Gamma spectroscopy - Prelab questions 1. What characteristics distinguish x-rays from gamma rays? Is either more intrinsically dangerous? 2. Briefly discuss dead time in a detector. What factors are important in the determination of dead time? 3. What scattering angle will maximise the amount of energy transferred to a scattered electron by a photon as the result of Compton scattering? (Equation 1.) Write down the expression in terms of E γ, m e and c 2. Calculate the maximum energy of the electron for the case of an incident 662 kev photon. (Consider your units.) 4. Show that as E γ, the energy of photons scattered through 180 approaches 255 kev, and is independent of E γ. 5. Figure 1 below shows Compton scattering where a gamma ray scattering off an electron emerges with reduced energy. Continue drawing the diagram. Where do the photon and electron travel to? Draw the same type of diagram for one of the other two processes listed in the theory section. hν hν' θ Φ e - Figure 1: Kinematics of Compton Scattering. 1
Gamma spectroscopy - Theory γ radiation is the product of another nuclear decay and/or reaction. A γ-ray source starts with decay modes (α decay, β decay or electron capture) that leave the daughter nucleus in an excited state. The daughter nucleus then decays to its ground state by γ emission. Figure 2 shows the production of a 14.41 kev γ from 57 Fe as a result of the decay of 57 Co. The half-life of 57 Co is 271 days so it is a suitable source of very low energy γ rays. The other sources in this experiment operate by similar decays and transitions. Three sources will be used to calibrate the detector: Caesium-137, Sodium-22 and Cobalt-60. The γ-ray energies these isotopes emit are given in Table 1. Source E γ (MeV) 137 Cs 0.032 0.662 22 Na 0.511 1.275 60 Co 1.173 1.332 Table 1: Radioactive sources used for the calibration, and the energies of the photons produced. Interaction of γ-rays with matter γ-rays are photons, interacting with matter via distinct processes. The result of all three processes is that an electron is produced which ionises the surrounding material, producing visible light, with an energy of order 1 ev. The processes are described below. Photoelectric effect A photon is absorbed by an atomic electron and imparts all of its energy to it. The electron is ejected from the atom with kinetic energy almost equal to the photon energy. Some energy is 57 Co EC 706.41 0.18% 366.74 136.47 14.41 99.82% 57 Fe Figure 2: Decay from 57 Co by electron capture (inverse β decay) to 57 Fe. The three most likely gamma decays are shown by the vertical downward arrows, denoting γ-rays of 136.47, 122.06, and 14.41 kev. γ-rays can also be emitted due to electron de-excitation from any of the other levels, but these are not shown. 2
lost in doing work to unbind the electron from the atom, and some energy absorbed through nuclear recoil. However, due to the relatively huge mass of the nucleus this recoil energy is negligible for our purposes. The kinetic energy of the ejected electron is thus E K = E γ - E B, where Eγ = hν and E B is the binding energy of that electron to the atom. The primary photoelectron then ionises the surrounding material producing light in the visible spectrum from the subsequent transitions. What may also occur is the emission of an X-ray due to the photoelectron originating from an inner shell. The probability of this process is proportional to Z 4, so a high-z material is most useful for detection purposes. This is why the NaI scintillator crystals in our detection system are doped with Tl. Elastic scattering (coherent scattering) Photons with energy between 0.1keV to 80 kev primarily interact through photoabsorption or elastic (coherent) scattering. As γ-ray energy increases, Compton (inelastic) scattering becomes the dominant interaction process. For light elements Compton scattering becomes the dominant process at lower energies (3 5 kev for hydrogen). Additional nuclear scattering and absorption occurs above MeV energies, including pair production, Delŭck scattering from the nuclear field, and nuclear resonant processes such as nuclear Thomson scattering. Compton scattering (Inelastic or incoherent scattering) In Compton scattering the photon undergoes scattering off of free electrons. In the detector material, the process affects the outermost electrons which are very loosely bound and as such may be considered free to a good approximation. The process is illustrated in Figure 3. hν hν' θ Φ Figure 3: Kinematics of Compton Scattering. The energy of the scattered electron depends on the angle of scattering and is given by e - E e = E γ α(1 cos θ) 1 + α(1 cos θ), (1) where E e is the scattered electron energy, E γ is the photon energy and α = E γ /(m e c 2 ), and where m e = 511 kev/c 2 is the rest mass energy of the electron. The scattered electron then ionises the surrounding material and liberates visible-light photons. The scattered γ undergoes more and more scattering events, scattering more electrons which produce more light, until the original photon expends its energy. Pair production (e + e ) If the incident photon energy is more than twice the rest mass energy of an electron, a third 3
process is possible. Pair production takes place in a region of a high Coulomb field, where the photon is converted into an electron-positron pair. The Tl nucleus provides the strong Coulomb field here. The electron ionises the surrounding material while the positron collides with an electron causing annihilation into two 511 kev photons, which may also be detected. The signal When detecting the photons as a function of energy, a structured spectrum appears. As all the energy is ultimately deposited by a photoelectric event in the detector, one obtains a peak (called the photopeak ) at the energy of the incident photon. Likewise, one obtains a peak at the energy of the incident photon after pair production. In that case, however, two escape peaks can be observed, at 511 kev and 1022 kev below the full energy peak, corresponding to one or two of the 511 kev photons escaping the detector. Electron energy after Compton scattering is continuous up to a maximum, seen as an edge (the Compton edge) which lies just below the photopeak. Figure 4: A typical response of a detector to monoenergetic γ-rays. Real peaks are likely to be more broad. Multiple Compton scattering will fill the gap between the Compton edge and the photopeak. The escape peaks appear only if the γ-ray energy is above 1.022 MeV. The resolution of the signal The resolution is defined as the width of the peaks in the spectrum, related to the full width half maximum (FWHM) - the width as measured at half the number of counts in the peak. The resolution is then Resolution (%) = F W HM E γ 100 This may be counter-intuitive to your understanding of resolution. We often attach positive connotations to high-resolution images meaning we can see more detail. However, for the energy resolution above, a small percentage is desirable, meaning we can identify the position of a peak to within x percent of the real position. 4
Gamma spectroscopy - Equipment The Detector System Gamma source Detector Amplifier Multi-channel analyser Nal crystal Pre-amp Digital oscilloscope Computer (a) Detector components (b) Figure 5: (a) Configuration of NaI(Tl) detector with photomultiplier tube. (b) Circuit for detecting γ rays and system calibration. The NaI(Tl) scintillator NaI(Tl) scintillators are used as γ detectors. Gamma rays interact with the NaI crystal as previously discussed. The thallium (Tl, Z = 81) dopant is critical, acting as a wavelength shifter for the produced photons. The probability of total internal reflection inside the detector is high as NaI crystal has a refractive index of 2. This means visible photons are unlikely to escape. Photons travel through the crystal to the far side of the detector where a glass window is optically coupled to a photomultiplier tube. The photomultiplier tube (PMT) Light in the crystal is guided to a phosphor screen at the entrance to the PMT. Each photon liberates an electron via the photoelectric effect. Electrons hit the cathode, which is the first dynode in the chain. Each dynode is under high voltage (V 1 kv) so each collision has the effect of creating an electron shower. The shower of electrons proceeds to the next dynode, producing yet another shower per electron. 10 to 14 dynode stages in a PMT provide a gain of 10 7 electrons at the anode per electron produced at the phosphor. The charge at the anode is then passed to the electronics. The multi-channel analyser (MCA) The signal from the PMT is amplified by the amplification system to produce a pulse proportional to the photon energy. The MCA compares the voltage of the signal with an internal voltage stepping up from zero volts. The number of steps defines the channel, displayed on the computer screen. The MCA you are using accepts pulses up to 5 V, and divides this into channels defined by the Analogue-to-Digital-Converter (ADC). There are 4096 channels here, so each channel defines a region of 12mV. 5
Gamma spectroscopy - Procedure Gain of the detection system and MCA 1. Starting with the detector, identify each piece of equipment and follow the wiring as per figure 5. Ensure that the wiring from the detector to the MCA is complete, and that the HV supply is connected to the detector. 2. It is very important to be careful with the proper order of turning on these pieces of equipment. Switching them on or off incorrectly will cause damage to the equipment. 3. Set the HV supply to 800V then turn on only the power switch. Wait until you hear a click and the orange light labelled STD BY RESET comes on; only then turn on the High Voltage switch. 4. Look at the list of particle energies and find which of your labelled radioactive sources has the highest energy emission. Place this source in front of the detector, and ensure a signal is present on the oscilloscope. Question 1 Describe the pulse showing on the oscilloscope and note how it fluctuates. Why do we see these fluctuations? 5. Recall that the MCA accepts voltages between 0 and 5 V. The gain on the amplifier should be set such that the pulses on the oscilloscope are in this range. 6. Open the program SpectLab on your computer. Connect the device though the connect menu in the device drop down box. Once connected, click go to start taking a spectrum. 7. Adjust the amplifier gain such that the photopeak for this source is to the far right of the spectrum. Note the amplifier settings. Question 2 What features of this test spectrum are prominent? What channels do they appear in? 8. The overall gain of the detection system is also affected by the HV applied to the photomultiplier tube. Investigate the effect of the HV supply on the gain by making gradual small changes to the applied voltage and observe the effect on the output voltages. Question 3 Do you notice any background counts, and do these change linearly with changes in the HV supply? 9. When one particle is detected the electronics are working to process the event. Another particle entering the detector during this time will not be detected - the detector is dead for this time. Dead time is recorded at the bottom of the MCA screen. Note a few dead time values and see if it fluctuates. Will dead time affect your experiment? 6
Measuring Spectra 1. Reset the HV to 800 V and allow a few minutes for the power supply to stabilise. Question 4 Given that the energies you will encounter in this experiment are no greater than 2 MeV spread over 4096 MCA channels, which source should we examine first? Is the only factor the highest energy emission, or are other factors important? Refer to table 2. 2. Collect a spectrum using the first source you wish to examine. Adjust the gain settings so that the spectrum will show a max of 2 MeV (see question 4). Once you are happy with your gain settings, keep them constant throughout the experiment. 3. Print your spectrum, making sure the axis along the bottom of your graph is Channel Number, not Energy, and record the channel numbers of all of the important features in the spectrum. 4. Identify and label features of your spectrum. Note down the channel numbers of the peak positions including error in a table. 5. Repeat these steps for the 22 Na and 60 Co sources. What relationship did you determine between energy and channel number? This is your detector calibration. Include error analysis. Question 5 Are there any features which appear only in certain spectra? What features are common to all spectra? Question 6 What are the half-lives of each of the elements we re using? Given your spectra, how often would you estimate we need to replace the sources? Identifying unknown features 1. Using the known energies of the photons in the three sources given in Table 1 plot a graph of energy against channel number. 2. Use this calibration to mark the energies of the Compton edges and pair production peaks in each of your spectra. Compare the measured energies with your expectations. 3. Now you can identify an unknown source. Collect the spectra of the unknown source and determine the energies of the peaks. Compare those energies to the ones listed in table 2 and identify the unknown source. You will need a decent number of points with high (meaning low) resolution. Which peaks will help you most with identifying the source - the lower or higher energetic ones? Question 7 What is the unknown source? Note which energies you are using as points of comparison. 7
Energy resolution 1. Measure the FWHM of the photopeaks in your spectra and calculate the resolution for each of the six known peaks, including errors as you go. 2. Plot a function of resolution against E γ and discuss your findings. Question 8 Is the resolution better at high energies or low energies? Reflect on the accuracy of your determination of the unknown source. Question 9 What function of the detection system causes the resolution observed? Remember how the detector works to observe a photon. Table 2: Table of various γ sources with the energies in kev. Source Cs 320 662 22 Na 511 1275 60 Co 1173 1332 125 Sb 35 176 428 463 601 607 636 131 I 80 284 364 637 723 133 Ba 53 81 273 303 356 32 142 Ba 77 232 225 364 425 600 894 949 1001 1078 1204 152 Eu 122 245 344 779 964 1086 1408 207 Bi 570 1064 1770 226 Ra 53 186 242 295 353 609 769 1120 1238 1378 1764 56 Ni 270 750 812 1562 75 Se 97 121 136 265 280 401 88 Y 898 1836 e + e 511 8