Alexander Fufaev
My name is Alexander FufaeV and here I will explain the following topic:

Compton Scattering: Collision of a Photon with an Electron

Important Formula

Formula: Compton Scattering
What do the formula symbols mean?

Wavelength difference

Unit
Wavelength difference is the difference between the wavelength of the photon before the collision and the wavelength of the photon after the collision.

Compton wavelength

Unit
Compton wavelength characterizes the collision particle. For example, an electron has a Compton wavelength of \( 2.426 \cdot 10^{-12} \mathrm{m} \).

Scattering angle

Unit
Scattering angle is the enclosed angle after the collision, between the momentum of the particle (or the momentum of the photon) and the horizontal x-axis.
Compton scattering (photon-electron scattering)
Table of contents
  1. Important Formula
  2. Collision of a photon with an electron at rest
  3. Electron is in motion before the collision
  4. Exercises with Solutions

In Compton scattering, photons of a certain wavelength \(\lambda\) are scattered by an electron. The scattered photon then has a different wavelength \(\lambda'\). Here we want to derive a formula for the wavelength of the scattered photon.

Collision of a photon with an electron at rest

A photon is scattered by an electron at rest.

Here we assume that the electron is at rest. Its momentum is therefore zero: \( \boldsymbol{P} ~=~ 0 \). If the electron is bound in an atom, then it should be very weakly bound. A photon with momentum \( \boldsymbol{p} \) is scattered at this electron. To investigate this scattering process, we consider energy conservation as well as momentum conservation.

Total momentum before collision:
The total momentum before the collision corresponds only to the momentum of the photon \( \boldsymbol{p} ~+~ \boldsymbol{P} ~=~ \boldsymbol{p}\), since the electron at rest has no momentum \(\boldsymbol{P}\) before the collision.

Total momentum after collision:
After the collision the photon has an unknown momentum \( \boldsymbol{p}' \). The photon collided with the electron, therefore the electron could have got a momentum \( \boldsymbol{P}' \) too.

Conservation of momentum, which states that the total momentum before the collision must be EQUAL to the total momentum after the collision: 1 $$\boldsymbol{p} ~=~ \boldsymbol{p}' ~+~ \boldsymbol{P}'$$

The energy of the photon before the collision is given by: 2 $$W_{\text p} ~=~ h \, f ~=~ \frac{h \, c}{\lambda}$$

Here \( \lambda \) is the wavelength of the photon before the collision. We assume the wavelength in the experiment as known because we choose it ourselves.

Total energy before collision:
What about the energy of the electron before the collision? In any case it is NOT zero, although the resting state of the electron may suggest that... According to special relativity, the electron - even at rest - has an energy; a so-called rest energy: 3 $$W_{\text e} ~=~ m_{e} \, c^2$$

Here \( m_{e} \) is the rest mass of the electron with the value: \( m_{e} ~=~ 9.1 ~\cdot~ 10^{-31} \, \mathrm{kg} \). The total energy before the collision is thus: 4 $$W_{\text p} ~+~ W_{\text e} ~=~ \frac{h \, c}{\lambda} ~+~ m_{e} \, c^2$$

Total energy after collision:
After the collision, the wavelength \( \lambda \) of the photon may have changed. We refer to the new wavelength of the photon as \( \lambda' \). A changed wavelength means a changed energy of the photon: 5 $$W_{\text p}' ~=~ \frac{h \, c}{\lambda'}$$

The electron has also changed its energy due to the collision. Besides the rest energy 3, which the electron already had before the collision, it may have got an additional kinetic energy, which you can notice if the electron is in motion after the collision.

The formula for classical kinetic energy \( \frac{1}{2} \, m \, v^2 \) is rather inappropriate here, because for the Compton scattering one usually uses photons with very high energy (X-rays and gamma rays). By the collision of the energetic photon and the resting electron, the electron can be brought to very high velocities, so that the formula for classical kinetic energy no longer applies. Therefore you have to calculate relativistically if you want to get useful results for the Compton scattering. That is, instead of using the classical formula, we use the relativistic total energy \(W_{\text e}'\), which already includes the rest energy and relativistic kinetic energy of the electron: 6 $$\begin{align}W_{\text e}' &~=~ \sqrt{m_{e}^2 \, c^4 ~+~ \boldsymbol{P}'^2 \, c^2} \\\\ &~=~ \sqrt{{W_{\text e}}^2 ~+~ \boldsymbol{P}'^2 \, c^2} \end{align}$$

Thus, the total energy of the photon and electron after the collision is the sum of 5 and 6: 7 $$W_{\text p}' ~+~ W_{\text e}' ~=~ \frac{h \, c}{\lambda'} ~+~ \sqrt{m_{e}^2 \, c^4 ~+~ \boldsymbol{P}'^2 \, c^2}$$

According to the conservation of energy, the total energy of the system before the collision must be equal to the total energy after the collision: 8 $$W_{\text p} ~+~ W_{\text e} ~=~ W_{\text p}' ~+~ W_{\text e}'$$

The relativistic total energy 7 of the electron gives us also the relationship between its energy and its momentum \( \boldsymbol{P}' \). In this way, we can combine conservation of momentum with conservation of energy. Rearrange the conservation of momentum 1 for \( \boldsymbol{P}' \): 9 $$\boldsymbol{P}' ~=~ \boldsymbol{p} ~-~ \boldsymbol{p}'$$

Since momentum \(\boldsymbol{P}'^2\) occurs in total energy 7, we square Eq. 9 to obtain a relation for \(\boldsymbol{P}'^2\) (we use a binomial formula for this): 10 $$\begin{align}\boldsymbol{P}'^2 & ~=~ \left( \boldsymbol{p} ~-~ \boldsymbol{p}'\right )^2 \\\\ & ~=~ \boldsymbol{p}^2 ~+~ \boldsymbol{p}'^2 ~-~ 2\boldsymbol{p}\cdot\boldsymbol{p}'\end{align}$$

The last summand contains the scalar product between \( \boldsymbol{p}\) and \(\boldsymbol{p}'\). We can write it as follows using the angle \(\theta\) between \( \boldsymbol{p}\) and \(\boldsymbol{p}'\): \( \boldsymbol{p} ~\cdot~ \boldsymbol{p}' ~=~ p \, p' \, \cos(\theta) \). Here \( p ~=~ |\boldsymbol{p}| \) and \( p' ~=~ |\boldsymbol{p}| \) are the magnitudes of the two momentum vectors. Moreover, \(\boldsymbol{P}'^2 ~=~ P'^2 \) holds. Let's use that in Eq. 10: 11 $$\begin{align}{P'}^2 & ~=~ \left( \boldsymbol{p} ~-~ \boldsymbol{p}'\right )^2 \\\\ & ~=~ {p'}^2 ~+~ p^2 ~-~ 2p' \, p \, \cos(\theta)\end{align}$$

Rearrange the total energy 6 of the electron for \( P'^2 \): 12 $$P'^2 ~=~ \frac{W_{\text e}'^2 ~-~ W_{\text e}^2}{c^2}$$

Substitute the squared momentum 11 into Eq. 12: 13 $$p'^2 ~+~ p^2 ~-~ 2p' \, p \, \cos(\theta) ~=~ \frac{W_{\text e}'^2 ~-~ W_{\text e}^2}{c^2}$$

Next, we use photon energies 2 and 5 to replace the photon momentum magnitudes with \( p = \frac{W_{\text p}}{c} \) and \(p' = \frac{W_{\text p}'}{c} \): 14 $$\frac{W_{\text p}'^2}{c^2} ~+~ \frac{W_{\text p}^2}{c^2} ~-~ 2\frac{W_{\text p}'}{c} \, \frac{W_{\text p}}{c} \, \cos(\theta) ~=~ \frac{W_{\text e}'^2 ~-~ W_{\text e}^2}{c^2}$$

Let us first multiply Eq. 14 by \( c^2 \), then rearrange the conservation of energy 8 for \( W_{\text e}' = W_{\text p} ~+~ W_{\text e} ~-~ W_{\text p}' \), and then insert it: 15 $$\begin{align}W_{\text p}'^2 ~+~ W_{\text p}^2 ~-~ 2W_{\text p}' \, W_{\text p} \, \cos(\theta) &~=~ W_{\text e}'^2 ~-~ W_{\text e}^2 \\\\ W_{\text p}'^2 ~+~ W_{\text p}^2 ~-~ 2W_{\text p}' \, W_{\text p} \, \cos(\theta) &~=~ \left( W_{\text p} ~+~ W_{\text e} ~-~ W_{\text p}' \right)^2 ~-~ W_{\text e}^2\end{align}$$

Multiply out the parenthesis in 15: 16 $$\begin{align}W_{\text p}'^2 ~+~ W_{\text p}^2 ~-~ 2W_{\text p}' \, W_{\text p} \, \cos(\theta) &~=~ W_{\text p}^2 ~+~ W_{\text e}^2 ~+~ W_{\text p}'^2 ~+~ 2W_{\text p}W_{\text e} \\ &~-~ 2W_{\text p}W_{\text p}' ~-~ 2W_{\text p}'W_{\text e} ~-~ W_{\text e}^2\end{align}$$

Some summands in 16 cancel out: 17 $$- 2W_{\text p}' \, W_{\text p} \, \cos(\theta) ~=~ 2W_{\text p} \,W_{\text e} ~-~ 2W_{\text p} \,W_{\text p}' ~-~ 2W_{\text p}' \, W_{\text e}$$

Bring \( 2W_{\text p}\,W_{\text p}' = 2W_{\text p}' \, W_{\text p} \) to the left side and factor it out: 18 $$\begin{align}2W_{\text p}' \, W_{\text p} ~-~ 2W_{\text p}' \, W_{\text p} \, \cos(\theta) &~=~ 2W_{\text p}\,W_{\text e} ~-~ 2W_{\text p}' \,W_{\text e} \\\\ 2W_{\text p}' \, W_{\text p} \, \left( 1 ~-~ \cos(\theta) \right) &~=~ 2W_{\text p} \, W_{\text e} ~-~ 2W_{\text p}' \, W_{\text e}\end{align}$$

Divide the whole equation by \( 2W_{\text p}' \, W_{\text p} \, W_{\text e} \): 19 $$\frac{1}{W_{\text e}} \, \left( 1 ~-~ \cos(\theta) \right) ~=~ \frac{1}{W_{\text p}'} ~-~\frac{1}{W_{\text p}}$$

Then we insert the photon energies 2 and 5. We also insert the energy \(W_{\text e}\) of the electron before the collision, which corresponds to the rest energy 3: 20 $$\frac{1}{m_{e} \, c^2 } \, \left( 1 ~-~ \cos(\theta) \right) ~=~ \frac{\lambda'}{h \, c} ~-~ \frac{\lambda}{h \, c}$$

Let's multiply the equation by the factor \( h \, c \) and we are done: 21 $$\lambda' ~-~ \lambda ~=~ \frac{h}{m_{e} \, c } \, \left( 1 ~-~ \cos(\theta) \right)$$

Sometimes the formula is also written with the wavelength difference \(\delta \lambda = \lambda' - \lambda \) and the Compton wavelength \(\lambda_{\text C} = \frac{h}{m_{e} \, c } \): 22 $$\Delta \lambda ~=~ \lambda_{\text C} \, \left( 1 ~-~ \cos(\theta) \right)$$

Electron is in motion before the collision

We have assumed in the derivation that the electron is at rest. If it is not at rest at the beginning, the derivation is a little more complicated. But the principle is the same as in derivation of the Compton formula for an electron at rest!

Example of an initial situation: A photon with momentum \( \boldsymbol{p} \) travels in positive \(x\) direction, while an electron having momentum \( \boldsymbol{P} \) before the collision travels in negative \(x\) direction. First, you set up the equations for energy and momentum and proceed similarly to the derivation above: $$\hbar \, \omega ~+~ \sqrt{\boldsymbol{P}^2 \, c^2 ~+~ m^2 \, c^4} ~=~ \hbar \, \omega' ~+~ \sqrt{\boldsymbol{P}'^2 \, c^2 ~+~ m^2 \, c^4}$$ $$\boldsymbol{p} ~+~ \boldsymbol{P} ~=~ \boldsymbol{p}' ~+~ \boldsymbol{P}'$$

Exercises with Solutions

Use this formula eBook if you have problems with physics problems.

Exercise #1: Velocity of the Atom after Absorption and Emission of a Photon

When an atom absorbs or emits a photon, the atom experiences a recoil due to momentum conservation.

A resting free atom of mass \( m_{\text A} \) absorbs/emits a photon of frequency \( f \). What is the recoil velocity of the atom after...

  1. ...absorption of this photon?
  2. ...emission of this photon?

Hint: Use momentum conservation: write down the total momentum (atom + photon) before and after the collision. The momentum of a photon is given by \( p_{\gamma} ~=~ \frac{h}{\lambda} \).

Solution to Exercise #1.1

To find the velocity \( v_{\text A}' \) that the resting atom has after it has absorbed a photon, you use the momentum conservation equation: 1 \[ p_{\text A} ~+~ p_{\gamma} ~=~ p_{\text A}' \]

Equation 1 states that the total momentum (the momentum of the atom \( p_{\text A} \) and the photon \( p_{\gamma} \)) before the collision must be equal to the total momentum after the collision.

Since the photon is absorbed (swallowed by the atom), the momentum of the photon after the collision is \( p_{\gamma}' ~=~ 0 \). Thus, the total momentum after the collision resides only in the momentum of the atom \( p_{\text A}' \). According to the task, the atom is at rest before absorption, so the initial momentum disappears: \( p_{\text A} ~=~ m_{\text A} \, v_{\text A} ~=~ m_{\text A}*0 ~=~ 0 \).

Substituting the photon momentum \( p_{\gamma} ~=~ \frac{h}{\lambda} ~=~ \frac{h \, f}{c} \), the momentum conservation becomes: 2 \[ \frac{h \, f}{c} ~=~ m_{\text A} \, v_{\text A}' \]

To find the velocity of the atom after absorbing the photon, rearrange equation 2: 3 \[ v_{\text A}' ~=~ \frac{h \, f}{m \, _{\text A}} \]

As seen from equation 2, after absorption, both the atom and the photon move in the same direction!

Solution to Exercise #1.2

Similar to Exercise #1.1, you first set up the momentum equation. However, in this case, the photon is not absorbed but emitted, meaning the photon was not there before (\( p_{\gamma} ~=~ 0 \)): 3 \[ p_{\text A} ~=~ p_{\text A}' ~+~ p_{\gamma}' \]

According to the Exercise, the atom is at rest, so \( p_{\text A} ~=~ 0 \). Therefore, the momentum equation 3 expands to: 4 \[ 0 ~=~ m_{\text A} \, v_{\text A}' ~+~ \frac{h \, f}{c} \]

Rearrange equation 4 to find the velocity of the atom after emitting the photon: \[ v_{\text A}' ~=~ ~-~ \frac{h \, f}{c \, m_{\text A}} \]

As seen from equation 4, after emission, both the atom and the photon move in opposite directions!