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\Work {The Compton effect} {The Compton effect} {The energy of $\gamma$-quanta scattered on graphite is measured as a function of scattering angle by means of a scintillation spectrometer. The rest energy of the particles participating in the scattering is determined.}

Scattering of $\gamma$-rays on condensed matter is a phenomenon that reveals the dual nature of radiation. Wave theory, which explains scattering of radiation with a large wavelength, fails to give a satisfactory description of $X$- and $\gamma$-ray scattering. In particular, the theory cannot explain why the scattered radiation observed by Compton contains an additional component with a  frequency lower than the frequency $\omega_0$ of incident radiation. 

The additional component can be explained by assuming that $\gamma$-radiation is a flow of quanta (photons) with energy $\hbar\omega$ and momentum $p=\hbar\omega/c$. The Compton effect, an increase in the wavelength of scattered radiation compared to the incident one, is then interpreted as a result of elastic collision between two particles, $\gamma$-quantum (photon) and free electron.

Consider a simple theory of the Compton scattering. Let an electron be at rest before the collision (the electron rest energy is $mc^2$) and $\gamma$-quantum have the initial energy $\hbar\omega_0$ and momentum $\hbar\omega_0/c$. After the collision the electron energy and momentum become $\gamma mc^2$ and $\gamma mv$, where \mbox{$\gamma =(1-\beta^2)^{-1/2}$}, $\beta=v/c$, and the $\gamma$-quantum emerges at some angle $\theta$ with respect to the initial direction. Energy and momentum of $\gamma$-quantum become $\hbar\omega_1$ and $\hbar\omega_1/c$ (see~\refFigure{1_2_1}).

Let us write down the laws of energy and momentum conservation for this process:
$$
mc^2+\hbar\omega_0=\gamma mc^2+\hbar\omega_1,
$$
$$   \frac{\hbar\omega_0}{c}=\gamma mv\cos\phi+\frac{\hbar\omega_1}{c}\cos\theta,
$$
$$   \gamma mv\sin\phi=\frac{\hbar\omega_1}{c}\sin\theta. $$

Solving these equations and replacing the frequencies $\omega_0$ and $\omega_1$ with the corresponding wavelengths, $\lambda_0$ and $\lambda_1$, one obtains the difference between the wavelength of scattered and incident radiation (the Compton shift): 
$$
\Delta\lambda=\lambda_1-\lambda_0=\frac{h}{mc}(1-\cos\theta)=\Lambda\sub{Ê}(1-\cos\theta),   \eqMark{1_2_1}
$$
where $\lambda_0$ and $\lambda_1$~are the wavelengths of $\gamma$-quantum before and after the collision. The quantity
$$
\Lambda\sub{Ê}=\frac{h}{mc}=2{,}42\cdot10^{-10}\;\cm
$$
is called the Compton wavelength of the electron. Equation~(\refEquation{1_2_1}) shows that the Compton shift is independent of the wavelength of incident quanta and the substance in which it is observed \fFigure {Vector diagram of $\gamma$-quantum scattering on electron}1_2_1 {4.06cm}{1.89cm}{PIC/l01_2_01.eps}  In the above derivation the electron is considered free, which is a good approximation for $\gamma$-quanta with energy of tens or even hundreds kiloelectronvolts (keV). An electron binding energy in a light atom does not exceed several keV and it is even less for the majority of electrons.

When a quantum of light is scattered on bound electrons a change of its momentum is transferred to atom as a whole. Since atomic mass is very large, the corresponding energy transfer is negligible, so only the initial energy is observed in the spectrum of scattered radiation. Thus, scattering of $\gamma$-quanta on bound electrons can be considered as an elastic collision between the quanta and atoms. In classical physics this process is known as Rayleigh scattering: incident light induces resonant oscillation of atomic electrons which radiate photons of the same frequency. If the energy of incident quanta is not too high ($1\div10\:$\keV), part of the electrons behave as free and the rest as bound. Then both types of scattering can be observed simultaneously.  
 
The cross-section of Rayleigh scattering is proportional to $Z^2$, where $Z$ is atomic number, while the cross-section of the Compton scattering is proportional to $Z$. This can be understood as follows. In the Compton scattering each electron can be considered independently since the scattering occurs on a single electron. In Rayleigh scattering the photons are radiated by all (or almost all) electrons of an atomic shell oscillating in phase. Their radiation is coherent, so one must sum the wave amplitudes rather than the intensities.

\fFigure {Cross-section of scattering of photons on carbon ($Z=6$) for photon energies between $10$\,eV and $1$\,MeV; $\sigma_\textrm{ph}$~is cross-section of photo-electric effect, $\sigma_\textrm r$~is cross-section of Rayleigh scattering, and $\sigma_\textrm{c}$~is cross-section of the Compton scattering}1_2_9 {4.16cm}{6.38cm}{PIC/L01_2_09.eps}

Cross-sections of the Compton and Rayleigh scattering have different dependence on the photon energy. The Rayleigh scattering cross-section rapidly decreases while the Compton cross-section  decreases insignificantly as the energy grows.

The diagram in~$\refFigure{1_2_9}$ illustrates the difference between the energy dependence of
the Compton $\sigma\sub{c}$ and Rayleigh $\sigma\sub{r}$ cross-section. Notice, that in the original Compton experiment two lines were observed, the shifted and the unshifted one, because the energy of $X$-rays was $\sim20\,\keV$ and for scattering on carbon $\sigma\sub{c}\simeq\sigma\sub{r}$. The photons scattered on carbon in our experiment have much higher energy $\sim600\,\keV$, for which $\sigma\sub{c}\gg \sigma\sub{r}$, so only the shifted component is observed. The cross-section of Rayleigh scattering on atom is inversely proportional to $\lambda^2$ due to interference of the radiation scattered by different parts of electron distribution.

Finally notice that in addition to scattering $\gamma$-quanta are also absorbed in matter, the absorption is due to photoelectric effect and creation of electron-positron pairs. The process of pair creation does not occur below a threshold of $2mc^2=1{.}02\;\MeV$, so it is not possible in our experiment. The photoelectric effect consists in knocking out an atomic electron, when the atom absorbs a photon. The photon momentum is shared between the atom and electron momenta, while the photon energy is partially transferred to the electron and partially to the energy of atomic excitation. The atom returns to its ground state almost instantaneously (in about $10^{-8}\;\s$). Its excitation energy is either being radiated as a soft photon or transferred to another electron which leaves the atom (the Auger effect). In both cases the excitation energy is usually absorbed by the neighboring atoms.

The main goal of the experiment is verification of Eq.~(\refEquation{1_2_1}) which should be rewritten in terms of $\gamma$-quantum energy to adapt it to our experiment. One can easily check that the corresponding expression is 
$$
\frac{1}{\epsilon(\theta)}-\frac{1}{\epsilon_0}=1-\cos\theta.   \eqMark{1_2_2} $$

Here $\epsilon_0=E_0/(mc^2)$ is the energy of incident $\gamma$-quanta in the units of electron rest energy, $\epsilon(\theta)$ is the energy of quanta scattered at an \mbox{angle $\theta$} in the same units, and $m$ is the electron mass.

\vspace{10pt} \so{\textbf{Experimental setup}} \vspace{5pt}

The experimental setup is shown in~\refFigure{1_2_2}. The source of radiation~\textit{1} is a sample of~$^{137}\mathrm{Cs}$ emitting $\gamma$-rays with the energy of~$662\;\keV$. The sample is enclosed in a thick-wall container with a collimator. The collimated narrow beam of $\gamma$-quanta strikes a graphite target~\textit{2} (a cylinder of $40\;\mm$ in diameter and $100\;\mm$ in height).

\cFigure {Experimental setup for observing scattering of $\gamma$-rays}1_2_2 {8.13cm}{4.01cm}{PIC/l01_2_02.eps}

The quanta scattered in the target are registered by a scintillation counter, its operation principle is discussed in the experiment~5.5.3. The counter consists of a photomultiplier tube~\textit{3} (PMT) and a scintillator~\textit{4}\Footnotemark \Footnotetext{Scintillator and PMT are discussed in detail in Appendices II, III.}. The scintillator is a crystal cylinder of $\mathrm{NaI(Tl)}$, $40\;\mm$ in diameter and $40\;\mm$ in height, its output window is in optical contact with the PMT photocathode. A signal from the PMT output is amplified and applied to a PC analyzer. The crystal and the PMT are enclosed in a light-proof casing attached to a horizontal lever which can rotate around the target. A rotation angle is measured with a dial~\textit{6}. 

The front part of the scintillator is placed inside a lead collimator~\textit{5} forming the input beam and protecting the scintillator from an extraneous radiation. Mostly this radiation is due to $\gamma$-quanta from the source penetrating $6$-cm walls of the container. This background increases significantly when the Compton scattering is studied at large angles because the distance between the scintillator and the source becomes small. 

Figure~\refFigure{1_2_3} shows the block diagram of the setup which includes: the PMT powered by a source of high-voltage~\textsl{SHV} (necessary to operate the PMT in the spectrometric mode), an amplifier-analyzer~\textsl{AA} which serves as an input interface of a PC operated from a keyboard~\textsl{KB}. The experimental data are updated on the display~\textsl{D} in real time, the final results can be printed as tables and charts by means of a printer~\textsl{P}.

The amplitude of an output electric pulse on the anode of PMT operating in the spectrometric mode \cFigure {Block diagram of experimental setup}1_2_3 {8.94cm}{4.13cm}{PIC/l01_2_03.eps} is proportional to the energy of a detected $\gamma$-quantum. A flash of light in a scintillator is not directly caused by a $\gamma$-quantum, it is due to fast electrons produced by the quantum in the scintillator crystal. Conversion of energy of a $\gamma$-quantum into a certain number of optic photons at the scintillator output includes three stages: generation of fast electrons, excitation of atoms and molecules by these electrons, and radiation of optic quanta by the excited atoms and molecules. There are three mechanisms of interaction between $\gamma$-quanta and matter: the Compton scattering, photoelectric effect, and creation of electron-positron pairs (this process is not allowed in our experiment since the energy of $\gamma$-quanta is below the threshold of $1{.}02\;\MeV$). A fast electron emerging in matter as a result of all these processes excites atoms and molecules along its path due to the Coulomb interaction. The number of the excited centers is proportional to the electron energy.

Photoelectric effect is the complete absorption of $\gamma$-quantum by an atom. The quantum knocks an electron out of internal atomic shell (usually the $K$-shell) which carries away almost all energy of the $\gamma$-quantum and disperses it in a crystal. As a result the amplitude of the light flashes turns out to be proportional to the total energy of the primary $\gamma$-quantum. The probability of photoelectric effect is proportional to the atomic number of an absorber as $Z^5$ (see chapter~V). For this reason the scintillator of $\gamma$-spectrometers are usually made of heavy elements (in our installation it is iodine). The Compton scattering of a $\gamma$-quantum in matter occurs on weakly bound electrons. In this process the quantum energy is transferred to electron only partially, the rest is carried away by the scattered $\gamma$-quantum.

Thus a monochromatic radiation generates a distribution of electric pulses at the PMT output (see~\refFigure{1_2_4}). The distribution has the so called photopeak due to photoelectric effect and a continuous distribution due to the Compton scattering. The photopeak is often called the full energy peak, its position is unambiguously determined by the energy of $\gamma$-rays.

We are interested in a photopeak position (the channel number) versus the rotation angle of the detector. The photopeak width is instrumental and depends on the specifications of scintillator crystal and photomultiplier used in the experiment. \hFigure {Amplitude distribution of pulses generated in NaI(Ti) scintillator by monochromatic $\gamma$-quanta}1_2_4 {6.23cm}{3.32cm}{PIC/l01_2_04.eps} To determine the energy of $\gamma$-quanta one studies the curve of energy losses in the crystal, i.e. the amplitude distribution of electric pulses at the PMT output. In our experiment the distribution is measured by a PC operating as the amplitude analyzer. 

When $\gamma$-quanta are detected at $0\degree$ angle (direct beam), there are only the primary $\gamma$-quanta  ($662\;\keV$). The scattered radiation is detected at larger angles, the radiation energy is shifted to lower values according to Eq.~(\refEquation{1_2_1}). Notice that for a large energy of $\gamma$-quanta ($E_{\gamma} \gtrsim mc^2$), as in our case, the probability of Thomson scattering is very small, so the non-shifted line is absent in the observed energy spectrum. As the angle grows, the photopeak moves away from the original position and its width increases.

To the photopeak left there is a continuous distribution of Compton electrons separated by a wide gap. This background persists for any scattering angle and hinders determination of the photopeak position, however the background is easily identified when the detected spectra are plotted on the same graph. 

Electric pulses coming from the PMT output are registered as follows. The amplifier-analyzer~\textsl{AA} assigns a number $i$ between $0$ and $1023$ to an incoming pulse according to its amplitude and the PC adds one unit to the $i$-th memory cell. In so doing the PC memory accumulates the pulses sorted out according to their amplitudes. The operation lasts about $20\;\mks$, during this period the system <<is not sensitive >> to the output PMT pulses, it is the dead time of the detector. Thus, no more than $50$ thousand pulses can be detected per second. The content of all $1024$ memory cells is displayed as a histogram on the PC screen; the abscissa corresponds to the pulse amplitude (the channel number) and the number of pulses stocked in a given channel is the ordinate. The photopeak position can be determined with the accuracy of $1\%$.

\vspace{10pt} \so{\textbf{Treatment of results}} \vspace{5pt}

Let us replace in Eq.~(\refEquation{1_2_2}) the energy of the quantum scattered at an angle $\theta$ with the channel number $N(\theta)$ corresponding to the photopeak position at a given $\theta$. Let $A$ be the (unknown) proportionality factor between $\epsilon(\theta)$ and $N(\theta)$, then
$$
\frac{1}{N(\theta)}-\frac{1}{N(0)}=A(1-\cos\theta).   \eqMark{1_2_3} $$

The experimental results are then plotted in coordinates $(1-\cos\theta)$ (the abscissa) and $1/N(\theta)$ (the ordinate). According to Eq.~(\refEquation{1_2_3}) the experimental points must lie on a straight line. The intercept of the line with the ordinate determines the best value of $N\sub{best}(0)$. This value incorporates not only the measured value of $N(0)$ but also the measurements made at other angles; the intersection with the line $\cos\theta=0$ allows one to find the improved value of $N\sub{best}(90)$. Comparing the results with the theory allows one to verify the relation between the energy and the scattering angle and to determine the rest energy of the particles responsible for the Compton scattering (presumably, electrons). To do this let us inspect Eq.~ (\refEquation{1_2_2}) again. Restoring the variable $E$ one can see that at $\theta=90\degree$ Eq.~(\refEquation{1_2_2}) becomes 
$$
mc^2\left(\frac{1}{E(90)}-\frac{1}{E(0)}\right)=1,
$$
or
$$
mc^2=E(0)\frac{E(90)}{E(0)-E(90)}=E_{\gamma}\frac{N(90)}{N(0)-N(90)}.   \eqMark{1_2_4} $$

Here $E(0)=E_{\gamma}$ is the energy of electrons scattered forward, i.e. the energy of $\gamma$-rays emitted by the source (in our case $^{137}\mathrm{Cs}$). Besides we take into account that the channel number corresponding to a photopeak is proportional to the energy of $\gamma$-quanta.

Now it should be clear that the values of $N\sub{best}(0)$ and $N\sub{best}(90)$ determined from the plot must be substituted into Eq.~(\refEquation{1_2_4}) instead of the experimental values. This allows one to diminish the role of random errors due to, for example, voltage instability which significantly influences the gain of the PMT amplifier and the electronics.

\vspace{10pt} \so{\textbf{Directions}} \vspace{5pt}

\begin{Enumerate}{tab} 
\Item. %1.\, 
Turn the electronics and PC on.

\Item. %2.\, 
Start the PC code and initiate the spectrometric mode (the code contains a detailed instruction).

\Item. %3.\, 
Check operation of the installation in this mode for a small period of exposure (about $1$ min):

a) measure the spectrum at $\theta=0\degree$;

b) set the angle at $\theta\simeq30\degree$, measure the spectrum, and verify that the photopeak moves to the left (to smaller energies);

c) make sure that you can correctly identify the photopeak position (the channel number). 

\Item. %4.\, 
Measure the spectra and determine the photopeak position for each angle $\theta$ by setting the scintillation counter at different angles to the $\gamma$-quanta beam and typing the angles into the PC. The measurements must be done with a step of~$10\degree$ in the range between $0\degree$ and $120\degree$. It should be taken into account that the intensity of the scattered quanta decays as the angle grows, so longer periods of exposure are required.

\Item. %5.\, 
Using the PC code one can display the obtained values ($\theta$, $N(0)$, $N(\theta)$) on the screen and verify Eq.~(\refEquation{1_2_3}) experimentally.

\Item. %6.\, 
Plot the experimental data with the errors in coordinates $1-\cos\theta$ (abscissa) and $1/N(\theta)$ (ordinate), see Eq.~(\refEquation{1_2_3}). Draw the best fitted straight line through the points.

\Item. %7.\, 
Using the plot and Eq.~(\refEquation{1_2_4}) determine the rest energy of the particles responsible for the Compton scattering of $\gamma$-quanta. What factors determine the error? Estimate the error.

\end{Enumerate} 

\begin{center}\so{\textsf{\small ËÈÒÅÐÀÒÓÐÀ}}\end{center} {\small

1. \textit{Ãîëüäèí\;Ë.\;Ë., Íîâèêîâà\;Ã.\;È.} Ââåäåíèå â àòîìíóþ ôèçèêó.\,---\,Ì.: Íàóêà, $1988$. Ãë.\;$1$.

2. \textit{Öèïåíþê\;Þ.\;Ì.} Êâàíòîâàÿ ìèêðî- è ìàêðîôèçèêà.\,---\,Ì.: Ôèçìàòêíèãà, $2006$. \textsection\textsection\;$5{.}4$, $6{.}5$. }