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Compton Scattering (Compton Effect)

Compton scattering is a fundamental process in physics that describes the interaction between a photon and an electron . This phenomenon occurs when a high-energy photon collides with an electron, resulting in the photon losing some of its energy. When a photon interacts with an electron, it transfers some of its energy to the electron. This transfer causes the photon to change its wavelength and direction, resulting in scattering.

Compton scattering played a crucial role in confirming the particle nature of light. This process was first explained by American physicist Arthur Holly Compton in 1923, who demonstrated that light behaves as both particles (photons) and waves.

Energy and momentum must be conserved when a photon collides with an electron. The Compton scattering equation relates the photon’s wavelength change to the scattering angle and the electron’s rest mass.

The equation is written as:

Δλ is the change in wavelength of the photon

λ’ is the wavelength of the scattered photon

λ is the wavelength of the incident photon

h is Planck’s constant

m e is the rest mass of the electron

c is the speed of light

The scattering angle θ represents the angle between the incident and the scattered photon’s direction. This equation demonstrates that a shift in wavelength occurs due to the interaction between the photon and electron.

The derivation of the Compton scattering formula involves applying conservation laws and principles of relativistic mechanics. Let us look at the image below and understand the various quantities involved.

p 1 and E are the photon’s initial momentum and energy, respectively.

p 2 and E’ are the photon’s final momentum and energy, respectively.

m e and p 3 are the electron’s rest mass and final momentum, respectively.

θ is the angle by which the photon scatters.

φ is the angle by which the electron recoils.

We assume that the photon initially moves along the x-direction so that only the x and the y components of the momentum will be considered in our derivation.

Conservation of momentum along the x- and y- directions gives rise to the following equations.

p 1 = p 2 cos θ + p 3 cos φ (x-direction)

p 2 sin θ = p 3 sin φ (y-direction)

According to the conservation of energy , the total energy before the interaction is given by E + m e c 2 . After the scattering, the photon loses energy and momentum while the electron gains the same. Conservation of energy gives us the following relativistic equation.

We aim to solve for E’ in terms of cos θ and eliminate φ in the process. Continuing from the energy equation, we get

Squaring both sides

We need to replace p 3 and use the momentum conservation equations.

(p 3 cos φ) 2 = (p 1 – p 2 cos θ) 2

(p 3 sin φ) 2 = (p 2 sin θ) 2

Adding the two equations, we get

p 3 2 = p 1 2 + p 2 2 – 2 p 1 p 2 cos θ

Back to the energy equation

Using the photon-energy equation E = pc, we can rewrite the energy equation as

Using the relationship between energy and momentum for a photon, E = hc/λ, where h is Planck’s constant and λ is the wavelength, we can rewrite the equation as

Which is the Compton wavelength shift equation for the photon.

The minimum change in wavelength occurs when θ = 0° or cos θ = 1, resulting in Δλ = 0. The maximum change in wavelength occurs when θ = 180° or cos θ = -1, resulting in Δλ = 2h/m e c. In this case, the photon transfers to the electron as much momentum as possible. The value of this maximum change is

The quantity h/m e c is known as the Compton wavelength of the electron and is equal to 2.43 x 10 -12 m.

Applications

Compton scattering finds applications in various fields, such as astronomy, medical imaging, and nuclear physics. It plays a crucial role in understanding X-ray diffraction and gamma-ray astronomy. It helps scientists study the properties of matter, understand the behavior of radiation, and gather information about the structure and composition of materials.

Inverse Compton Scattering

When the electron is not at rest but has an energy greater than the typical photon energy, there can be a transfer of energy from the electron to the photon. This process is called inverse Compton scattering to distinguish it from direct Compton scattering, in which the electron is at rest, and the photon gives some energy to the electron. This energy exchange results in the emission of a new photon with a higher energy than the incoming photon. This phenomenon is the reverse of the Compton scattering process. It is applied in nuclear physics and astrophysics.

Compton Effect vs. Photoelectric Effect

Both Compton and photoelectric effects obey the conservation of momentum and energy. Here is a table highlighting the key differences between the Compton effect and the photoelectric effect :



	Scattering of photons by electrons	Ejection of electrons by photons
	Involves the interaction of X-rays or gamma rays with electrons	Involves the interaction of photons (typically ultraviolet or ) with electrons
	Results in the scattering of incoming photons with a change in their energy (wavelength)	Leads to the emission of electrons from a material surface when photons are absorbed
	The change in photon energy is proportional to the incident photon’s energy and the scattering angle.	The photoelectric effect is most efficient when incident photons have energy above a material-specific threshold, known as the work function.
	Important in astrophysics for understanding gamma-ray production and X-ray sources	Essential in the development of photodetectors, solar cells, and photomultiplier tubes

Compton Scattering – Hyperphysics.phy-astr.gsu.edu
Compton Effect – Eng.libretexts.org
Compton Effect – Wanda.fiu.edu
What the Compton Effect Is and How It Works in Physics – Thoughtco.com
Compton Scattering – Nuclear-power.com

Article was last reviewed on Friday, September 29, 2023

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6.3 The Compton Effect

Learning objectives.

By the end of this section, you will be able to:

Describe Compton’s experiment
Explain the Compton wavelength shift
Describe how experiments with X-rays confirm the particle nature of radiation

Two of Einstein’s influential ideas introduced in 1905 were the theory of special relativity and the concept of a light quantum, which we now call a photon. Beyond 1905, Einstein went further to suggest that freely propagating electromagnetic waves consisted of photons that are particles of light in the same sense that electrons or other massive particles are particles of matter. A beam of monochromatic light of wavelength λ λ (or equivalently, of frequency f ) can be seen either as a classical wave or as a collection of photons that travel in a vacuum with one speed, c (the speed of light), and all carrying the same energy, E f = h f . E f = h f . This idea proved useful for explaining the interactions of light with particles of matter.

Momentum of a Photon

Unlike a particle of matter that is characterized by its rest mass m 0 , m 0 , a photon is massless. In a vacuum, unlike a particle of matter that may vary its speed but cannot reach the speed of light, a photon travels at only one speed, which is exactly the speed of light. From the point of view of Newtonian classical mechanics, these two characteristics imply that a photon should not exist at all. For example, how can we find the linear momentum or kinetic energy of a body whose mass is zero? This apparent paradox vanishes if we describe a photon as a relativistic particle. According to the theory of special relativity, any particle in nature obeys the relativistic energy equation

This relation can also be applied to a photon. In Equation 6.17 , E is the total energy of a particle, p is its linear momentum, and m 0 m 0 is its rest mass. For a photon, we simply set m 0 = 0 m 0 = 0 in this equation. This leads to the expression for the momentum p f p f of a photon

Here the photon’s energy E f E f is the same as that of a light quantum of frequency f , which we introduced to explain the photoelectric effect:

The wave relation that connects frequency f with wavelength λ λ and speed c also holds for photons:

Therefore, a photon can be equivalently characterized by either its energy and wavelength, or its frequency and momentum. Equation 6.19 and Equation 6.20 can be combined into the explicit relation between a photon’s momentum and its wavelength:

Notice that this equation gives us only the magnitude of the photon’s momentum and contains no information about the direction in which the photon is moving. To include the direction, it is customary to write the photon’s momentum as a vector:

In Equation 6.22 , ℏ = h / 2 π ℏ = h / 2 π is the reduced Planck’s constant (pronounced “h-bar”), which is just Planck’s constant divided by the factor 2 π . 2 π . Vector k → k → is called the “wave vector” or propagation vector (the direction in which a photon is moving). The propagation vector shows the direction of the photon’s linear momentum vector. The magnitude of the wave vector is k = | k → | = 2 π / λ k = | k → | = 2 π / λ and is called the wave number . Notice that this equation does not introduce any new physics. We can verify that the magnitude of the vector in Equation 6.22 is the same as that given by Equation 6.18 .

The Compton Effect

The Compton effect is the term used for an unusual result observed when X-rays are scattered on some materials. By classical theory, when an electromagnetic wave is scattered off atoms, the wavelength of the scattered radiation is expected to be the same as the wavelength of the incident radiation. Contrary to this prediction of classical physics, observations show that when X-rays are scattered off some materials, such as graphite, the scattered X-rays have different wavelengths from the wavelength of the incident X-rays. This classically unexplainable phenomenon was studied experimentally by Arthur H. Compton and his collaborators, and Compton gave its explanation in 1923.

To explain the shift in wavelengths measured in the experiment, Compton used Einstein’s idea of light as a particle. The Compton effect has a very important place in the history of physics because it shows that electromagnetic radiation cannot be explained as a purely wave phenomenon. The explanation of the Compton effect gave a convincing argument to the physics community that electromagnetic waves can indeed behave like a stream of photons, which placed the concept of a photon on firm ground.

The schematics of Compton’s experimental setup are shown in Figure 6.11 . The idea of the experiment is straightforward: Monochromatic X-rays with wavelength λ λ are incident on a sample of graphite (the “target”), where they interact with atoms inside the sample; they later emerge as scattered X-rays with wavelength λ ′ . λ ′ . A detector placed behind the target can measure the intensity of radiation scattered in any direction θ θ with respect to the direction of the incident X-ray beam. This scattering angle , θ , θ , is the angle between the direction of the scattered beam and the direction of the incident beam. In this experiment, we know the intensity and the wavelength λ λ of the incoming (incident) beam; and for a given scattering angle θ , θ , we measure the intensity and the wavelength λ ′ λ ′ of the outgoing (scattered) beam. Typical results of these measurements are shown in Figure 6.12 , where the x -axis is the wavelength of the scattered X-rays and the y -axis is the intensity of the scattered X-rays, measured for different scattering angles (indicated on the graphs). For all scattering angles (except for θ = 0 ° ), θ = 0 ° ), we measure two intensity peaks. One peak is located at the wavelength λ , λ , which is the wavelength of the incident beam. The other peak is located at some other wavelength, λ ′ . λ ′ . The two peaks are separated by Δ λ , Δ λ , which depends on the scattering angle θ θ of the outgoing beam (in the direction of observation). The separation Δ λ Δ λ is called the Compton shift .

Compton Shift

As given by Compton, the explanation of the Compton shift is that in the target material, graphite, valence electrons are loosely bound in the atoms and behave like free electrons. Compton assumed that the incident X-ray radiation is a stream of photons. An incoming photon in this stream collides with a valence electron in the graphite target. In the course of this collision, the incoming photon transfers some part of its energy and momentum to the target electron and leaves the scene as a scattered photon. This model explains in qualitative terms why the scattered radiation has a longer wavelength than the incident radiation. Put simply, a photon that has lost some of its energy emerges as a photon with a lower frequency, or equivalently, with a longer wavelength. To show that his model was correct, Compton used it to derive the expression for the Compton shift. In his derivation, he assumed that both photon and electron are relativistic particles and that the collision obeys two commonsense principles: (1) the conservation of linear momentum and (2) the conservation of total relativistic energy.

In the following derivation of the Compton shift, E f E f and p → f p → f denote the energy and momentum, respectively, of an incident photon with frequency f . The photon collides with a relativistic electron at rest, which means that immediately before the collision, the electron’s energy is entirely its rest mass energy, m 0 c 2 . m 0 c 2 . Immediately after the collision, the electron has energy E and momentum p → , p → , both of which satisfy Equation 6.19 . Immediately after the collision, the outgoing photon has energy E ˜ f , E ˜ f , momentum p ˜ → f , p ˜ → f , and frequency f ′ . f ′ . The direction of the incident photon is horizontal from left to right, and the direction of the outgoing photon is at the angle θ , θ , as illustrated in Figure 6.11 . The scattering angle θ θ is the angle between the momentum vectors p → f p → f and p ˜ → f , p ˜ → f , and we can write their scalar product:

Following Compton’s argument, we assume that the colliding photon and electron form an isolated system. This assumption is valid for weakly bound electrons that, to a good approximation, can be treated as free particles. Our first equation is the conservation of energy for the photon-electron system:

The left side of this equation is the energy of the system at the instant immediately before the collision, and the right side of the equation is the energy of the system at the instant immediately after the collision. Our second equation is the conservation of linear momentum for the photon–electron system where the electron is at rest at the instant immediately before the collision:

The left side of this equation is the momentum of the system right before the collision, and the right side of the equation is the momentum of the system right after collision. The entire physics of Compton scattering is contained in these three preceding equations––the remaining part is algebra. At this point, we could jump to the concluding formula for the Compton shift, but it is beneficial to highlight the main algebraic steps that lead to Compton’s formula, which we give here as follows.

We start with rearranging the terms in Equation 6.24 and squaring it:

In the next step, we substitute Equation 6.19 for E 2 , E 2 , simplify, and divide both sides by c 2 c 2 to obtain

Now we can use Equation 6.21 to express this form of the energy equation in terms of momenta. The result is

To eliminate p 2 , p 2 , we turn to the momentum equation Equation 6.25 , rearrange its terms, and square it to obtain

The product of the momentum vectors is given by Equation 6.23 . When we substitute this result for p 2 p 2 in Equation 6.26 , we obtain the energy equation that contains the scattering angle θ : θ :

With further algebra, this result can be simplified to

Now recall Equation 6.21 and write: 1 / p ˜ f = λ ′ / h 1 / p ˜ f = λ ′ / h and 1 / p f = λ / h . 1 / p f = λ / h . When these relations are substituted into Equation 6.27 , we obtain the relation for the Compton shift:

The factor h / m 0 c h / m 0 c is called the Compton wavelength of the electron:

Denoting the shift as Δ λ = λ ′ − λ , Δ λ = λ ′ − λ , the concluding result can be rewritten as

This formula for the Compton shift describes outstandingly well the experimental results shown in Figure 6.12 . Scattering data measured for molybdenum, graphite, calcite, and many other target materials are in accord with this theoretical result. The nonshifted peak shown in Figure 6.12 is due to photon collisions with tightly bound inner electrons in the target material. Photons that collide with the inner electrons of the target atoms in fact collide with the entire atom. In this extreme case, the rest mass in Equation 6.29 must be changed to the rest mass of the atom. This type of shift is four orders of magnitude smaller than the shift caused by collisions with electrons and is so small that it can be neglected.

Compton scattering is an example of inelastic scattering , in which the scattered radiation has a longer wavelength than the wavelength of the incident radiation. In today’s usage, the term “Compton scattering” is used for the inelastic scattering of photons by free, charged particles. In Compton scattering, treating photons as particles with momenta that can be transferred to charged particles provides the theoretical background to explain the wavelength shifts measured in experiments; this is the evidence that radiation consists of photons.

Example 6.8

Compton scattering.

This gives the scattered wavelength:

The largest shift is

Significance

Check your understanding 6.8.

An incident 71-pm X-ray is incident on a calcite target. Find the wavelength of the X-ray scattered at a 60 ° 60 ° angle. What is the smallest shift that can be expected in this experiment?

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Publication date: Sep 29, 2016
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Photons and Matter Waves

The Compton Effect

Samuel J. Ling; Jeff Sanny; and William Moebs

Learning Objectives

By the end of this section, you will be able to:

Describe Compton’s experiment
Explain the Compton wavelength shift
Describe how experiments with X-rays confirm the particle nature of radiation

$\lambda$

Momentum of a Photon

${m}_{0},$

Therefore, a photon can be equivalently characterized by either its energy and wavelength, or its frequency and momentum. (Figure) and (Figure) can be combined into the explicit relation between a photon’s momentum and its wavelength:

${p}_{f}=\frac{h}{\lambda }.$

${\stackrel{\to }{p}}_{f}=\hslash \stackrel{\to }{k}.$

$\lambda \prime .$

Compton Shift

${\stackrel{\to }{p}}_{f}$

${E}_{f}+{m}_{0}{c}^{2}={\stackrel{˜}{E}}_{f}+E.$

${\stackrel{\to }{p}}_{f}={\stackrel{\to }{\stackrel{˜}{p}}}_{f}+\stackrel{\to }{p}.$

We start with rearranging the terms in (Figure) and squaring it:

${\left[\left({E}_{f}-{\stackrel{˜}{E}}_{f}\right)+{m}_{0}{c}^{2}\right]}^{2}={E}^{2}.$

Now we can use (Figure) to express this form of the energy equation in terms of momenta. The result is

${\left({p}_{f}-{\stackrel{˜}{p}}_{f}\right)}^{2}+2{m}_{0}c\left({p}_{f}-{\stackrel{˜}{p}}_{f}\right)={p}^{2}.$

With further algebra, this result can be simplified to

$\frac{1}{{\stackrel{˜}{p}}_{f}}-\frac{1}{{p}_{f}}=\frac{1}{{m}_{0}c}\left(1-\text{cos}\theta \right).$

This formula for the Compton shift describes outstandingly well the experimental results shown in (Figure) . Scattering data measured for molybdenum, graphite, calcite, and many other target materials are in accord with this theoretical result. The nonshifted peak shown in (Figure) is due to photon collisions with tightly bound inner electrons in the target material. Photons that collide with the inner electrons of the target atoms in fact collide with the entire atom. In this extreme case, the rest mass in (Figure) must be changed to the rest mass of the atom. This type of shift is four orders of magnitude smaller than the shift caused by collisions with electrons and is so small that it can be neglected.

Compton Scattering

$30\text{°}$

This gives the scattered wavelength:

$\lambda \prime =\lambda +\text{Δ}\lambda =\left(71+0.325\right)\phantom{\rule{0.2em}{0ex}}\text{pm}=71.325\phantom{\rule{0.2em}{0ex}}\text{pm}.$

The largest shift is

${\left(\text{Δ}\lambda \right)}_{\text{max}}={\lambda }_{c}\left(1-\text{cos}{180}^{0}\right)=2\left(2.43\phantom{\rule{0.2em}{0ex}}\text{pm}\right)=4.86\phantom{\rule{0.2em}{0ex}}\text{pm}.$

Significance

The largest shift in wavelength is detected for the backscattered radiation; however, most of the photons from the incident beam pass through the target and only a small fraction of photons gets backscattered (typically, less than 5%). Therefore, these measurements require highly sensitive detectors.

$60\text{°}$

In the Compton effect, X-rays scattered off some materials have different wavelengths than the wavelength of the incident X-rays. This phenomenon does not have a classical explanation.
The Compton effect is explained by assuming that radiation consists of photons that collide with weakly bound electrons in the target material. Both electron and photon are treated as relativistic particles. Conservation laws of the total energy and of momentum are obeyed in collisions.
Treating the photon as a particle with momentum that can be transferred to an electron leads to a theoretical Compton shift that agrees with the wavelength shift measured in the experiment. This provides evidence that radiation consists of photons.
Compton scattering is an inelastic scattering, in which scattered radiation has a longer wavelength than that of incident radiation.

Conceptual Questions

Discuss any similarities and differences between the photoelectric and the Compton effects.

Answers may vary

Which has a greater momentum: an UV photon or an IR photon?

Does changing the intensity of a monochromatic light beam affect the momentum of the individual photons in the beam? Does such a change affect the net momentum of the beam?

Can the Compton effect occur with visible light? If so, will it be detectable?

Is it possible in the Compton experiment to observe scattered X-rays that have a shorter wavelength than the incident X-ray radiation?

Show that the Compton wavelength has the dimension of length.

At what scattering angle is the wavelength shift in the Compton effect equal to the Compton wavelength?

right angle

What is the momentum of a 589-nm yellow photon?

What is the momentum of a 4-cm microwave photon?

$1.66\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\text{−}32}\phantom{\rule{0.2em}{0ex}}\text{kg}·\text{m/s}$

In a beam of white light (wavelengths from 400 to 750 nm), what range of momentum can the photons have?

$3.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-24}\text{kg}·\text{m/s}$

Find the momentum and energy of a 1.0-Å photon.

$6.63\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\text{−}23}\phantom{\rule{0.2em}{0ex}}\text{kg}·\text{m/s};$

82.9 fm; 15 MeV

$2.5\text{-}µ\text{m}$

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Introduction to Quantum Physics

Compton scattering, 1.4. compton scattering #, 1.4.1. introduction #.

Compton scattering shows that the wavelength of an electromagnetic (EM) wave increases when it is scattered from an electron. The only way to explain this is to treat the incident EM was as a particle and apply the conservation of momentum. We will first review the experimental evidence then we will show that the change in the EM waves wavelength is modelled by treating the wave as a particle and applying the conservation of momentum. In Compton’s original experiment x-rays are scattered from a carbon target. The scattered x-rays are then passed into a Bragg spectrometer, on the left of figure 1.9 , where the wavelength of the scattered x-ray is measured.

Fig. 1.9 Schematic diagram of Compton’s experiment. Compton scattering occurs in the graphite (carbon) target on the left. The slit passes X-ray photons scattered at a selected angle. The wave length of a scattered photon is measured using Bragg scattering in the crystal on the right in conjunction with ionization chamber. #

1.4.2. Experimental Results #

The data collected in the experiment showed that as the angle between the incident and scattered wave increases the wavelength of the scattered x-ray also increased. This is shown in figure 1.10 for scattering angles $0^\circ$ , $45^\circ$ , $90^\circ$ and $135^\circ$ .

A schematic representation of the Compton scattering data recorded by Arthur Compton. A peak is observed at 70 pico meters for 0 degrees scattering, as the 45 degrees scattering the peak separates into two components, one that remains at about 70 picometers and one about 72 picometers. As the scattering angle increases to 90 and then 135 degrees the separation between the first and second peak increases.

Fig. 1.10 A schematic representation of the Compton scattering data recorded by Arthur Compton. #

At $0^\circ$ only one peak is seen, with increased scattering angle this separates into two distinct peaks. The right peak is the Compton scattered x-ray from a loosely bound electron, it behaves similar to a free electron which is illustrated in figure 1.11 . It is clear that the wavelength after scattering is larger than the wavelength of the incident x-rays.

Fig. 1.11 A schematic representation of Compton scattering. An x-ray is incident from the left of the figure on a stationary electron. The x-ray Compton scatters from the stationary electron, the process is inelastic and some energy is transferred from the x-ray to the electron. Hence, the energy of the electron increases and the energy of the x-ray photon decreases. #

After scattering from a stationary electron the energy of an x-ray photon decreases. Hence, for an incident x-ray photon with wavelength $\lambda$ the scattered x-ray photon will have longer wavelength denoted by $\lambda^\prime$ .

1.4.3. Theory #

The Compton scattering formula is

Where $\lambda^\prime$ is the wavelength of the scattered x ray photon, $\lambda$ is the wavelength of the incident x ray photon, $h$ is Planck’s constant, $m$ is the mass of the scattered electron (note this can in practise be any particle but in this case it is an electron), $c$ is the speed of light, and $\theta$ is the scattering angle between the incident and the scattered photon.

Equation (1.4) is further evidence of the particle nature of EM radiation. It can be derived from the conservation of linear momentum (with a little application of relativity to account for the rest mass energy).

The left peak observed in each plot in figure 1.10 is the scattering from more tightly bound electrons. In contrast to the loosely bound electrons which scatter out of the carbon behaving similar to a free electron the more tightly bound electron remain bound to their carbon atom. In this case the mass $m$ in equation (1.4) is the mass of the atom which recoils hence the shift in the observed wavelength is $10^{-5}$ small due to the increased mass.

../_images/compton_scattering_vectors.png

Fig. 1.12 The vector representation of electron and photons momentum in a Compton scattering event. The incident photon has momentum $p_{\gamma}$ , the scattered photon has momentum $p_{\gamma^\prime}$ , and the scattered electron has momentum $p_e$ . Note that the initial momentum of the electron is zero hence does not contribute. #

To derive equation (1.4) start by considering the conservation of momentum. From the diagram in figure 1.12 the vector addition gives

Where the last step was achieved using the cosine rule and the $p$ ’s are now the magnitudes of the vectors rather than vectors.

Next we consider the conservation of energy. To perform this step we need to consider the possibility that the electron will recoil at a very large velocity. Instead of simply considering the kinetic energy we will consider the full relativistic energy given by

where $E$ is the total energy, $p$ is the momentum, $m$ is the mass (in relativity you will later learn this is called the rest mass, for now just think of this as the mass and use the values you are familiar with, e.g. $9.11\times10^{-31}$ kg for an electron) and of course $c$ is the speed of light. Before the x-ray photon scatters from the electron the momentum is zero so the equation reduces to $E=mc^2$ , which you should be familiar with from your A-level physics course, after scattering it will have gained some momentum and we will denote the energy as $E_e$ . The energy of the photon before scattering we will give as $E_\gamma$ and after as $E_{\gamma\prime}$ . The conservation of energy gives

Which we can interpret as the sum of the energies before the scattering is equal to the sum of the energies after the scattering. Rearranging the equation for energy (1.6) to make the momentum of the electron the subject gives.

Substitute (1.7) into (1.8) we obtain

Multiply equation (1.5) by $c^2$ and equating to equation (1.9) gives

Using the relationship between a photons energy and momentum $E=pc$ we can simplify the above equation to

Equation (1.4) , the Compton scattering equation, can be obtained from equation (1.10) by substitution of $E=\frac{hc}{\lambda}$ .

I will leave is as an exercise for you to determine a similar relationship for the scattering angle of the electron relative to the scatteing angle of the photon.

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Compton effect

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Michigan State University - Department of Physics and Astronomy - The Compton Effect
Engineering LibreTexts - Compton Effect
National Center for Biotechnology Information - PubMed Central - Characterization of Compton-scatter imaging with an analytical simulation method
Portland State University - X-Rays and the Compton Effect
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University fo Central Florida Pressbooks - University Physics Volume 3 - The Compton Effect

Compton effect , increase in wavelength of X-rays and other energetic electromagnetic radiations that have been elastically scattered by electrons ; it is a principal way in which radiant energy is absorbed in matter. The effect has proved to be one of the cornerstones of quantum mechanics , which accounts for both wave and particle properties of radiation as well as of matter. See also light: Early particle and wave theories .

The American physicist Arthur Holly Compton explained (1922; published 1923) the wavelength increase by considering X-rays as composed of discrete pulses, or quanta , of electromagnetic energy . The American chemist Gilbert Lewis later coined the term photon for light quanta. Photons have energy and momentum just as material particles do; they also have wave characteristics, such as wavelength and frequency. The energy of photons is directly proportional to their frequency and inversely proportional to their wavelength, so lower-energy photons have lower frequencies and longer wavelengths. In the Compton effect, individual photons collide with single electrons that are free or quite loosely bound in the atoms of matter. Colliding photons transfer some of their energy and momentum to the electrons, which in turn recoil. In the instant of the collision , new photons of less energy and momentum are produced that scatter at angles the size of which depends on the amount of energy lost to the recoiling electrons.

Italian-born physicist Dr. Enrico Fermi draws a diagram at a blackboard with mathematical equations. circa 1950.

Because of the relation between energy and wavelength, the scattered photons have a longer wavelength that also depends on the size of the angle through which the X-rays were diverted. The increase in wavelength, or Compton shift, does not depend on the wavelength of the incident photon. The wavelength of the photon after (λ′) and before (λ) the scattering event differs by λ′ − λ = ( h / m c )(1 − cos θ). Here h is Planck’s constant , m is the rest mass of the electron , c is the speed of light , and θ is the angle through which the photon is scattered.

The Compton effect was discovered independently by the Dutch physical chemist Peter Debye in early 1923.

What the Compton Effect Is and How It Works in Physics

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M.S., Mathematics Education, Indiana University
B.A., Physics, Wabash College

The Compton effect (also called Compton scattering) is the result of a high-energy photon colliding with a target, which releases loosely bound electrons from the outer shell of the atom or molecule. The scattered radiation experiences a wavelength shift that cannot be explained in terms of classical wave theory, thus lending support to Einstein's photon theory. Probably the most important implication of the effect is that it showed light could not be fully explained according to wave phenomena. Compton scattering is one example of a type of inelastic scattering of light by a charged particle. Nuclear scattering also occurs, although the Compton effect typically refers to the interaction with electrons.

The effect was first demonstrated in 1923 by Arthur Holly Compton (for which he received a 1927 Nobel Prize in Physics). Compton's graduate student, Y.H. Woo, later verified the effect.

How Compton Scattering Works

The scattering is demonstrated is pictured in the diagram. A high-energy photon (generally X-ray or gamma-ray ) collides with a target, which has loosely-bound electrons in its outer shell. The incident photon has the following energy E and linear momentum p :

E = hc / lambda p = E / c

The photon gives part of its energy to one of the almost-free electrons, in the form of kinetic energy , as expected in a particle collision. We know that total energy and linear momentum must be conserved. Analyzing these energy and momentum relationships for the photon and electron, you end up with three equations:

x -component momentum
y -component momentum

... in four variables:

phi , the scattering angle of the electron
theta , the scattering angle of the photon
E e , the final energy of the electron
E ', the final energy of the photon

If we care only about the energy and direction of the photon, then the electron variables can be treated as constants, meaning that it's possible to solve the system of equations. By combining these equations and using some algebraic tricks to eliminate variables, Compton arrived at the following equations (which are obviously related, since energy and wavelength are related to photons):

1 / E ' - 1 / E = 1 /( m e c 2 ) * (1 - cos theta ) lambda ' - lambda = h /( m e c ) * (1 - cos theta )

The value h /( m e c ) is called the Compton wavelength of the electron and has a value of 0.002426 nm (or 2.426 x 10 -12 m). This isn't, of course, an actual wavelength, but really a proportionality constant for the wavelength shift.

Why Does This Support Photons?

This analysis and derivation are based on a particle perspective and the results are easy to test. Looking at the equation, it becomes clear that the entire shift can be measured purely in terms of the angle at which the photon gets scattered. Everything else on the right side of the equation is a constant. Experiments show that this is the case, giving great support to the photon interpretation of light.

Edited by Anne Marie Helmenstine, Ph.D.

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Research Highlight
Published: 28 March 2023

In retrospect

A century of Compton scattering

Kirsty E. McGhee 1

Nature Reviews Physics volume 5 , page 322 ( 2023 ) Cite this article

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Quantum physics

Wave–particle duality is a fundamental tenet of quantum physics that physicists took a while to come to terms with. The wave nature of light had been long accepted, but it was only at the turn of the 20th century that a possible particle interpretation started to be considered. In 1923, Arthur Compton published the results of an X-ray scattering experiment that could only be explained by considering light as discrete particles. He later recalled, “when I presented my results at a meeting of the American Physical Society, it initiated the most hotly contested scientific controversy that I have ever known.” By the late 1920s, wave–particle duality was accepted and, in 1927, Compton won the Nobel Prize for his work.

Today, Compton scattering is a useful tool for physicists. In condensed matter, Compton scattering can be used to probe the electronic states in ferro- and ferrimagnetic materials. When the electromagnetic field of circularly polarized, high-energy rays interacts with the magnetic moment of the electrons, the rays are Compton scattered. If the crystal magnetization is then reversed and the scattered rays measured once more, one can determine the momentum distribution of the spin-up and spin-down electrons. In astrophysics, inverse Compton scattering is used to study black holes and galaxy clusters. In this case, low-energy rays scatter from high-energy electrons to produce high-energy X-rays or gamma rays. The energy shift can then be used to build up a picture of the astronomical bodies that surround us. For example, cosmic microwave background (CMB) rays are inverse-Compton-scattered by galaxy clusters and this shift in the CMB spectrum can be used to map the presence of such galaxies. Contrary to the original effect, in inverse Compton scattering, it is the electrons that lose energy as opposed to the photons.

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Original article

Compton, A. H. A Quantum theory of the scattering of X-rays by light elements. Phys. Rev. 21 , 483–502 (1923)

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McGhee, K.E. A century of Compton scattering. Nat Rev Phys 5 , 322 (2023). https://doi.org/10.1038/s42254-023-00576-1

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Derivation of Physics Formula
Derivation Of Compton Effect

Derivation of Compton Effect

What is Compton Effect?

The Compton effect is defined as the effect that is observed when x-rays or gamma rays are scattered on a material with an increase in wavelength. Arthur Compton studied this effect in the year 1922. During the study, Compton found that wavelength is not dependent on the intensity of incident radiation. It is dependent on the angle of scattering and on the wavelength of the incident beam. It is given in the following mathematical form:

Ө: the angle at which radiation scattered

m 0 : rest mass of an electron

h/m 0 c : Compton wavelength of the electron

λ s and λ 0 : radiation spectrum peaks.

Derivation of Compton effect equation

Considering the elastic collide between a photon and an electron, the following is the derivation:

hν 0 : energy of photon

Therefore, above is the Compton effect equation and $\begin{array}{l}\frac{h}{m_{0}c}\equiv \lambda _{c}\end{array} $ is Compton wavelength of an electron.

Difference Between Compton Effect and Photoelectric Effect


This is the effect caused by the inelastic scattering of high-energy photons that are bound to free electrons.	This is the effect caused by the weakly bound electrons that are ejected from the surface of the material when electromagnetic radiation interacts with the electrons.
Arthur Compton explained the effect.	Albert Einstein explained the effect.
The energy associated with the free electrons is mid-energy.	The energy associated with the electrons is low-energy.
The wavelength of the scattered photon is higher than that of the incident photon.	The wavelength is not observed as the photon disappears after interacting with the electrons.

This is a step by step explanation on the derivation of the Compton effect equation derivation. Stay tuned with BYJU’S to know more other related Physics topics.

Frequently Asked Questions – FAQs

What is an x-ray, what is a gamma-ray, what is meant by radiation, what is meant by the photoelectric effect, who discovered the compton effect, what is meant by the compton effect.

The Compton effect is the enlargement of wavelengths of X-rays and other EM waves that have been scattered by electrons. It is a fundamental way in which radiating energy is absorbed by matter.

What is wavelength?

A wavelength is a length between successive troughs or crests.

Which symbol is used to represent the Compton wavelength?

Compton wavelength is represented by the letter λ (Lambda).

What is a photon?

A photon is a fundamental particle that is the force carrier of electromagnetic force. It is massless and travels at the speed of light 299792458 m/s (in vacuum).

Who explained the photoelectric effect?

Albert Einstein explained the photoelectric effect.

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Lab Experiment 4: Compton Scattering

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Demonstrate how the gamma-ray energy varies following Compton scattering.
Demonstrate the possible range of scattered gamma-ray energies.

Equipment Required:

Theoretical Overview:

Scattered energy as a function of angle

As previously discussed in Experiment 1, in Compton scattering a photon of energy Eγ and momentum Eγ/c scatters inelastically from an electron of mass m 0 c 2 . Some of the energy of the photon is transferred to the electron, and the photon is scattered through an angle θ with reduced energy Eγ′.

The Compton scattering formula can be written as:

This may be rearranged as the equation for a straight line:

In this experiment Eγ is measured as a function of θ for incident gamma rays of 662 keV from a 137 Cs source. Figure 4-1 illustrates the experimental setup, where the detector is positioned at about θ = 40°.

Figure 4-1: Collimated 137 Cs source and NaI detector positioned on the scattering table.

Note: During this experiment you should work in keV, where the mass of the electron is 511 keV/c 2 .

The expected energies are shown in Figure 4-2:

Figure 4-2: The energy of the Compton-scattered photon as a function of scattering angle for an initial photon energy of 662 keV.

Angular distribution

The scattering probability as a function of angle is called the differential cross section, and for Compton scattering it is given theoretically by the Klein-Nishina formula (the units are square meters per steradian, m 2 /sr):

r 0 = 2.82 x 10 -15 m, the classical electron radius, and for 137 Cs

A plot of the differential cross section, dσ/dΩ, is shown in Figure 4-3. The units are barns per steradian, b/sr , where 1 barn = 10 -28 m 2 .

Figure 4-3: The differential cross section of Compton scattering for a photon energy of 662 keV.

Experiment 4 Guide:

1. Use the ProSpect software to connect to the detector. Configure the MCA settings and bias voltage as recommended in Experiment 1. Make sure that the detector is correctly calibrated for energy.

2. Ensure the collimated 137 Cs source fixture is aligned at the 1800 mark and is pointing toward the aluminum scattering pillar. During this experiment, keep the source fixture fixed at this location. Caution: Be sure to be aware of your surroundings. Do not have the source pointed at yourself or any other people working in the laboratory.

3. Place the detector with the collimated detector shielding on the scattering table. Starting with the 00 marking, orientate the detector assembly toward the aluminum scattering pillar. Acquire a spectrum until a noticeable peak is visible. Note: To increase the count rate, you can complete this experiment without the add-on detector collimator slit. For better angular resolution, use the detector collimator slit in the vertical orientation and extend the counting time until good statistics are achieved.

4. Repeat Step 3 , moving the detector assembly to different angles up to 1600 degrees.

5. For each measurement determine the centroid energy of the scattered peak.

6. In a spreadsheet, plot the energy of the scattered peak as a function of angle and compare the results with Figure 4-1.

7. Plot the reciprocal of the scattered gamma-ray energy (1/E′ ) as a function of ( 1 - cos θ ). Use the graph and Equation 4-1 to determine the original gamma-ray energy and the rest mass of the electron. Check that these match the values provided above.

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Lab Experiment 10: Positron Annihilation

By using coincidence counting techniques and the Angular Correlation table, students explore the geometrical behavior of positron annihilation events.

Lab Experiment 11: Mathematical Efficiency Calibration

Mathematical modeling is increasingly used instead of source based efficiency calibration for improvement in cost, flexibility, and safety. In this experiment, students generate efficiency calibrations using the LabSOCS efficiency calibration software and compare against traditional source based calibrations.

Lab Experiment 5: Half-Life Measurement

Students calculate the half-life of a short-lived nuclide using multi-channel scaling acquisition.

Lab Experiment 8: Gamma-Ray Efficiency Calibration

Using both a NaI detector and an HPGe detector, the concept of detection efficiency is explored.

Lab Experiment 12: True Coincidence Summing

Students observe true coincidence summing and quantify the effect on observed count rate using LabSOCS mathematical efficiency software.

Lab Experiment 1: Gamma-Ray Detection with Scintillators

In this introduction to gamma-ray detection, students will identify photoelectric effect, Compton scattering, and pair production in a spectrum and perform an energy calibration using known reference sources.

Lab Experiment 9: Gamma-Ray Coincidence Counting Techniques

Counting with multiple detectors correlated in time can yield incredible information about fundamental nuclear structures. In this experiment, students learn these techniques by acquiring and interpreting time-stamped list mode data for synchronized detectors.

Lab Experiment 2: Counting Statistics and Error Prediction

Students will perform a series of background and gamma-ray measurements with a NaI detector and apply statistical principles to these measurements.

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Students will measure the effective attenuation of a set of materials with varying densities and photon absorption cross sections.

Lab Experiment 6: Signal Processing with Digital Signal with Electronics

Using the built-in Digital Signal Oscilloscope feature of the LYNX II, students observe the effects of changing signal processing parameters using several different acquisition modes.

Lab Experiment 7: High-Resolution Gamma-Ray Spectroscopy with HPGe Detectors

Semiconductor gamma-ray detection is introduced and students compare HPGe resolution to NaI detector resolution.

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The Compton Effect or Compton Scattering in Physics
The Compton effect (also called Compton scattering) is the result of a high-energy photon colliding with a target, which releases loosely bound electrons from the outer shell of the atom or molecule. The scattered radiation experiences a wavelength shift that cannot be explained in terms of classical wave theory, thus lending support to Einstein's photon theory.
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Derivation of Compton Effect

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Who explained the photoelectric effect?

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Lab Experiment 4: Compton Scattering

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