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.

Previous Topic: Numerical Problems on Einstein’s Photoelectric Equation

For More Topics in Physics Click Here

Science > Physics > Photoelectric Effect >Applications of Photovoltaic Cell

The post Applications of Photovoltaic Cell appeared first on The Fact Factor.

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

Previous Topic: Einstein’s Photoelectric Equation (Theory)

Next Topic: Applications of Photoelectric Effect

Science > Physics > Photoelectric Effect >Numerical Problems on Einstein’s Photoelectric Equation

The post Numerical Problems on Einstein’s Photoelectric Equation appeared first on The Fact Factor.

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.

Previous Topic: Numerical Problems on Photoelectric Effect

Next Topic: Numerical Problems on Photoelectric Equation

Science > Physics > Photoelectric Effect >Einstein’s Photoelectric Equation

The post Einstein’s Photoelectric Equation appeared first on The Fact Factor.

In this article, we shall study to calculate, Energy of incident photon, threshold wavelength and threshold frequency of metal.

Example – 01:

The energy of a photon is 2.59 eV. Find its frequency and wavelength.

Given: Energy of photon = E = 2.59 eV = 2.59 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

To Find: Frequency of photon = ν =? Wavelength = λ = ?

Solution:

We have E = h ν

∴   ν = E/h = (2.59 x 1.6 x 10^-19) / (6.63 x 10^-34) = 6.244 x 10¹⁴ Hz

Now c = ν λ

∴ λ = c/ν = (3 x 10⁸) / ( 6.244 x 10¹⁴) = 4.805 x 10^-7 m

∴ λ = 4805 x 10^-10 m = 4805 Å

Ans: The frequency of photon is 6.244 x 10¹⁴ Hz and its wavelength is 4805 Å

Example – 02:

The energy of a photon is 1.0 x 10^-8 J. Find its frequency and wavelength.

Given: Energy of photon = E = 1.0 x 10^-18 J, speed of light = c = 3 x 10⁸ m/s, Planck’s constant = h = 6.63 x 10^-34 Js

To Find: Frequency of photon = ν =? Wavelength = λ = ?

Solution:

We have E = h ν

∴   ν = E/h = (1.0 x 10^-18) / (6.63 x 10^-34) = 1.508 x 10¹⁵ Hz

Now c = ν λ

∴ λ = c/ν = (3 x 10⁸) / ( 1.508 x 10¹⁵) = 1.989 x 10^-7 m

∴ λ = 1989 x 10^-10 m = 1989 Å

Ans: The frequency of photon is 1.508 x 10¹⁴ Hz and its wavelength is 1989 Å

Example – 03:

The energy of a photon is 300 eV. Find its wavelength.

Given: Energy of photon = E = 300 eV = 300 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

To Find: Wavelength = λ =?

Solution:

We have E = h ν = hc/λ

∴ λ = hc / E = (6.63 x 10^-34)(3 x 10⁸)/(300 x 1.6 x 10^-19) = 4.144 x 10^-9 m

∴ λ = 41.44 x 10^-10 m = 41.44 Å

Ans: The wavelength of photon is 41.44 Å

Example – 04:

Find the energy of a photon in eV if its wavelength is 10 m

Given: Wavelength of photon = λ = 10 m, speed of light = c = 3 x 10⁸ m/s, Planck’s constant = h = 6.63 x 10^-34 Js

To Find: Wavelength = λ =?

Solution:

E = hc/λ = (6.63 x 10^-34)(3 x 10⁸)/(10) = 19.89 x 10^-27 J

∴ E = (19.89 x 10^-27)/(1.6 x 10^-19) = 1.243 x 10^-7 eV

Ans: The energy of the photon is 1.243 x 10^-7 eV

Example – 05:

Find the energy of a photon whose frequency is 5.0 x 10¹⁴ Hz

Given: Frequency of photon = ν = 5.0 x 10¹⁴ Hz, Planck’s constant = h = 6.63 x 10^-34 Js

To Find: Energy of photon = E =?

Solution:

We have E = h ν

∴ E = (6.63 x 10^-34) x (5.0 x 10¹⁴)= 3.315 x 10^-29 J

Ans: The energy of the photon is 3.315 x 10^-29 J

Example – 06:

The photoelectric work function of silver is 3.315 eV. Calculate the threshold frequency and threshold wavelength of silver.

Given: Work function of silver = Φ = 3.315 eV = 3.315 x 1.6 x 10^-19 J, speed of light = 3 x 10⁸ m/s, Planck’s constant = h = 6.63 x 10^-34 Js

To Find: Threshold frequency of silver = ν_o =? Threshold wavelength of silver = λ_o = ?

Solution:

We have Φ = h ν_o

∴   ν_o = Φ/h = (3.315 x 1.6 x 10^-19)/(6.63 x 10^-34) = 8 x 10¹⁴ Hz

Now c = ν_o λ_o

∴ λ_o = c/ν_o = (3 x 10⁸)/( 8 x 10¹⁴) = 3.750 x 10^-7 m

∴ λ_o= 3750 x 10^-10 m = 4805 Å

Ans: The threshold frequency of silver is 8 x 10¹⁴ Hz and its threshold wavelength is 3750 Å

Example – 07:

A light of wavelength 4800 Å can just cause photoemission from a metal. What is the photoelectric work function for metal in eV?

Given: Threshold wavelength = λ_o = 4800 Å = 4800 x 10^-10 m, speed of light = c = 3 x 10⁸ m/s, Planck’s constant = h = 6.63 x 10^-34 Js

To Find: Work function of silver = Φ =?

Solution:

We have Φ = h ν_o = hc/λ_o

∴    Φ = (6.63 x 10^-34) x (3 x 10⁸) / (4800 x 10^-10) = 4.144 x 10^-19 J

∴    Φ = (4.144 x 10^-19) / (1.6 x 10^-19) = 2.59 eV

Ans: The photoelectric work function of the metal is 2.59 eV

Example – 08:

The photoelectric work function of a metal is 2 eV. Calculate the lowest frequency radiation that will cause photoemission from the surface.

Given: Work function of silver = Φ = 2 eV = 2 x 1.6 x 10^-19 J, Planck’s constant = h = 6.63 x 10^-34 Js

To Find: Threshold frequency of silver = ν_o =?

Solution:

We have Φ = h ν_o

∴   ν_o = Φ/h = (2 x 1.6 x 10^-19)/(6.63 x 10^-34) = 4.827 x 10¹⁴ Hz

Ans: The threshold frequency of metal is 4.827 x 10¹⁴.

Example – 09:

The photoelectric work function of platinum is 6.3 eV and the longest wavelength that can eject photoelectron from platinum is 1972 Å. Calculate the Planck’s constant.

Given: Work function of platinum = Φ = 6.3 eV =6.3 x 1.6 x 10^-19 J, Threshold wavelength of silver = 1972 Å = 1972 x 10^-10 m, speed of light = c = 3 x 10⁸ m/s,

To Find: Planck’s constant = h =?

Solution:

We have Φ = h ν_o = hc/λ_o

∴   h = Φλ_o/c = (6.3 x 1.6 x 10^-19) x (1972 x 10^-10) / (3 x 10⁸) = 6.625 x 10^-34 Js

Ans: The value of Planck’s constant is 6.625 x 10^-34 Js

Example – 10:

The photoelectric work function of metal is 1.32 eV. Calculate the longest wavelength that can cause photoelectric emission from the metal surface.

Given: Work function of silver = Φ = 1.32 eV = 1.32 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

To Find: Threshold wavelength of metal = λ_o =?

Solution:

We have Φ = h ν_o = hc/λ_o

∴ λ_o = hc/Φ =(6.63 x 10^-34) x (3 x 10⁸) / (1.32 x 1.6 x 10^-19) = 9.418 x 10^-7 m

∴ λ_o= 9418 x 10^-10 m = 9418 Å

Ans: The threshold wavelength is 9418 Å

Example – 11:

The photoelectric work function of metal is 5 eV. Calculate the threshold frequency for the metal. If a light of wavelength 4000 Å is incident on this metal surface, will photoelectron will be ejected?

Given: Work function of silver = Φ = 5 eV = 5 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, wavelength of incident light = λ = 4000 Å = 4000  x 10^-10 m

To Find: Threshold wavelength of metal = λ_o =?

Solution:

We have Φ = h ν_o

∴   ν_o = Φ/h = (5 x 1.6 x 10^-19)/(6.63 x 10^-34) = 1.2 x 10¹⁵ Hz

Now c = ν λ

∴ ν = c/λ = (3 x 10⁸) / ( 4000 x 10^-10) = 7.5 x 10¹⁴ Hz

The frequency of incident light is less than the threshold frequency.

No photoelectrons will be emitted from the metal surface.

Ans: The threshold frequency is 1.2 x 10¹⁵ Hz and no photoelectron will be emitted.

Example – 12:

The photoelectric work function of a metal is 2.4 eV. Calculate the incident frequency, the threshold frequency for the metal. If a light of wavelength 6800 Å is incident on this metal surface, will photoelectron will be ejected?

Given: Work function of silver = Φ = 2.4 eV= 2.4 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, wavelength of incident light = λ = 6800 Å = 6800  x 10^-10 m

To Find: Threshold wavelength of metal = λ_o =?

Solution:

We have Φ = h ν_o

∴   ν_o = Φ/h = (2.4 x 1.6 x 10^-19)/(6.63 x 10^-34) = 5.79 x 10¹⁴ Hz

Now c = ν λ

∴ ν = c/λ = (3 x 10⁸) / ( 6800 x 10^-10) = 4.41 x 10¹⁴ Hz

The frequency of incident light is less than the threshold frequency.

No photoelectrons will be emitted from the metal surface.

Ans: The incident frequency is 4.41 x 10¹⁴ Hz and the threshold frequency is 5.79 x 10¹⁴ Hz, and no photoelectron will be ejected.

Example – 13:

The photoelectric work function of a metal is 3 eV. Calculate the threshold frequency for the metal. If light of wavelength 6000 Å is incident on this metal surface, will photoelectron will be ejected?

Given: Work function of silver = Φ = 3 eV = 3 x 1.6 x 10^-19 J, speed of light = 3 x 10⁸ m/s, planck’s constant = h = 6.63 x 10^-34 Js, wavelength of incident light = λ = 6000 Å = 6000 x 10^-10 m

To Find: Threshold wavelength of metal = λ_o =?

Solution:

We have Φ = h ν_o

∴   ν_o = Φ/h = (3 x 1.6 x 10^-19)/(6.63 x 10^-34) = 7.24 x 10¹⁴ Hz

Now c = ν λ

∴ ν = c/λ = (3 x 10⁸)/( 6000 x 10^-10) = 5 x 10¹⁴ Hz

The frequency of incident light is less than the threshold frequency.

No photoelectrons will be emitted from the metal surface.

Ans: The threshold frequency is 7.24 x 10¹⁴ Hz,

and no photoelectron will be ejected.

Previous Topic: Photoelectric Effect (Theory)

Next Topic: Einstein’s Photoelectric Equation

Science > Physics > Photoelectric Effect > Numerical Problems on Photoelectric Effect

The post Numerical Problems on Photoelectric Effect appeared first on The Fact Factor.

It is found that when the light of very short wavelength (or high frequency) is incident on a certain metallic surface of photosensitive material, electrons are emitted by the surface. Most of the metals emit electrons when ultra-violet light is incident on them.  However, alkali metals like sodium, potassium, etc. emit electrons even when ordinary light falls on them. These materials are known as photosensitive materials. In this article, we shall study, the photoelectric effect.

This phenomenon of emission of electrons by a certain metal surface when radiation of suitable frequency is incident on it is called photoelectric effect. The electrons emitted from the metal surface are called photo-electrons. And the current is called photoelectric current or photocurrent. This phenomenon was first discovered by Hertz in 1887.

Experimental Study of Photoelectric Effect:

Circuit Arrangement:

The experimental arrangement to study the photoelectric effect is as shown in the diagram.  The apparatus consists of an evacuated glass tube. Two metal electrodes plate P and collector C, are enclosed in this tube. This glass tube is provided with a side quartz window. The plate A is connected to the positive terminal of a battery and the cathode is connected to the negative terminal of the battery. An ultra-violet light of variable frequency is made incident on the plate through the window. The plate acts as a photoelectron emitter. When the ultra-violet light of suitable frequency is incident on the plate, it emits photoelectrons which starts flowing through the external circuit and constitute the photoelectric current.

Study of the Effect of the Frequency of the Incident Light on Photoelectric Effect:

A suitable potential difference is applied between plate P and collector C. Light of constant intensity but the variable frequency is allowed to fall on plate P. The frequency of the incident light is slowly increased.  It is found that there would be no photoelectric current up to certain limiting frequency.  This frequency is called threshold frequency ( ν ₀).  The value of the threshold frequency depends on the material of the plate P.

The minimum frequency of incident radiation for which photoelectrons are just emitted from the photosensitive material is called threshold frequency. The variation of photoelectric current with the frequency of incident radiation is as shown in the following graph.

This shows that the photo-electric effect depends upon the frequency of the incident light (radiation). The threshold frequency is different for different materials of the emitter and it is a characteristic property of that material.

Study of the Effect of Intensity of Incident Light on Photoelectric Effect:

A suitable potential difference is applied between plate P and collector C. Light of constant frequency having frequency more than the threshold frequency of material of emitter P but of the variable intensity is allowed to fall on plate P. The intensity of the incident light is slowly increased.  As the intensity of incident light increases the photoelectric current increases. This shows that the photo-electric effect depends upon the intensity of the incident light (radiation).

The variation of photoelectric current with the intensity of incident radiation is as shown in the following graph.

Study of the Effect of Potential Difference Between the Emitter and Collector on Photoelectric Effect:

Light of constant frequency having frequency more than the threshold frequency of material of emitter P and constant intensity is allowed to fall on plate P. A variable potential difference is applied between plate P and collector C. The positive potential difference applied to the emitter P and collector C initially slowly increased zero, the photoelectric current increases. At a certain positive potential of plate A, all electrons emitted are collected by plate A and the photoelectric current becomes maximum. If we increase the positive potential of the plate, the photoelectric current remains constant. This maximum value of photoelectric current is called saturation current.

The variation of photoelectric current with the potential difference between the plate and collector is shown in the following graphs.

If the positive potential difference applied to the emitter P and collector C is slowly decreased to zero.  As the potential difference applied between plate P and collector C decreases the photoelectric current decreases.

This shows that all photo-electrons are not emitted with the same kinetic energy and the maximum kinetic energy of the photoelectrons depends upon the frequency of incident light.

If the potential of the plate is reduced below zero and it is made more and more negative, a point will be reached when the photoelectric current reduces to zero. The magnitude of this retarding potential for which photoelectric current is zero is called stopping potential Vo. The value of stopping potential depends on the kinetic energy of photoelectrons.

We have,

Threshold Frequency and Threshold Wavelength:

It is found that there would be no photoelectric current up to certain limiting frequency.  This frequency is called threshold frequency (ν_o).  The value of the threshold frequency depends on the material of the plate P.

The minimum frequency of incident radiation for which photoelectrons are just emitted from a photosensitive material is called threshold frequency. The corresponding wavelength of the radiation is called the threshold wavelength. The maximum wavelength of incident radiation for which photoelectrons are just emitted from the photosensitive material is called threshold wavelength (λ_o).

Concept of Saturation Current:

The positive potential difference applied to the emitter P and collector C slowly increased zero, the photoelectric current increases. At a certain positive potential of plate A, all electrons emitted are collected by plate A and the photoelectric current becomes maximum. If we increase the positive potential of the plate, the photoelectric current remains constant. This maximum value of photoelectric current is called saturation current. For constant frequency of radiation, as the plate potential increases the saturation current increases.

Concept of Stopping Potential:

If the potential of the plate is reduced below zero and it is made more and more negative, a point will be reached when the photoelectric current reduces to zero. The magnitude of this retarding potential for which photoelectric current is zero is called stopping potential V_o. The value of stopping potential depends on the kinetic energy of photoelectrons. We have,

Where,  e = Charge on electron

V_o = Stopping potential

m = Mass of electron

v_max= Maximum velocity of photoelectron

For the constant potential of the plate, as the plate frequency of incident radiation increases the stopping potential increases.

Variation of Stopping potential with Frequency:

The following graph shows the variation of stopping potential with the frequency of incident radiations for two metals say A and B.

Conclusions:

Stopping potential at threshold frequency is zero and the stopping potential varies directly with the frequency of incident radiation.

Characteristics of the Photoelectric Effect:

• 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 (ν_o) known as the threshold frequency.   The threshold frequency is different for different Substances,
• 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 photoelectric current is directly proportional to the intensity of incident light.
• Photo-electrons are emitted with different velocities.  The maximum velocity (and hence maximum K.E.) of a photo-electron depends upon the frequency of incident light and does not depend upon its intensity.
• The maximum K.E. of the photo-electron increases with an increase in the frequency of incident light.
• 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.
• In the case of alkali metals like sodium, potassium, etc. first ionization potential (ionization enthalpy) is low and hence emit electrons even when ordinary light falls on them. These materials are known as photosensitive materials.

Note:         

Frequencies of radio waves are low, they are lower than threshold frequencies of alkali metals (having lowest threshold frequencies). Hence no photoemission is possible with radio waves.

Previous Topic: Numerical Problems on Specific Charge (e/m ratio)

Next Topic: Numerical Problems on Photoelectric Effect

Science > Physics > Photoelectric Effect > Photoelectric Effect

The post Photoelectric Effect appeared first on The Fact Factor.

In this article, we shall see problems based on Thomson’s experiment to calculate the specific charge (e/m).

Example – 01:

In Thomson’s experiment, a beam of electrons travelling at 6.8 x 10⁷ m/s is bent into a circular path of radius 4 cm in a magnetic field of induction 10^-2 Wb/m² normal to its path. Find the specific charge (e/m).

Given: The speed of electron = v = 6.8 x 10⁷ m/s, Radius of circular path = r = 4 cm = 4 x 10^-2 m, Magnetic induction = B = 10^-2 Wb/m².

To Find: e/m =?

Solution:

Ans: The value of e/m is 1.7 x 10¹¹ C/kg

Example – 02:

In Thomson’s experiment, a beam of electrons travelling at 10.2  x 10⁷ m/s is bent into a circular path of radius 6 cm in a magnetic field of induction 10^-2 Wb/m² normal to its path. Find the specific charge (e/m).

Given: The speed of electron = v = 10.2 x 10⁷ m/s, Radius of circular path = r = 6 cm = 6 x 10^-2 m, Magnetic induction = B = 10^-2 Wb/m².

To Find: e/m =?

Solution:

Ans: The specific charge ratio e/m is 1.7 x 10¹¹ C/kg

Example – 03:

In Thomson’s experiment, a beam of electrons travelling at 2.652 x 10⁷ m/s is bent into a circular path in a magnetic field of induction 0.5 Wb/m² normal to its path. Find the radius of circular path if e/m = 1.76 x 10¹¹ C/kg.

Given: The speed of electron = v = 2.652 x 10⁷ m/s, Magnetic induction = B = 0.5 Wb/m², e/m = 1.76 x 10¹¹ C/kg.

To Find: Radius of circular path = r =?

Solution:

Ans: The radius of the circular path is 3.01 x 10^-4 m

Example – 04:

In Thomson’s experiment, an electron is accelerated from rest in an electric field through a P.D. of 100 V. It then enters normally in the magnetic field of induction 10^-3 Wb/m². Find the radius of curvature of the path.

Given: P.D. = V = 100 V, Magnetic induction = B = 10^-3 Wb/m², e/m = 1.76 x 10¹¹ C/kg.

To Find: Radius of circular path = r =?

Solution:

Let v be the speed of the electron

Ans: The radius of the circular path is 3.37 cm

Example – 05:

In Thomson’s experiment, an electron is accelerated from rest in an electric field through a P.D. of 4000 V. It then enters normally in the magnetic field of induction 10^-2 Wb/m². Find the velocity of electron and the radius of curvature of the path. Mass of electron = 9.1 x 10^-31 kg, charge on electron = 1.6 x 10^-19 C.

Given: P.D. = V = 4000 V, Magnetic induction = B = 10^-2 Wb/m², mass of electron = m = 9.1 x 10^-31 kg, charge on electron = e = 1.6 x 10^-19 C.

To Find: Radius of circular path = r =?

Solution:

Let v be the speed of the electron

Ans: Velocity of electron is 3.75 x 10⁷ m/s

The radius of the circular path is 0.0213 m

Example – 06:

In Thomson’s experiment, an electron is accelerated from rest in an electric field through a P.D. of 2000 V. It then enters normally in the magnetic field of induction 3 x 10^-4 Wb/m². Find the velocity of electron and the radius of curvature of the path. Mass of electron = 9.1 x 10^-31 kg, charge on electron = 1.6 x 10^-19 C.

Given: P.D. = V = 4000 V, Magnetic induction = B = 10^-2 Wb/m², mass of electron = m = 9.1 x 10^-31 kg, charge on electron = e = 1.6 x 10^-19 C.

To Find: Radius of circular path = r =?

Solution:

Let v be the speed of the electron

Ans: Velocity of electron is 2.625 x 10⁷ m/s

The radius of the circular path is 0.503 m

Example – 07:

In Thomson’s experiment, an electron with kinetic energy 2000 eV enters normally in the magnetic field of induction 0.02 Wb/m². Find the velocity of electron and the radius of curvature of the path. Mass of electron = 9.1 x 10^-31 kg, charge on electron = 1.6 x 10^-19 C.

Given: Kinetic energy = 2000 eV = 2000 x 1.6 x 10^-19 J, Magnetic induction = B = 10^-2 Wb/m², mass of electron = m = 9.1 x 10^-31 kg, charge on electron = e = 1.6 x 10^-19 C.

To Find: Radius of circular path = r =?

Solution:

Let v be the speed of the electron

Ans: Velocity of electron is 2.625 x 10⁷ m/s

The radius of the circular path is 7.54 mm

Example – 08:

A beam of electrons travelling at 8.5 x 10⁶ m/s is bent into a circular path in a magnetic field of induction 0.1 Wb/m² normal to its path. Find the radius of circular path if e/m = 1.7 x 10¹¹ C/kg. Find its angular speed in the magnetic field in the number of revolution.

Given: The speed of electron = v = 8.5 x 10⁶ m/s, Magnetic induction = B = 0.1 Wb/m², e/m = 1.7 x 10¹¹ C/kg.

To Find: Radius of circular path = r =?, number of revolutions per second = n = ?

Solution:

We have v  = r ω

ω = v/r = 8.5 x 10⁶ / 5 x 10^-4 =  1.7 x 10¹⁰ rad/s

ω = 2πn

n = ω/2π = 1.7 x 10¹⁰ / 2 x 3.142 = 2.71 x 10⁹  r.p.s.

Ans: The radius of the circular path is 5 x 10^-4 m

The angular speed is 2.71  x 10⁹ r.p.s.

Example – 09:

In Thomson’s experiment, electrons accelerated from rest through a P.D. of 2500 V are collimated into a fine beam to pass through a space between two metal plates where crossed electric and magnetic field is applied. If the magnetic induction is 6 x 10^-3 Wb/m² and the electric intensity is 1.8 x 10⁵ V/m, the electrons beam pass undeviated along a path perpendicular to both electric and magnetic field. Find the e/m ratio.

Given: P.D. applied = V = 2500 V, Magnetic induction = B = 6 x 10^-3 Wb/m², Electric intensity = E = 1.8 x 10⁵ V/m.

To Find: e/m =?

Solution:

Ans: The specific charge ratio e/m is 1.8 x 10¹¹ C/kg

Example – 10:

In Thomson’s experiment, electrons accelerated from rest through a P.D. of 1000 V are collimated into a fine beam to pass through a space between two metal plates where crossed electric and magnetic field is applied. If the magnetic induction is 10^-4 Wb/m² and the electric intensity is 2000 V/m, the electrons beam pass undeviated along a path perpendicular to both electric and magnetic field. Find the e/m ratio and speed of the electron.

Given: P.D. applied = V = 1000 V, Magnetic induction = B = 10^-4 Wb/m², Electric intensity = E = 2000 V/m.

To Find: e/m =?

Solution:

v = E/B = 2000 / 10^-4 = 2 x 10⁷ m/s

Ans: The specific charge ratio e/m is 2 x 10¹¹ C/kg

Velocity of electron is 2 x 10⁷ m/s

Example – 11:

In Thomson’s experiment, electrons accelerated from rest through a P.D. of 1000 V enter the space between two metal plates between which a P.D. of 250 V is maintained. The separation between the plates is 5 cm. What magnetic field should be applied perpendicular to the electric field so that the beam remains at right angles to both the fields passes through undeviated. e/m = 1.76 x 10¹¹ C/kg.

Given: P.D. applied = V = 1000 V, P.D. across plate = V_p = 250 V, separation between plates = 5 cm = 5 x 10^-2 m, e/m = 1.76 x 10¹¹ C/kg

To Find: Magnetic induction = B =?

Solution:

E = V_p/d = 250/5 x 10^-2   = 5 x 10³ V/m

Ans: The applied magnetic field is 2.67 x 10^-4 Wb/m²

Example – 12:

In Thomson’s experiment, electrons accelerated from rest through a P.D. of 2000 V enter the space between two metal plates between which a P.D. of 800 V is maintained. The separation between the plates is 1 cm. What magnetic field should be applied perpendicular to the electric field so that the beam remains at right angles to both the fields passes through undeviated. e/m = 1.76 x 10¹¹ C/kg.

Given: P.D. applied = V = 2000 V, P.D. across plate = V_p = 800 V, separation between plates = 1 cm = 1 x 10^-2 m, e/m = 1.76 x 10¹¹ C/kg

To Find: Magnetic induction = B =?

Solution:

E = V_p/d = 800/1 x 10^-2   = 8 x 10⁴ V/m

Ans: The applied magnetic field is 3.02 x 10^-3 Wb/m²

Example – 13:

In Thomson’s experiment, a beam of electrons enters the space between two parallel metal plates with a velocity of 2.5 x 10⁷ m/s to encounter a magnetic induction field B = 1.8 x 10^-3 Wb/m² at right angles to its path. What is the radius of curvature of path of the electrons? Find the P.D. to be applied between the plates which are 0.5 cm apart so that the electrons beam passes undeviated. Given e/m = 1.76 x 10¹¹ C/kg

Given: velocity of electron = v = 2.5 x 10⁷ m/s, Magnetic induction = B = 1.8 x 10^-3 Wb/m², separation between plates = d – 0.5 cm = 0.5 x 10^-2 m, e/m = 1.76 x 10¹¹ C/kg

To Find: radius of curve path = r =? P.D. across plates = Vp =?

Solution:

Now v = E/B

E = v x B = 2.5 x 10⁷ x 1.8 x 10^-3 =  4.5 x 10⁴  V/m

Now, E = V/d

V = E x d = 4.5 x 10⁴  x 0.5 x 10^-2 = 225 V

Ans: 225 V

Example – 14:

In Thomson’s experiment to find e/m, the deflector plates were 1.2 cm apart and a P.D. of 820 V was maintained between them. A magnetic field of induction 2.4 x 10^-3 Wb/m² is applied at right angles to the electric field and the direction of motion of the electron beam to produce null deflection. On switching off the electric field, the electron beam traces a circle of radius 7 cm. Find the velocity and e/m of electrons.

Given: P.D. applied = V = 820 V, The separation between plates = d = 1.2 cm = 1.2 x 10^-2 m, Magnetic induction = B = 2.4 x 10^-3 Wb/m², Radius of circular path of electron = r = 7 cm = 7 x 10^-2 m.

To Find: velocity of electron =? e/m =?

Solution:

E = V/d = 820/1.2 x 10^-2 = 6.833 x 10⁴ V/m

Now, the velocity of the electron

v = E/B = 6.833 x 10⁴ / 2.4 x 10^-3 = 2.847 x 10⁷ m/s

Ans: The velocity of the electron is 2.847 x 10⁷ m/s

The value of e/m is 1.695 x 10¹¹ C/kg

Example – 15:

In Thomson’s experiment to find e/m, A crossed electric field of strength 8000 N/C and a magnetic field of induction 2 x 10^-3 Wb/m² is applied at right angles to the electric field and the direction of motion of the electron beam to produce null deflection. On switching off the electric field, the electron beam traces a circle of radius 11.4 mm. Find the velocity and e/m of electrons.

Given: Electric intensity = 8000 N/C, Magnetic induction = B = 2.4 x 10^-3 Wb/m², Radius of circular path of electron = r = 11.4 mm = 11.4 x 10^-3 m.

To Find: velocity of electron =?, e/m =?

Solution:

velocity of electron v = E/B = 8000 / 2 x 10^-3 = 4 x 10⁶ m/s

Ans: The velocity of the electron is 4 x 10⁶ m/s

The value of e/m is 1.754 x 10¹¹ C/kg

Example – 16:

In Thomson’s experiment, a beam of electrons passes undeviated through cross electric and magnetic field. If E = 1.5 x 10³ V/m and B = 0.4 Wb/m², find velocity of electron if its direction is perpendicular to both E and B.

Given: E = 1.5 x 10³ V/m, B = 0.4 Wb/m².

To Find: Velocity of electron = v =?

Solution:

V = E/B = 1.5 x 10³  / 0.4= 3.75 x 10³ m/s

Ans: The velocity of electron is 3.75 x 10³ m/s

Previous Topic: Thomson’s Experiment to Find Specific Charge (e/m Ratio)

Next Topic: Photoelectric Effect

Science > Physics > Photoelectric Effect > Problems on Specific Charge

The post Problems on Specific Charge Ratio appeared first on The Fact Factor.

In this article, we shall study Thomson’s experiment to find the velocity of the electron in cathode rays and the specific charge ratio of an electron.

Concept of Work Function:

When an electron tries to come out of a metal surface (atom) the remaining part of the surface (atom) acquires a positive charge. This positive charge on the remaining part of the surface pulls back the negatively charged outgoing electron. Thus certain minimum energy is to be supplied to the electron from the external agency so that the electron can escape the surface of the metal. This energy is called the work function. It is denoted by the symbol ∅. Its practical unit is eV and the S. I. unit is joule (J).

This minimum energy can be supplied to the electron by following physical processes.

• Thermionic emission: In this process, the metal is heated and sufficient thermal energy is supplied to the electrons to get emitted from the surface.
• Field Emission: By applying a very strong electric field of order 10⁸ V/m, the emission of an electron can be achieved.
• Photoelectric Emission: When the light of a certain frequency illuminates a metal surface, the emission of an electron can be achieved.

Discovery of Electron:

J.J. Thomson discovered cathode rays during his experiment of discharge of electricity through gases at low pressure. Further studies by J.J. Thomson, W. Crooke, and J. Perin proved that the cathode rays are made up of identical negatively charged particles. These particles were called as electrons. Further J.J. Thomson in his experiments found the velocity and specific charge i.e. e/m ratio of the electron.
He proved that the specific charge or e/m ratio is the same irrespective of the material of cathode and irrespective of the nature of the gas in the discharge tube. From these observations, he concluded that electrons are fundamental particles of an atom of all substances

Effect of Electric Field on Cathode Rays Particles:

When cathode rays are passed through an electric field created by applying a potential across the plates P₁ and P₂, it is found that the cathode rays particles get deflected towards the positive plate. This indicates that the cathode rays particles carry a negative charge on them.

Let ‘e’ be the charge on the particle and let E’ ‘be the intensity of the electric field applied. Then the electrical force acting on the particle is given by

F_E = eE

Under the action of this electric force, the particle moves in the parabolic path in the electric field. The particle leaves the electric field in the straight line in the direction tangent to the parabola at the point of leaving.

Effect of Magnetic Field on Cathode Rays Particles:

When cathode rays are passed through a magnetic field created by applying a strong magnetic field, it is found that the cathode rays particles get deflected in a circular path.

Let ‘e’ be the charge on the particle and let ‘B’ be the intensity of the magnetic field (magnetic induction) applied. Let ‘v’ be the velocity of cathode-ray particle in the field. Then the magnetic force acting on the particle is given by

F_M = Bev

Under the action of this magnetic force, the particle moves in a circular path in the magnetic field. The particle leaves the magnetic field in the straight line in the direction tangent to the circular path at the point of leaving.

Thomson’s Experiment to Determine Specific Charge (e/m Ratio):

The velocity of the electron was first found by J.J. Thomson in 1897. Using the same experiment he found the specific charge ratio.

Construction:

The apparatus consists of a discharge tube containing gas at a very low pressure of about 0.01 mm of mercury. The discharge tube has a cathode at one end and a fluorescent screen at the other. The anode consists of a cylinder with a fine bore. The electric field can be applied between the plates P₁ and P₂. Plate P₁ is positive and plate P₂ is negative. A magnetic field can be applied perpendicular to the electric field and perpendicular to the plane of the diagram and into the plane.

Working :

Cathode emits electrons and they are collimated by cylindrical fine bore anode. The velocity of electrons depends on the potential difference between the cathode and the anode. When no field is applied. The electrons move in a straight line and form spot at B at the centre of the screen.

Effect of Electric Field:

The cathode rays are passed through only electric field created by applying a potential across the plates P₁ and P₂, it is found that the cathode rays particles get deflected towards the positive plate and strike the screen at A. Let e be the charge on the particle and let E be the intensity of electric field applied. Then the electrical force acting on the particle is given by

F_E = eE

Under the action of this electric force, the particle moves in a parabolic path in the electric field. The particle leaves the electric field in the straight line in the direction tangent to the parabola at the point of leaving.

Effect of Magnetic Field :

The cathode rays are passed through the only magnetic field created by applying a strong magnetic field, it is found that the cathode rays particles get deflected in a circular path and strike the screen at C. Let e be the charge on the particle and let B be the intensity of magnetic field (magnetic induction) applied. Let v be the velocity of the cathode-ray particle in the field. Then the magnetic force acting on the particle is given by

F_M = Bev

Under the action of this magnetic force, the particle moves in a circular path in the magnetic field. The particle leaves the magnetic field in the straight line in the direction tangent to the circular path at the point of leaving.

Determination of Velocity:

Thomson applied both the fields simultaneously and adjusted its value such that the spot remains at the centre of the screen at B. At this point

F_E = F_M
∴    e E = Bev
∴  v = E/B

Using this formula the velocity of the electron in the discharge tube can be determined.

Determination of e/m Ratio (Specific Charge)by Applying Electric Field Only:

If D.C. potential of V is applied between cathode and anode then the potential energy of electron entering the field is given by eV. In the field this potential energy gets converted into kinetic energy

This is the expression for determining e/m ratio

Determination of e/m Ratio (Specific Charge) by Applying Magnetic Field Only:

When the only magnetic field is applied the electron moves in a circular orbit. Necessary centripetal force is provided by the magnetic force due to the magnetic field. Let r be the radius of the circular path of an electron.

This is the expression for determining e/m ratio (Specific Charge)

Next Topic: Numerical Problems on Specific Charge

Science > Physics > Photoelectric Effect > Thomson’s Experiment

The post Specific Charge Ratio of Electron appeared first on The Fact Factor.