# 6. Compton Scattering¶

## 6.1. Background¶

When a photon is scattered by a charged particle, such as an electron, it imparts some of its energy and momentum to the particle, which then recoils. The photon comes away from the collision with less energy and therefore longer wavelength than it had before the collision. In 1922, Compton applied the laws of conservation of energy and momentum to a collision between a photon and a free electron to obtain an expression for the change in wavelength as a function of the scattering angle, $$\theta$$ , the photon Fig. 6.1:

Here

$\begin{split}k_\mu ^i = \left( {\begin{array}{*{20}{c}} {{{\vec k}_i}} \\ {{E_0}} \end{array}} \right),\left| {{{\vec k}_i}} \right| = {E_0}\end{split}$

is the 4-vector of the incoming photon,

$\begin{split}k_\mu ^f = \left( {\begin{array}{*{20}{c}} {{{\vec k}_f}} \\ {{E_f}} \end{array}} \right),\left| {{{\vec k}_f}} \right| = {E_f}\end{split}$

is the 4-vector of the scattered photon and

$\begin{split}p_\mu ^f = \left( {\begin{array}{*{20}{c}} {{p_f}} \\ {{E_R}} \end{array}} \right),{E_R} = \sqrt {p_f^2 + {M^2}}\end{split}$

is the 4-vector of the recoiling particle.

Energy conservation leads to the following expression:

(6.1)${E_0} + M = {E_f} + \sqrt {p_f^2 + {M^2}}$

and momentum conservation :

(6.2)$\begin{split}\begin{array}{*{20}{c}} {{E_0} = {E_f}\cos (\theta ) + {p_f}\cos (\psi )} \\ {0 = {E_f}\sin (\theta ) - {p_f}\sin (\psi )} \end{array}\end{split}$

From equations (6.1) and (6.2) one can solve for $$E_f$$ as a function of $$\theta$$ and obtains:

(6.3)${E_f} = \frac{{{E_0}}}{{1 + \frac{{{E_0}}}{M}\left( {1 - \cos (\theta )} \right)}}$

Using the relation $$E = \frac{{hc}}{\lambda }$$ equation can be used to calculate the change in wavelength of the photon leading to:

(6.4)$\Delta \lambda = {\lambda _f} - {\lambda _i} = \frac{h}{{mc}}\left( {1 - \cos (\theta )} \right)$

The expression $$\frac{h}{mc}$$ is called the Compton wavelength.

## 6.2. Experimental Procedure¶

Fig. 6.2 shows your experimental setup. The source is strong 137Cs source that is housed in a lead cylinder with an opening that you instructor needs to open and close. Target is an aluminum cylinder which can be removed. The detector is a NaI(Tl) scintillation detector that you probably have used previously and should be setup as described in the Scintillator experiment. Close to the target is an angle scale where you can read and set the scattering angle.

The goal of this experiment is to measure the energy of the scattered photons as a function of their scattering angle, to verify the expression from equation and to determine the mass of the recoiling particle.

### 6.2.1. Taking Data¶

1. Set the MCA live time to 300s. Set the detector at 0 degree and remove the target. Push the scintillator towards the target (this should not require any force) in such a way that about 1 inch sticks out of the surrounding lead ring. Place a 60Co source on the detector and take a spectrum. This spectrum will show a very prominent peak for 137Cs and the two peaks of 60Co. You will use this spectrum to calibrate the instrument. Do not change anything in the detector electronics once you measured the calibration spectrum. Now you are ready to take your data.

2. Start at and angle of 20 degrees and take a spectrum with and without target. At this angle I would recommend you accumulate events for 4 minutes.

3. Continue this procedure in steps of 20 degrees until you have reached 120 degrees. See Section 6.2.2

Remember : save all the spectra as SPE (ASCII) files. In that way you can analyze them later using the LabTools and working with MCA data package.

## 6.3. Analysis¶

Finish the derivation of equation (6.3)!

Using your calibration spectrum, determine the energy calibration (Look at the scintillator lab or documentation working with MCA data). Apply this calibration to all your spectra. From all spectra taken with the target subtract the corresponding spectrum taken without a target. Make sure the both spectra have been taken with the same lifetime, otherwise correct for the lifetime before you take the difference. This leaves you with an energy spectrum of the scattered photons only. Determine the energy of the photo peak (fit a Gaussian to the photo peak). This gives you the energy of the scattered photons. (Hint: please look in the Collection of useful tools for Python code examples that are helpful for your analysis e.g. selecting data and peak fitting)

In order to verify that equation (6.3) holds, plot the ratio $$\frac{E_0}{E_f}$$ versus $$(1-cos(\theta))$$. Then fit a line through the data. Check the offset and the slope. From the slope you can determine the mass of the recoiling particle. If the photons scatter of electrons then this should be the electron mass. How close are you ?