Radiation Detection Laboratory & Comp. Physics 2 Last compiled August 24, 2017
Contents 1 Introduction 4 1.1 Prelab questions............ 5 1 Experiment 1 - Beta spectroscopy 8 1.1 Background theory.......... 8 1.2 Constituent particles and the Kurie variable................... 10 1.3 Determining the energy of Beta particles hitting the detector......... 14 1.4 Beta spectroscopy - Equipment.... 16 1.4.1 The magnetic spectrometer, Geiger- Müller tube and dead time.. 16 1.5 Beta spectroscopy - Procedure.... 18 1.5.1 Setting up the electronics... 19 1.5.2 Taking measurements..... 20 1.5.3 Analysis............ 22 2 Experiment 2 - Gamma spectroscopy 25 2.1 Background theory.......... 25 2.1.1 Interaction of Gamma-rays with matter............. 26 2.1.2 The signal........... 31 2.2 Gamma spectroscopy - Equipment.. 33 2.3 Gamma spectroscopy - Procedure.. 35 2.3.1 The sources.......... 35
2.3.2 Gain of the detection system and MCA........... 35 2.3.3 Measuring spectra...... 38 2.3.4 Identifying unknown features 39 3 Useful data 41
1 Introduction α γ B β Figure 1: The motion of three key radiation types in a magnetic field It is likely that most previous radiation experiments you ve performed investigated one of the following: radioactive decay rates, effective shielding for radioactive sources, or maybe the differences between types of radioactive particles. Today in these experiments we move past the basics of radiation statistics 1 and instead focus on radiation production, detection, and using radiation as a form of spectroscopy. It is important to first understand how gamma rays and beta particles are produced. You will be using that knowledge to understand how these particles are detected, and then use their detection to determine physical properties. 1 and assume you know how to handle radiation sources properly...
You might notice on examining the two setups that they don t have many elements in common. This is both because we are investigating two different types of radiation, and different properties of each. As a brief historical note, the study of radiation really only began in 1896. In 1930, Pauli proposed the existence of an extremely light particle called a neutrino after experiments demonstrated a continuous, rather than single-peaked, electron energy spectrum during beta decay. Something had to be removing energy from the system, but also be extremely light and hard to detect. This lead to an experimental observation of the neutrino in 1956. 1.1 Prelab questions 1. What characteristics distinguish beta particles from gamma rays? What are the charge and mass of each type of particle? 2. Briefly discuss the idea of dead time in a radiation detector. What properties of the detector mostly influence the length of the dead time? 3. Consider the following diagram of Compton Scattering, and the equation describing the proportionality between the energy of the final state
electron and incident photon: γ 2= hv' θ γ 1= hv Φ Figure 2: Kinematics of Compton Scattering. e - E e E γ(1 cos θ) 1 + E γ (1 cos θ), (1) a) What angle will maximise the amount of energy transferred to the scattered electron (E e ) by the incident photon (E γ )? b) Inside the detector, what happens to γ 2 and the electron after the first Compton scattering event? 4. The nuclear β ± decay process is expressed as: { A ν ZM N A Z 1M N±1 + e ± + ν. (2) Write this equation down and describe the process and each term. Is mass conserved? 5. How will you experimentally measure the velocity of the electron, v, and its momentum, p?
6. Looking at the equation for the Kurie variable provided in the background theory. (a) What is the value of E 0 (kev), the difference in energy between initial and final nuclear state for this source? (b) What magnetic field does this correspond to? (use previous answer plus equation 11, and determine a numerical value. (c) What current produces this field? (use previous plus equation 13.) Again, determine a numerical value. 7. Use equations??, 14 and 15 in the theory section to derive an expression for T in terms of B. This will allow you to know which energy beta particles are hitting the detector at any given magnetic field.
1 Experiment 1 - Beta spectroscopy 1.1 Background theory The β decay process involves a nucleus decaying, and leaves a daughter nucleus with the same mass but different neutron/proton number (Z ± 1, N 1). Along with the daughter nucleus, there is the emission of either an electron or positron (the β particle), plus a neutrino/anti-neutrino. This process is mediated by the weak force, and has an interesting history relating directly to the prediction of the neutrino. The underlying processes are: p n + e + + ν (3) n p + e + ν. (4) Now imagine you don t know there s a neutrino there (as they aren t easily detectable, this is understandable); when scientists first observed Beta decay, they thought just an electron/positron was emitted. By conservation of energy and momentum, all the electrons would have the same energy an energy determined by the difference in the rest mass energies of the initial and final nuclear states: E 0 = E initial E final. (5)
However, an early beta particle energy spectrum (1935) measured from Radium (Figure 3) showed a continuous spectrum: electrons were emitted with energies from 0 to E 0. Two-body kinematics could not account for this behaviour; there therefore had to be a third, as yet unobserved, particle taking some energy away. Taking into account the contribution from an anti-/ neutrino, nuclear β ± decay can be expressed as: A ZM N A Z 1M N±1 + e ± + { ν ν (6) Figure 3: Beta spectrum from F. A. Scott s 1935 paper Energy Spectrum of the Beta Rays of Radium E.
The decay can only occur if the conditions are energetically favourable, with the binding energy of M higher than that of the daughter nucleus M. While the mass number is conserved, the change particle type causes a slight difference in the rest masses of the parent and daughter nuclei. 1.2 Constituent particles and the Kurie variable. Figure 4 shows a Feynman diagram of beta decay. In terms of constituent particles, you see that this is due to the transition from up to down quark. This is mediated by a W-boson, which, when then decays into positron and neutrino. Luckily for us, we are able to assume that at the energy scale we are working at that the decay process occurs at a single point and that the mother and daughter particles will essentially be stationary in the initial and final state 2. We call this assumption that of a contact interaction. The Feynman diagram in Figure 4 is not just a pretty picture! When you study graduate-level Quantum field theory you can actually go through and calculate a value from this called a matrix element, M if. Squar- 2 Think conservation of momentum...
p + { d u u d u d { n W + e + Figure 4: Feynman diagram of β + decay. Note that the particles in the brackets are quarks, and the final state particles (n, e +, ν e ) are on the right. ν e ing this value gives a measure of the probability that the interaction will take place as shown. Of course this is complicated, and we won t be getting you to do this here. Instead we will introduce an experimental way of dealing with these abstract notions of probability. Let us begin by assuming that the electron is emitted with a momentum between p and p + dp: The number of final states with electron momentum in this range (from the total range of physically possible final states) is given by: ρ f = 16π2 h 6 c 3 p2 (E E 0 )dp
where E is the electron kinetic energy. 3 The probability per unit time that the decay will produce an electron with momentum between p and p + dp is defined as: w(p)dp = 4π2 h M if 2 ρ f Fermi s Golden Rule As β decay is a weak interraction, we can consider M if, the matrix element from before, to be approximately constant.to account for the interaction of the electron with the electrostatic field of the nucleus, the Coulomb correction factor, F (Z, p), also called the Fermi function, is introduced. The maximum amount of energy initially available to the electron must be greater than E 0 because the electron loses energy as it moves away from the nucleus. This is a dynamical correction to M if 2. We write this correction factor as: 2πη F (Z, p) = 1 e 2πη, where η = Ze2 4πɛ 0 hv and v is the electron speed. Combining these contributions, we find: w(p)dp = 64π4 h 7 c 3 M if 2 p 2 (E E 0 )F (Z, p)dp. (7) 3 Bear with me- this mathematical notation is just useful for describing the derivation, and you don t need to understand where it comes from
Like we said before, we don t want to go about actually calculating M if. Instead we can now define the Kurie variable K(Z, p) to be K(Z, p) = w(p) p 2 F (Z, p) = 8π h 7/2 c 3/2 M if (E E 0 ). (8) Looking at equation (8) and remembering that our matrix element is approximately constant, we expect a region where K(Z, p) plotted against E will produce a straight line with negative slope. This region is called a Kurie Plot. The Kurie plot is only valid for a narrow range of energy levels, because we have made a number of assumptions in getting to this point. In non-linear regions, the plot is not a true representation of the situation being modelled. As w(p) can be related to the number of counts at each β energy, we can define the experimental Kurie variable as: K(Z, p) = N(p) p 2 F (Z, p). (9) We will use this in our experiment to determine a value for E 0.
1.3 Determining the energy of Beta particles hitting the detector. A particle of mass m, charge q, and velocity v, moving in a uniform magnetic field B, experiences a force F where F = qv B. (10) If B is perpendicular to v then the particle will move in a circle of radius r where mv 2 r = qvb (11) with momentum p = qrb. We can therefore choose the momentum of the charged particles by changing the magnetic field given that the path taken from source to detector is fixed. As we know from Maxwell s equations, electric and magnetic fields are related. Ampere s law states Bdl = µ 0 I enc, (12) where B is the magnetic field, µ 0 the magnetic constant and I enc the net free current. This shows that a magnetic field can be generated by either a changing electric field or by electrical current. An increasing current implies an increasing magnetic field. The linear relationship between the current and the magnetic
Figure 5: Graph showing the linear relationship between current and magnetic field. There are some things wrong with this graph - how would you improve it? field can be seen in figure. 5. We will use Figure 5 and its trend-line to find the magnetic field through the chamber using your current readings: B = 0.1315I + 0.0031. (13) Look at the situation carefully and you should be able to see that changing B will allow you to select which kinetic energy of the beta particles will hit the detector. We should then derive a relationship between magnetic field strength B (solely dependent on your current through the electromagnet), and the energy of the particles hitting the detector T. This is to be done in the pre-lab questions.this will allow you to know which energy β particles are hitting the detector at
any given magnetic field. You should note that the relativistic consideration of momentum gives us pc = E 2 m 2 c 4 (14) where E the total energy is given by E = T + mc 2 (15) and T is the particle s kinetic energy. 1.4 Beta spectroscopy - Equipment 1.4.1 The magnetic spectrometer, Geiger-Müller tube and dead time To measure the energy of the electrons in the experiment we use a variable magnetic field to guide our electrons to a Geiger-Müller tube. An easily ionisable gas is contained within a thinwalled chamber. After entering the detector, a β ionises the gas. Each collision liberates a secondary electron which is accelerated towards the anode, together with the original β. An applied electric field in the chamber provides extra energy to the electrons as they travel towards the anode, and the mean free
(a) Figure 6: (a) Magnetic spectrometer showing the configuration with the 147 Pm source and detector. The radius of the path the β traverses is r = 3.80 ± 0.01 cm. (b) Operation of a Geiger-Müller tube. (b) path of electron in the gas is very small, so the electrons undergo many collisions. This process is an electron cascade which results in a large detectable charge at the anode. Similarly, the positively charged ions (what are the ions in this case?) move towards the cathode but much more slowly as they are significantly heavier. If another β enters the detector before these ions are discharged at the cathode no cascade is possible as recombination of the ions and electrons will take place. The tube is thus useless until the first event
has been cleared from the chamber. This is known as dead time, denoted by τ. It comes about as a combination of the detector and electronics, which require a finite time to collect the charge from the detector coming from each detected particle. The dead time for the detector you will be using has been measured, at a count rate of 1000 counts/sec, to be 20.0±0.5 µs. To a good approximation, the dead time may be considered constant for the spectra you will measure. The Geiger tube has a dead time τ, and N counts are observed in a time t, giving the true number of events N as: N = N 1 τ t N (16) 1.5 Beta spectroscopy - Procedure First, make sure you are familiar with all of the equipment. Identify both the test and measurement chamber, the electromagnet, the source and follow the wires to understand the electronics.
1.5.1 Setting up the electronics You first need to degauss the measurement chamber. To do this: 1. Plug the magnet into the degaussing supply and turn the supply up to 11. 2. Turn on the supply, then slowly decrease the supply to zero. 3. Perform this degaussing procedure two or three times to ensure the process has been effective. Question 1 Explain the end result of degaussing the system. Why is this necessary? Once the electromagnets are effectively degaussed: 4. Connect the DC supply to the magnet. 5. Ensure that the vacuum pump is connected to the chamber and switch on the pump and gauge. After a few minutes the pressure in the chamber should be below 10 1 Torr. 6. Connect up the circuit shown in figure 7.
Detector Amplifier SCA Counter High voltage Oscilloscope Timer Figure 7: Electronics for the detection of β particles. 7. Set the electronics to the following settings and then switch on the NIM bin: High voltage unit : Try 700 V initially, but if counts are too low, move up to 780 V. Amplifier: Course gain = 16; fine gain = 4; input = positive; output = unipolar. SCA: ULD = 10 V; LLD = 2 V. Counter: Discriminator = 3.0 V. Timer: 100 s 1.5.2 Taking measurements 1. Start measuring the number of counts per 100 s in the current range 0 to 400 ma, taking suitable steps across the range (check with your demonstrator). Remember NEVER to backtrack, always increasing the current, and allowing for a few seconds before counting each time you change the current to allow the field to stabilise.
2. Note: if it s absolutely necessary for you to backtrack, you can degauss with the proper chamber inserted. However, you aren t able to verify the degaussing as there s no space for the Hall probe. 3. If you do need to degauss the system again, you must also start your measurements again. Question 2 Why would backtracking render the calibration useless? 4. As the data are taken, observe the output pulses from the amplifier on the oscilloscope. Question 3 Draw, on scaled axes, the shape of the output. At what current does the output appear smallest? 5. Check, for a current of 100 ma, that the output does not exceed the upper level on the SCA. Discuss the shape of the pulses in terms of the operation of the detector. Question 4 Are the pulses consistent with the quoted dead time? Could the dead time have changed since it was first measured? What do you think is the main limiting factor for the dead time?
Question 5 In which direction are the β particles moving as a result of the magnetic field? Draw a diagram. 1.5.3 Analysis 1. After taking your measurements, increase the current to 500 ma and take one more data point. This will serve as the background measurement, which must be subtracted from each of your previous measurements. Question 6 Why is this a suitable current at which to take such a background reading? Why not higher/lower? 2. You should now have a table of the measurements along with the appropriate uncertainties. The analysis of these data requires a series of calculations and great care must be taken to ensure the transformations are done correctly. The steps you make should be made clear; be sure to comment on all aspects of the analysis. 3. Use Excel to correct the number of counts for background and dead time, and calculate the electron kinetic energies from the deduced momenta. Plot the corrected counts against the kinetic energy (in kev). Explain the shape of
the spectrum, discussing its features in terms of both the physics and the apparatus used. 4. From this spectrum, and again using Excel, transform the appropriate region of the spectrum into a Kurie plot. Some notes about our experimental Kurie plot: As we are trying to find a value for E 0, we can set the Fermi function, F (Z, p) = 1. The Fermi function takes into account the effect of a negative charge being close to the nucleus, and experiencing a pull which will slow it down. This effect is most noticeable at lower energies, and minimal at higher energies (close to E 0 ). This leaves our experimental Kurie variable as: N K(Z, p) = p. (17) 2 Using the relativistic momentum from earlier you should be able to obtain an expression for K where the only variable is n, the number of counts. Plotting the Kurie variable against particle energy should give a straight line, with an x intercept at E 0. Remember, however, due to our sim-
plification of the Fermi function, this relationship will only hold true for values approaching E 0 so look at the graph, select the region where this assumption holds true, and determine a value for E 0. Question 7 How does your deduced value for E 0 compare with the quoted value for the decay of 147 Pm? Question 8 Is the experiment sensitive enough to allow for a determination of the neutrino mass? Discuss.
2 Experiment 2 - Gamma spectroscopy 2.1 Background theory γ radiation is often the product the decay of an excited final state nucleus of another nuclear decay and/or reaction. This daughter nucleus then decays to its ground state by γ emission. Figure 8 shows the production of a 14.41 kev γ from 57 Fe as a result of the decay of 57 Co. 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 Europium-152. The γ- ray energies these isotopes emit are given in Table 1. The half-life of each of these isotopes is many years, so they are suitable as a source of very low energy γ rays. Source E γ (MeV) 137 Cs 0.032 0.662 22 Na 0.511 1.275 152 Eu 0.122 0.245 0.344 0.779 0.964 1.086 1.408 Table 1: Radioactive sources used for the calibration, and the energies of the photons produced.
2.1.1 Interaction of Gamma-rays with matter To detect γ-rays, we need them to interact with matter. Each of the peaks in figure 9 are due to interactions with matter. Starting from the highest energy peak, we can describe the peaks and the processes as follows: The photopeak The highest energy individual peak for a given spectrum is its photopeak. This peak is pro- 706.41 kev 57 Co 0.18% Electron capture 366.74 136.47 14.41 99.82% 57 Fe Figure 8: 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.
C ou n ts Backscatter peak Annihilation peak Secondary photon continuum Continuum Double escape peak Primary photon continuuum Single escape peak Compton edge X-ray escape peaks Photopeak Eγ - Ec 511 Eγ - 1022 Eγ - 511 Ec Eγ - Ex Eγ Energy (kev) Figure 9: 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. duced when γ-rays deposit (almost) all of their energy in the detector through the photoelectric effect. A photon is absorbed by an atomic electron, and the electron is ejected with kinetic energy (almost) equal to the photon energy. Some energy is lost in doing work to unbind the electron from the atom, and a negligible amount of energy is absorbed through nuclear recoil. Depending on which shell the electron was excited from, light in the visible spectrum, or oc-
casionally x-rays, can be detected as this electron ionises surrounding material. The crosssection of a photoelectric interaction is proportional to Z 4, so a high-z material will return a higher number of photopeak counts. The scintillator crystals in our detection system are doped with thallium (Z = 81) for this reason. The x-ray escape peak This peak is due to the photoelectric x-ray being detected. The original photon has lost some energy to the atom (overcoming the higher innershell binding energy) and so the x-ray escape peak is slightly below the full energy photopeak. The Compton edge In this case, the photon undergoes multiple scattering off of electrons. In the detector material, the process affects the outermost electrons which are very loosely bound and so we consider free to a good approximation. The scattered γ can undergo numerous scattering events, in turn scattering more electrons, which themselves produce light. This produces a range of energies in the detector up to a maximum, which is defined by:
E e = E γ α(1 cos θ) 1 + α(1 cos θ), (18) with E e the scattered electron energy, E γ the photon energy and α = E γ /(m e c 2 ), with m e = 511 kev/c 2, the rest mass energy of the electron. The process is illustrated below: γ 1= hv γ 2= hv' Figure 10: Kinematics of Compton Scattering. θ Φ e - Single and double escape peak If the incident photon energy is more than twice the rest mass energy of an electron, the photon can be converted into an electron-positron pair. The positron will then re-combine with an electron, producing two 511 kev photons 4. 4 This is the rest mass m e c 2 of an electron/positron
If one of these photons escapes the detector, the energy deposited is the initial photon energy minus 511 kev. If both photons escape, the energy deposited is the initial photon energy minus 1022 kev. These give rise to the single and double escape peaks, respectively. Annihilation peak Alternatively, if the electron-positron recombination occurs outside of the detector, only one 511 kev photon is able to reach the detector. In this case we observe an annihilation peak at 511 kev, where no other energy reaches the detector. The backscattering peak This peak shows as a result of a photon Compton scattering against an obstruction on the way to the detector. It is then left with energy equivalent to the photopeak minus the Compton edge value. Lastly, it is important to note that beyond the photopeak, there will likely be peaks that appear as combinations of the other peaks. They are treated as single events because the detector is a finite size and occasionally treats two events as a single reading.
2.1.2 The signal When detecting these 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, two pair production or escape peaks can be observed. One or both of the 511 kev photons may escape the detector, producing peaks at 511 kev and 1022 kev below the full energy peak. Compton scattering produces electron with continuous energy values up to a maximum, with the maximum observed as the Compton edge, just below the photopeak. The resolution of the detector For a given energy spectrum, we can make comments on the resolution of the detector used. But resolution in this context isn t the same as the resolution you might be used to when talking about visual displays (e.g.: of your phone). Rather, the resolution we are interested in is defined as the relative resolution of energies in the spectrum with reference to the full width (at) half maximum (FWHM). The full width half maximum is the width of a peak at half of its maximum height. The resolution of each peak is then defined as:
Figure 11: Note that energy and channel number are proportional to each other on the x- axis. Resolution(%) = E FWHM E 0 100 (19) Notice that the resolution is given as a percentage. Here, a small percentage is desirable, as a smaller resolution means each peak can be more easily individually resolved. (Each peak is a sharp line in ideal conditions). The resolution of a given detector is important to know, and you will be asked to produce a graph of energy resolution vs. energy at the end of this experiment.
2.2 Gamma spectroscopy - Equipment Gamma source Detector Amplifier Multi-channel analyser Nal crystal Pre-amp Digital oscilloscope Computer (a) Detector components Figure 12: The detector system (a) Configuration of NaI(Tl) detector with photomultiplier tube. (b) Electronic components for detecting γ rays. 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) (b)
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.
2.3 Gamma spectroscopy - Procedure 2.3.1 The sources Firstly, notice where your demonstrator has placed the sources. They should be in a bag inside a leadlined chest. Take them out one at a time, and note the elements you have. Three will be identified, and one will be covered in lead, with only a small gap to lead γ rays through. This is your unknown source. When not in use, place all of the sources in the lead box. 2.3.2 Gain of the detection system and MCA You will begin by turning on the detector and setting it to show the correct γ energy range. 1. Starting with the detector, identify each piece of equipment and follow the wiring as per figure 12. 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. Is a signal is present on the oscilloscope? Question 9 Describe the pulse on the oscilloscope and note how it fluctuates. Why do we see these fluctuations? 5. Remember that the MCA accepts voltages between 0 and 5 V. The gain on the amplifier should be set so the pulses on the oscilloscope are observed across 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 10 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 11 Do you notice any background counts, and do these change linearly with changes in the HV supply? 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?
2.3.3 Measuring spectra You can now move on to collecting spectra for the individual isotopes. 1. Reset the HV to 800 V and allow a few minutes for the power supply to stabilise. Question 12 Which source should we examine first? Why? Refer to table 1. 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. Make sure the axis along the bottom of your graph is Channel Number, not Energy, and print your spectrum. 4. Note on the printed spectrum the important features, and channel numbers and channel error of the various peaks and features. 5. Repeat these steps for the other two sources. 6. Using the known energies of the photons in the three sources given in Table 1 plot a graph of energy against channel number.
7. 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. What numerical relationship did you determine between energy and channel number? Include error analysis. Question 13 What features are common to all spectra? Are there any features which appear only in certain spectra? Question 14 What are the half-lives of each of the elements we re using? Do you think the number of counts are dependent only on the mass of the source? 2.3.4 Identifying unknown features Now you can identify the lead-encased source! 1. Collect the spectra of the unknown source and determine the energies of the peaks, with error. 2. Measure the FWHM of the photopeaks in your unknown spectra and calculate the resolution for a range of peaks, including errors as you go.
3. Plot a function of resolution against E γ and discuss your findings. How accurately can you resolve each peak for a given energy? 4. Now you have a good idea of errors in the system, compare your detected energies to the ones listed in table 2 and try to identify the unknown source. 5. It s best to use a decent number of points, and preferably peaks with good resolution. Question 15 What is the unknown source? Were the low or high energy peaks closer to the given values? Question 16 What function of the detection system causes the resolution observed? Remember how the detector works to observe a photon. This should demonstrate to you the power of both a properly calibrated detector and the importance of recognising resolution limits in the process of identifying radioactive sources.
Table 2: Table of various γ source emission energies. Source Emission energy (kev) 137 Cs 320 662 22 Na 511 1275 152 Eu 122 245 344 779 964 1086 1408 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 226 Ra 53 186 242 295 353 609 769 1120 1238 1378 1764 207 Bi 570 1064 1770 56 Ni 270 750 812 1562 60 Co 1173 1332 75 Se 97 121 136 265 280 401 88 Y 898 1836 e + e 511 3 Useful data
Quantity E 0, for the decay of 147 Pm Planck s Constant, h Speed of Light, c Permittivity of free space, ε 0 Electron Charge, e Electron-Volt, ev Electron Rest Mass, E e Electron Rest Mass, E e Value 224.5 ± 0.4 kev 6.6256 10 34 Js 2.997925 10 8 ms 1 8.85416 10 12 Fm 1 1.6022 10 19 C 1.6022 10 19 J 511 kev c 2 9.1096 10 31 kg