- Introductory comments- To start the computer MCA controller- To use the peak locator to find the centroid of a peak- To Print the Scatterplot- To clear your dataRR Oct 2001SS Dec 2001 Physics 342 LaboratoryScattering of Photons from Free Electrons: Compton ScatteringObjective: To measure the energy of high energy photons scattered from electrons in abrass rod as a function of the scattering angle.References: 1. A.H. Compton, Phys. Rev. 21, 715 (1923) and Phys. Rev. 22, 409 (1923). 2. A.C. Melissinos, Experiments in Modern Physics, Academic Press, New York, 1966,p. 252-65. 3. K. Krane, Modern Physics, 2nd Ed., Wiley and Sons, New York, 1996, p. 83-87.Apparatus: A low energy (22 keV) and weak (15C) portable Cd109 source; a higherenergy (662 keV) strong (5 mC) Cs137 source in a cylindrical lead shield; a NaI(T)scintillator mounted on a Photo-Multiplier Tube (PMT); an iron shield surrounding thePMT; a Canberra 1024-channel PC based Multi-Channel Analyzer (MCA); a high voltage(1.5 kV) power supply; a brass cylindrical rod for use as a scattering target; a carriage torotate the scintillator and PMT assembly at a fixed distance around the target.Introduction In 1923, Compton considered the problem of high energy photons (-rays) scatteringfrom solids. Experimentally, he found that low energy (few MeV  - ray) monochromaticphotons scattered by metals change their frequency and that the frequency changedepends on the scattering angle. This proved to be problematic, since at that time, lightscattering was understood in terms of diffraction in which the scattered (diffracted) wavedoes NOT change frequency. Compton’s experiments and his theoretical analysis of themcame to be know as Compton scattering. Historically, his experiments are importantbecause they provided further compelling evidence that photons do behave as particleswhich obey conservation of momentum and energy laws. Compton was awarded theNoble prize in 1927 for his seminal work.Compton’s experiment can be understood by considering the interaction of the incidentphotons with the electrons that comprise a metal. Because metals are good conductors ofelectricity, some fraction of the electrons associated with each atom in the metal can beconsidered to be free. If the quantized nature of electromagnetic radiation is taken into1account (electromagnetic radiation consists of photons, each of which has the sameenergy E=h), and relativistic kinematics are used to describe the scattering process, thechange of wavelength is understandable as a straightforward consequence of total energyand momentum conservation during a scattering process in which an incoming photonloses some of its energy to a free electron having a mass me. The basic kinematic diagramillustrating this interaction is sketched in Fig. 1.Figure 1: A schematic energy diagram showing the interaction between a photon and afree electron. The incident  has an energy E and a momentum p=h/. The energy of theelectron before scattering is just the rest mass energy of the electron, Ee=mec2; themomentum of the electron before scattering is zero.The scattered  has an energy E' anda momentum p'=h/'. The scattered electron has an energy Ee' and momentum 2221'' mcEcpee.For a beam of incident photons, each of which has the same energy E=h, there will bephotons emerging at various angles  with respect to the incident photon direction. Theenergy E' of a photon emerging in a given direction can be calculated using relativistickinematics and has a value given by  cos1/1'2cmEEEe(1)where  is the angle between the direction of the emerging (scattered) photon and theincident photon, me is the rest mass of the electrons, and c is the velocity of light. From Eq. (1) it can be seen that in order to obtain a large Compton shift (i.e. a largevalue for E – E’), the incident photons should have an energy E of the same order of2magnitude as the electron rest energy mec2 (mec2 = 511 keV). In this experiment 662 keV( - ray) photons from a Cs137 source will be Compton-scattered by a cylindrical rodmade of brass. Experimental Considerations The primary source of photons in this experiment is a 5 mC cesium (Cs137) emitting662 keV photons. The source is kept in a lead cylinder for shielding. When you are readyto take data, remove the end piece of the shielding cylinder. A narrow channel drilledthrough the center of the cylinder permits only photons which travel along it to scatter offthe target. The direction of the channel defines =0, the direction of the unscatteredbeam. This procedure of creating a parallel beam of particles from an otherwise isotropic sourceis called collimation. The emerging photons impinge on the target rod, and some smallfraction will be scattered. Some of the scattered photons will enter the NaI(T) crystalattached to a photomultiplier tube (PMT) (see Fig. 2).Figure 2: A schematic diagram of the experimental apparatus.The energy lost in collisions with the I-atoms in the NaI crystal results in the emission oflight photons from excited T atoms which are intentionally incorporated into the NaIcrystal. These photons in turn eject electrons from the photo-cathode of the PMT. Theelectrons are accelerated through 1300 volts of electrostatic potential in such a way thatthey produce a cascade of electrons from electrodes (called dynodes) placed inside thetube. This electron current pulse is then transformed into a voltage pulse which is pulse-shaped so that its height is proportional to the energy of the incoming . The pulse height3is then measured and sorted by a computer using a specially designed multi-channelanalysis (MCA) board. If the source is so intense that two of these pulses arrive at the same time within the PMT,a spurious overlapping signal is produced. This case must be avoided as there is no way tolearn about the energy of each of these pulses. The time separation between two pulsesmust be larger than certain value (called dead time) in order for these two pulses to beresolved separately. In order to avoid this problem, you will be measuring the energydistribution of the source at a scattering angle of 5 instead of the normally chosen angleof 0, By moving the detector slightly off-center from the source, the intensity of the beam