Compton Wavelength: Equation and Compton Scattering

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Compton wavelength is the wavelength of the particle that is equal to the wavelength of a photon having the same mass. It is a quantum mechanical property of a particle. The scattering of the photons due to the presence of the charged particles in the form of the electrons and positrons is termed, Compton. Compton wavelength is generally denoted through the Greek symbol λ. It is measured through the SI unit of length i.e. meter.

Table of Content

What is Compton Wavelength?
Equation of the Compton Wavelength
What is Compton Scattering?
What is Compton Effect?
Things to remember
Previous Year Questions
Sample Questions

Key Terms: Compton, Compton Wavelength, Compton Scattering, Photon, Wavelength of electrons, Electron, Proton, Wavelength

What is Compton Wavelength?

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Compton wavelength is a property of any particle whose wavelength is similar to that of the photon.

However, the energy of the photon must match the mass of the corresponding particle.
Without this additional condition, the function of Compton wavelength might not work.
The derivation of the Compton effect for any of the particles is simply derived using the Compton wavelength of the electrons.
While considering any particle, the value of the Compton wavelength for it shall remain unchanged.
Although, this value varies as per different particles.

The value of the Compton wavelength of an electron as per the CODATA 2014 is given below

Particulars	Details
Compton wavelength λ	2.4263102367(11) × 10^-12 meter
Compton wavelength λ	0.0242 angstrom

The value of Compton wavelength is constant for a particle. Therefore, the value varies from particle to particle. To derive the Compton effect equation for a particle, the Compton wavelength of an electron is used.

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Compton Wavelength – Related Topics
Potential due to an Electric Dipole	Davisson and Germer Experiment	Electron Emission
Experimental Study of Photoelectric Effect	Hadron	Quantum Theory Of Light

Compton Wavelength Equation

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As the Compton wavelength is used to express the relation between the wavelength of a particle and the wavelength of the photon of the same mass, the equation of the Compton wavelength is expressed as follows.

\(\begin{array}{l}\lambda = \frac{h}{mc}\end{array}\)

Now, here,

λ refers to the value of the Compton wavelength of a particle which shall be measured using the meter.
h refers to the Planck constant.
m refers to the value of the mass of the particle which is measured using the unit of a gram.
Whereas, c symbolizes the speed of light in a vacuum measured using a meter per second.

Compton Wavelength Example

Let us take a photon that collides with an electron.

Initially, the momentum and kinetic energy of the photon is p0’ and E0’.
When the photon collides with an electron along the x-axis, the kinetic energy decreases to E1’ and momentum to p1’.
The angle formed by the scattering photon with the x-axis is Θ.

An electron, on the other hand, gains momentum ‘p’ and kinetic energy ‘k’ to form an angle with the x-axis, Ф.

The momentum remains conserved at all axis. Applying the law of conservation of momentum along the X-axis and Y axis we get,

Along the X-axis

p0 = p1cosΘ + pCosФ
p0 - p1 CosΘ = p CosФ ---- (1)

Along the Y-axis

Along the y-axis, p0. Therefore,

0 = p1 Sin Θ - p Sin Ф
p1 Sin Θ = p Sin Ф ---- (2)

Now, squaring equations (1) & (2), we get:

From equation 1:

(p0 - p1 cosΘ)² = p² cos²Ф
= p0² + p1²cosΘ + 2 p0 p1cosΘ = p² cos²Ф ---- (3)

From equation 2:

p1Sin²Θ = p²Sin²Ф …..(4)
And, adding equations (3) and (4), we get:
po²+ p1² cosΘ + 2p0 p1cosΘ + p1sin²Θ = p²Cos²Ф + p²Sin²Ф
po²+ p1²+ 2p0 p1cosΘ = p² ---- (5)

Applying the law of conservation of energy:

E0 = E1 + k
E0 - E1 = k
cp0 - cp1 = k

⇒ p0 - p1 = k/c

The momentum of a photon is given by: p = h/λ = hf/λf = E/c

So, we represent the energy of a photon as; E = pc

The relation among the relativistic energy, momentum, and the rest mass of the electron after the collision is:

(k + mc²)² = p²c² + m²c⁴

⇒ k² + m²c⁴ + 2kmc² = p²c² + m²c⁴

⇒ k² + 2kmc² = p²c²

⇒ k²/c² + 2km = p² ……(6)

From eq (5) in (6): (p0 - p1)² + 2km = p² …..(7)

From eq (5) in (6): (p0 - p1)² + 2km = p0² + p1² - 2p0p1 cosΘ

On solving this, we will get the equation as:

1/p1 - 1/p0 = 1/mc (1 - cosΘ)

Now, putting p = λ/h

⇒ λ0 - λ1 = h/mc (1 - Cos Θ)

Therefore, the Compton wavelength value of a given electron is 0.243 Å.

What is Compton Scattering?

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An experiment in 1912 that was conducted by Max Von Laue concluded that the prevailing X-rays always perform diffraction but this diffraction takes place in the presence of light only. Therefore, he believed that in such a case, the X-rays shall be compared to the light which has a wave nature.

However, even before, in 1905, Einstein claimed light to be a bunch of small packets or quanta which are known as photons. Therefore, both the x-rays and the light must have a particle nature.
Further, Arthur Holly Compton, in his experiment, named Compton scattering claimed that the interaction of the X-rays to that of the electrons with each other resembles the two colliding balls. Thus, defining the Compton scattering.

What is Compton Effect?

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The Compton effect can be expressed as the following using the given equation.

λ₁ - λ₀ = h/mc(1 - cosΘ)

Here,

λ1 - λ0 refers to the difference between the wavelength

m refers to the rest mass

c refers to the speed of light, i.e., 3 x 10³ m/s

h refers to Planck’s constant that is, 6.626 x 10³ J/s

Compton Effect

Compton Effect

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Chapter Related Links
Radiation	Relativity	NCERT Solutions of Dual Nature
Millikan's Oil Drop Experiment	Einstein’s Photoelectric Equation	Dual Nature of Radiation and Matter
Planck Equation	Effects of Radiation	Dual Nature of Radiation and Matter MCQ
Quark	De broglie Wavelength Formula	Dual Nature of Radiation & Matter Important Questions

Things to Remember

Compton wavelength is the wavelength of the electromagnetic wave that is taken into consideration in relation to the energy of the photon.
The photon energy must match the rest energy of the electron to make this phenomenon work.
However, the occurrence of the electron Compton wavelength is seen only after the annihilation of the electron and its antiparticle, positron
Annihilation, refers to the complete obliteration of the object which takes place due to the collision of electron and positron, followed by the emitting of the gamma rays.

Previous Year Questions

If its disintegration constant is 3.7×10⁴sec⁻¹, the mass of X is…. [ AP EAPCET 1998 ]
The emission of photoelectrons is directly proportional to…. [NEET 1997]
The de-Broglie wavelength associated with the electrons would be… [ NEET 2011]
By photoelectric effect, Einstein proved…. [NEET 2000]
The energy of most energetic photoelectron is…. [JIPMER 1999]
When monochromatic radiation of intensity II falls on a metal surface…. [NEET 2010]
An example for the best source of monochromatic light is… [JKCET 2019]
The wavelength of a 1 keV photon is 1.24 nm… [KEAM]
In a photoelectric experiment, if both the intensity and frequency of the incident… [KEAM]
If the kinetic energy of the particle is increased to 16 times its previous value… [KEAM]

Sample Questions

Ques. Explain the meaning of Compton along with the illustration of the Compton effect through its equation. (3 marks)

Ans. The scattering of the photons due to the presence of the charged particles in the form of the electrons and positron is termed, Compton.

The Compton effect can be expressed as the following using the given equation.

λ1 - λ0 = h / mc (1 - Cos Θ)

Here,

λ1 - λ0 refers to the difference between the wavelength

m refers to the rest mass

c refers to the speed of light, i.e., 3 x 10? m / s

h refers to the Planck’s constant that is, 6. 626 x 10?³? J / s

Ques. Explain the relation between the electrons and photons in terms of Compton wavelength. (3 marks)

Ans. In simple words, when we consider the electron Compton wavelength, the wavelength of the electromagnetic wave is taken into consideration in relation to the energy of the photon. Another remarkable feature of this concept says that the photon energy must match the rest energy of the electron to make this phenomenon work.

However, the occurrence of the electron Compton wavelength is seen only after the annihilation of the electron and its antiparticle, positron

Ques. What is the concept of annihilation in relation to the Compton wavelength? (3 marks)

Ans. Annihilation, here, refers to the complete obliteration of the object which takes place due to the collision of electron and positron, followed by the emitting of the gamma rays.

Application

when we consider the electron Compton wavelength, the wavelength of the electromagnetic wave is taken into consideration in relation to the energy of the photon.
Another remarkable feature of this concept says that the photon energy must match the rest energy of the electron to make this phenomenon work.
However, the occurrence of the electron Compton wavelength is seen only after the annihilation of the electron and its antiparticle, positron

Ques. Give a brief explanation of the emergence of the concept of Compton scattering. (3 marks)

Ans. An experiment of 1912 that was conducted by Max Von Laue involved the explanation that the prevailing X - rays always perform diffraction but this diffraction takes place in the light only. Therefore, he believed that in such a case, the X - rays shall be compared to the light which has a wave nature.

However, even before, in 1905, Einstein claimed the light to be a bunch of small packets or quanta which are known as photons. Therefore, both the x – rays and the light must have a particle nature.

Further, Arthur Holly Compton, in his experiment, named Compton scattering claimed that the interaction of the X - rays to that of the electrons with each other, resembles the two colliding balls. Thus, defining the Compton scattering.

Ques. What is Compton wavelength? (3 marks)

Ans. The Compton wavelength is a property of any particle whose wavelength is similar to that of the photon. However, the energy of the photon must match the mass of the corresponding particle. Without this additional condition, the function of Compton wavelength might not work. The derivation of the Compton effect for any of the particles is simply derived using the Compton wavelength of the electrons.

While considering any particle, the value of the Compton wavelength for it shall remain unchanged. Although, this value of Compton wavelength varies as per different particles.

Ques. Explain Compton wavelength. Give the values of Compton wavelength. (3 marks)

Ans. The Compton wavelength is a property of any particle whose wavelength is similar to that of the photon. However, the energy of the photon must match the mass of the corresponding particle.

The value of the Compton wavelength of an electron as per the CODATA 2014 is given below

Particulars	Details
Compton wavelength λ	2.4263102367(11) × 10^-12 meter
Compton wavelength λ	0.0242 angstrom

Ques. What is Compton? Give its equation. (3 marks)

Ans. Compton wavelength is used to express the relation between the wavelength of a particle and the wavelength of the photon of the same mass, the equation of the Compton wavelength is expressed as follows.

λ = h / mc

Now, here,

λ refers to the value of the Compton wavelength of a particle
h refers to the Planck constant.
m refers to the value of the mass of the particle
Whereas, c symbolizes the speed of light in a vacuum

Ques. What is Compton wavelength? (3 marks)

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Compton Wavelength: Equation and Compton Scattering

What is Compton Wavelength?

Compton Wavelength Equation

Compton Wavelength Example

What is Compton Scattering?

What is Compton Effect?

Things to Remember

Previous Year Questions

Sample Questions

CBSE CLASS XII Related Questions

1.
Two thin lenses of focal length \( f_1 \) and \( f_2 \) are placed in contact with each other coaxially. Prove that the focal length \( f \) of the combination is given by \[ f = \frac{f_1 f_2}{f_1 + f_2}. \]

2.
A tank is filled with a liquid to a height of \( 12.5 \, \text{m} \). The apparent depth of a needle lying at the bottom of the tank is measured to be \( 9.0 \, \text{m} \). Calculate the speed of light in the liquid.

3.
Suppose a pure Si crystal has \( 5 \times 10^{28} \) atoms per \( \text{m}^3 \). It is doped with \( 5 \times 10^{22} \) atoms per \( \text{m}^3 \) of Arsenic. Calculate majority and minority carrier concentration in the doped silicon. (Given: \( n_i = 1.5 \times 10^{16} \, \text{m}^{-3} \))

4.
The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:

6.
Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.

Comments

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Compton Wavelength: Equation and Compton Scattering

What is Compton Wavelength?

Compton Wavelength Equation

Compton Wavelength Example

What is Compton Scattering?

What is Compton Effect?

Things to Remember

Previous Year Questions

Sample Questions

CBSE CLASS XII Related Questions

1.Two thin lenses of focal length \( f_1 \) and \( f_2 \) are placed in contact with each other coaxially. Prove that the focal length \( f \) of the combination is given by \[ f = \frac{f_1 f_2}{f_1 + f_2}. \]

2.A tank is filled with a liquid to a height of \( 12.5 \, \text{m} \). The apparent depth of a needle lying at the bottom of the tank is measured to be \( 9.0 \, \text{m} \). Calculate the speed of light in the liquid.

3.Suppose a pure Si crystal has \( 5 \times 10^{28} \) atoms per \( \text{m}^3 \). It is doped with \( 5 \times 10^{22} \) atoms per \( \text{m}^3 \) of Arsenic. Calculate majority and minority carrier concentration in the doped silicon. (Given: \( n_i = 1.5 \times 10^{16} \, \text{m}^{-3} \))

4.The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:

6.Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Two thin lenses of focal length \( f_1 \) and \( f_2 \) are placed in contact with each other coaxially. Prove that the focal length \( f \) of the combination is given by \[ f = \frac{f_1 f_2}{f_1 + f_2}. \]

2.
A tank is filled with a liquid to a height of \( 12.5 \, \text{m} \). The apparent depth of a needle lying at the bottom of the tank is measured to be \( 9.0 \, \text{m} \). Calculate the speed of light in the liquid.

3.
Suppose a pure Si crystal has \( 5 \times 10^{28} \) atoms per \( \text{m}^3 \). It is doped with \( 5 \times 10^{22} \) atoms per \( \text{m}^3 \) of Arsenic. Calculate majority and minority carrier concentration in the doped silicon. (Given: \( n_i = 1.5 \times 10^{16} \, \text{m}^{-3} \))

4.
The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:

6.
Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.