λ – hc/λ_o

But K.E. _max = eV_s

∴ eV_s = hc/λ – hc/λ_o

∴ eV_s = hc(1/λ – 1/λ_o)

∴ (1.6 x 10^-19) x V_s = (6.63 x 10^-34)(3 x 10⁸)(1/(160 x 10^-9) – 1/(240 x 10^-9))

∴ (1.6 x 10^-19) x V_s = (6.63 x 10^-34)(3 x 10⁸)x 10⁹ x (1/160  – 1/240 )

∴ (1.6 x 10^-19) x V_s = 19.89 x 10^-17 x (240 -160)/(160 x240 )

∴ (1.6 x 10^-19) x V_s = 19.89 x 10^-17 x 80/(160 x240 )

∴ (1.6 x 10^-19) x V_s = 4.143 x 10^-19

∴ V_s = 4.143 x 10^-19/  (1.6 x 10^-19) = 2.59 eV

Ans: the stopping potential is 2.59 eV

Example – 09:

Calculate the change in stopping potential when the wavelength of light incident on photoelectric surface is reduced from 4000 Å to 3600 Å.

Given: Initial wavelength = λ₁ = 4000 Å = 4000 x 10^-10 m = 4 x 10^-7 m, Final wavelength = λ₂ = 3600 Å = 3600 x 10^-10 m = 3.6 x 10^-7  m, speed of light = c = 3 x 10⁸ m/s, Planck’s constant = h = 6.63 x 10^-34 Js, Charge on electron = e = 1.6 x 10^-19 C.

To Find: Stopping potential = V_s =?

Solution:

By Einstein’s photoelectric equation

K.E. _max = hν – Φ = hc/λ – Φ

But K.E. _max = eV_s

∴ eV_s1 = hc/λ₁ – Φ ……….. (1)

∴ eV_s2 = hc/λ₂ – Φ ……….. (2)

Subtracting equation (1) from (2)

∴ eV_s2 – eV_s1 = (hc/λ₂ – Φ) – (hc/λ₁ – Φ)

∴ e (V_s2 – V_s1) = hc/λ₂ – hc/λ₁

∴ e (V_s2 – V_s1) = hc(1/λ₂ – 1/λ1₂)

∴ (1.6 x 10^-19) (V_s2 – V_s1) = (6.63 x 10^-34) (3 x 10⁸)(1/(3.6x 10^-7) – 1/(4 x 10^-7))

∴ (1.6 x 10^-19) (V_s2 – V_s1) = (6.63 x 10^-34) (3 x 10⁸) x 10⁷ x (1/3.6 – 1/4)

∴ (1.6 x 10^-19) (V_s2 – V_s1) = 19.89 x 10^-19 x (4 – 3.6)/1(3.6 x 1/4)

∴ (1.6 x 10^-19) (V_s2 – V_s1) = 19.89 x 10^-19 x 0.4/1(3.6 x 1/4)

∴ (1.6 x 10^-19) (V_s2 – V_s1) = 5.525 x 10^-20 

∴ (V_s2 – V_s1) = 5.525 x 10^-20/(1.6 x 10^-19) = 0.3453 V

Ans: the change in stopping potential is 0.3453 eV

Example – 10:

When light of frequency 2.2 x 10¹⁵ Hz is incident on a metal surface, photoelectric current can be stopped by a retarding potential of 6.6 V. For light of frequency 4.6 x 10¹⁵ Hz the reverse potential is 16.5 V. Find h

Given: Initial frequency = ν₁ = 2.2 x 10¹⁵ Hz, initial stopping potential = V_s1 =6.6 V, Final frequency = ν₂ = 4.6 x 10¹⁵ Hz, Final stopping potential = V_s2 = 16.5 V, Charge on electron = e = 1.6 x 10^-19 C.

To Find: Planck’s constant = h =?

Solution:

By Einstein’s photoelectric equation

K.E. _max = hν – Φ

But K.E. _max = eV_s

∴ eV_s = hν – Φ

∴ eV_s1 = hc/λ₁ – Φ ……….. (1)

∴ eV_s2 = hc/λ₂ – Φ ……….. (2)

Subtracting equation (1) from (2)

∴ eV_s2 – eV_s1 = (hν₂ – Φ) – (hν₁ – Φ)

∴ e (V_s2 – V_s1) = hν₂ – hν₁

∴ e (V_s2 – V_s1) = h(ν₂ – ν₂)

∴ (1.6 x 10^-19) (16.5 – 6.6) = h (4.6 x 10¹⁵ – 2.2 x 10¹⁵)

∴ 1.6 x 10^-19 x 9.9 = h (2.4 x 10¹⁵)

∴ h = (1.6 x 10^-19 x 9.9) / (2.4 x 10¹⁵)

∴ h = 6.6 x 10^-34 Js

Ans: the value of Planck’s constant is 6.6 x 10^-34 Js

Example – 11:

When light of frequency 2 x 10¹⁵ Hz is incident on a metal surface, photoelectric current can be stopped by a retarding potential of 6 V. For light of frequency 10¹⁵ Hz the reverse potential is 2 V. Find Planck’s constant, work function and threshold frequency.

Given: Initial frequency = ν₁ = 2 x 10¹⁵ Hz, initial stopping potential = V_s1 = 6 V, Final frequency = ν₂ = 10¹⁵ Hz, Final stopping potential = V_s2 = 2 V, speed of light = c = 3 x 10⁸ m/s, Charge on electron = e = 1.6 x 10^-19 C.

To Find: Planck’s constant = h =?

Solution:

By Einstein’s photoelectric equation

K.E. _max = hν – Φ

But K.E. _max = eV_s

∴ eV_s = hν – Φ

∴ eV_s1 = hν₁ – Φ ……….. (1)

∴ eV_s2 = hν₂ – Φ ……….. (2)

Subtracting equation (2) from (1)

∴ eV_s1 – eV_s2 = (hν₁ – Φ) – (hν₂ – Φ)

∴ e (V_s1 – V_s2) = hν₁ – hν₂

∴ e (V_s1 – V_s2) = h(ν₁ – ν₂)

∴ (1.6 x 10^-19) (6 – 2) = h (2 x 10¹⁵ – 10¹⁵)

∴ 1.6 x 10^-19 x 4 = h (1 x 10¹⁵)

∴ h = (1.6 x 10^-19 x 4)/1 x 10¹⁵)

∴ h = 6.4 x 10^-34  Js

From equation (1) we have

eV_s1 = hν₁ – Φ

∴ 1.6 x 10^-19 x 6 = 6.4 x 10^-34 x 2 x 10¹⁵ – Φ

∴ 9.6 x 10^-19 = 12.8 x 10^-19 – Φ

∴ Φ = 12.8 x 10^-19 – 9.6 x 10^-19

∴ Φ = 3.2 x 10^-19 J

∴ Φ = (3.2 x 10^-19) / (1.6 x 10^-19) = 2 eV

Φ = hν_o

ν_o = Φ /h = (3.2 x 10^-19 )/(6.4 x 10^-34) = 5 x 10¹⁴ Hz

Ans: the value of Planck’s constant is 6.4 x 10^-34 Js,

work function = 2 eV and Threshold frequency = 5 x 10¹⁴ Hz

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In this article, we shall, derive Einstein’s photoelectric equation and study its use to verify the characteristics of the photoelectric effect of light.

Wave Nature of Light:

Christian Huygen’s proposed that the light is propagated in the form of a wave. But this theory has a serious drawback. It was not able to explain the propagation of light in a vacuum. This drawback was removed by Maxwell he proposed that light is an electromagnetic wave and for the propagation of electromagnetic waves no material medium is required. Thus the wave nature of light was established.

Wave theory was able to explain all the phenomena associated with the propagation of light. But it failed to explain the energy distribution and modern phenomenon like photoelectric effect, Crompton effect, etc.

Particle Nature of Light:

Max Planck proved that the propagation of light or energy takes place in the form of packets of energy called quanta. Quantum of light is called a photon and thus he established the particle nature of light. Using particle or quantum nature of radiation we can explain the phenomenon of photoelectric effect and Crompton effect.

Planck’s Quantum Theory:

The quantum theory was proposed by Max Planck. According to this theory, radiation from a source is not emitted continuously, but it is emitted in packets or bundles of energy.  These packets are called quanta or photons. If the radiation is of frequency ν, each quantum has energy where h is Planck’s constant.

Thus energy of photon = E = hν

The energy is emitted in a discontinuous manner.  This is contrary to the classical theory which assumes that emission of energy is a continuous process.

Particle Nature of Electromagnetic Radiations:

In the interaction of radiation with matter, the radiation behaves as if it is made up of particles. These particles are called photons. Each photon has energy which is given by

E = hν = hc/λ

All photons of light of particular frequency (Wavelength) has the same amount of energy associated with them. The increase in the intensity of light increases the number of photons per second through a given area, but the energy of each photon will be the same. Photons are electrically neutral and are unaffected by electric or magnetic fields. Photons travel in a straight line with the speed of light ‘c’ but show diffraction in certain conditions.

The momentum of each photon is given by

The wavelength of photon changes with the media, hence they have different velocities in different media. The rest mass of a photon is zero. Its kinetic mass is given by

In photon particle collision (such as a photon-electron collision) the total energy and momentum are conserved. However, the number of photons may not be conserved in a collision. The photon may be absorbed or a new photon may be created.

Einstein’s Photoelectric Equation:

On the basis of Planck’s quantum theory, Einstein derived an equation for the photoelectric effect known as Einstein photoelectric equation. Einstein assumed that

• Light consists of photons or quanta of energy, energy in each photon is hν. Where h is the Planck’s constant and ν is the frequency of light
• Each incident photon collides with an electron inside an atom and gives all its energy to the electron.
• Part of this energy is used by the electron to come out of the surface of the metal and the remaining part is the kinetic energy with which the electron is emitted.
• The minimum energy required by an electron to come out of the surface of the metal is called the photoelectric work function (∅_o) of the metal.
• The remaining energy (hν – ∅_o) is the maximum kinetic energy of the electron with which a photoelectron will be ejected.

Thus, Maximum kinetic energy of electron = energy of photon – work function

Let ‘m’ be the mass of electron and v_max be the maximum velocity of photo-electron by which it will be ejected.

This equation is known as Einstein’s photoelectric equation

Photoelectric Work Function:

In the photoelectric effect, the most loosely attached electron of an atom of photosensitive material is removed. The minimum energy required to free an electron from the given surface is called the photoelectric work function (∅_o) of the material of the surface. The work function is a characteristic property of the metal surface.

Mathematically work function is given by

∅_o = h ν_o

Where ν_o = Threshold frequency and h = Planck’s constant.

Explanation of the Existence of Threshold Frequency on the Basis of Einstein’s Photoelectric Equation:

For a given metallic surface, photo-electrons are emitted only when the frequency of incident light is greater than or equal to a certain minimum frequency (no) known as the threshold frequency. The threshold frequency is different for different Substances,

By Einstein’s photoelectric equation

Where ν_o  = Threshold frequency and h = Planck’s constant and 

ν =  frequency of incident radiation

The kinetic energy is always non-negative quantity i.e. it may either be positive or zero thus

Which indicates that for the photoelectric effect, the frequency of incident radiation or incident photon should be equal to or greater than the threshold frequency. The attractive force acting on probable photoelectrons in different atoms is different. Therefore the threshold frequency is different for the different substances.

Explanation of the Effect of Intensity on the Basis of Einstein’s Photoelectric Equation:

If the frequency of incident light is less than the threshold frequency, photoelectrons are not emitted, however large the intensity of incident light may be.

The number of photo-electrons emitted per second is directly proportional to the intensity of incident light.  Thus the photoelectric current is directly proportional to the intensity of incident light. If the intensity of light is more, the number of incident photons on the surface are more. Due to the increased number of photoelectron the rate of photoemission increases, hence the strength of photoelectric current increases. Thus we can conclude that the photoelectric effect (current) is directly proportional to the intensity of incident radiation.

Explanation of the Possible Maximum Kinetic Energy on the Basis of Einstein’s Photoelectric Equation:

By Einstein’s photoelectric equation

Where ν_o  = Threshold frequency and h = Planck’s constant and 

ν =  frequency of incident radiation

This equation does not contain the term of intensity, thus we can say that the maximum kinetic energy of photoelectron is independent of the intensity of incident radiation but depends upon the frequency of incident radiation. This equation indicates that the maximum kinetic energy of the electron depends upon the frequency of incident radiation. And if the frequency of incident radiation is increased kinetic energy of photoelectron also gets increased.

Explanation of Instantaneity of Photoelectric Effect on the Basis of Einstein’s Photoelectric Equation:

The photoelectric effect is an instantaneous process. There is no time lag between the incidence of light and the emission of the photo-electrons in other words, the surface begins to emit photo-electrons as soon as light falls on it.  Also the emission of photo-electrons stops the moment incident light is cut off.

When radiation is incident on the photo-emitting surface at that instant, the whole energy of the photon is transferred to a single electron in one go. Thus the electron gets emitted without any time lag and the photoelectric effect is the instantaneous process.

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