Medical Imaging Group Study Worksheet 2014/15 1. Compton Cameras Compton Cameras are devices that exploit the fact that the energy of a Compton scattered photon is related to the angle through which it is scattered. They usually comprise a thin scatter detector with very good position resolution and a separate segmented absorber to measure the energy of the scattered photon. Knowing the energy of the Compton scattered photon and the precise location of the origin of the scatter it is possible to project a cone backwards from the point of scatter toward the source. The source is constrained to lie somewhere on the surface of the cone. It is a cone rather than a line due to the ambiguity in the direction of the incoming photon. At first sight this doesn’t sound very useful. However, a second scattered photon will generate another cone and the intersection between the two cones reveals the location of the source. In principle, the ambiguity in the photon direction in the back projection could be eliminated if the direction of the recoil electron could be measured in the scattering detector. E
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Scattering plane Absorption plane
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2. Compton Scattering Figure 1. A schematic diagram of a Compton Camera illustrating the process of back projection to locate a source. The photon enters from the left and Compton scatters through an angle θ in the first detector and is then absorbed via the photoelectric effect in the second detector. In this diagram the photon scatters into the plane of the paper. Back projecting several photons locates the position of the source. 2
You should make yourself familiar with the ways in which electromagnetic radiation interacts with matter. The two main types of interaction that are of interest for this study are Compton scattering and the photoelectric effect. For completeness, there is also Rayleigh scattering and pair production, although the latter is not observed at energies below 1.022 MeV. Compton’s formula relates the angle of scatter to the change in wavelength between the outgoing and incoming photon. This can be rearranged to give an expression for the Page 1 of 4 energy of the outgoing photon in terms of the incoming energy and angle of scatter. The probability with which photons of a given energy are scattered through a certain angle can be calculated from the Klein-­‐Nishina formula. This is normally written as the differential cross section dσ/dΩ per electron. Be warned that the formula given in Krane [1] on page 200 is incorrect. (However, please note that Krane is otherwise an excellent book.) The correct formula can be found on Wikipedia [2]. The total Compton cross section can be found by integrating over all solid angle. Exercise 1. Use the trapezium rule to perform a numerical integration of the Klein-­‐Nishina formula over solid angle to obtain the total Compton cross section for 662 keV gamma rays. You can do this in EXCEL for example. Note that the accuracy of your answer will depend on the number of panels you choose for the integration. You can compare your answer with the analytic solution given by equation 7.17 in Krane. (Hint: you need to find an expression for a unit of solid angle, dΩ, in spherical coordinates.) [9 marks] 3. Gamma-­‐ray Attenuation Gamma rays will be attenuated in the scattering detector both by Compton scattering and by the photoelectric effect. For efficiency we want to ensure that the Compton cross section is large in the scattering detector and that the photoelectric cross section is large in the second detector. Cross sections are energy dependent. The 4
photoelectric cross section also varies strongly with the Z of the absorber (~ Z ), whereas Compton scattering is equally likely off any electron and so the cross section per atom is proportional to Z. 2
Attenuation coefficients are often quoted in units of cm /g. These can be converted into macroscopic cross 3
sections by multiplying by the density of the absorber. If the density is in units of g/cm , macroscopic cross –1
sections have units of cm . The macroscopic cross section is otherwise known as the linear attenuation coefficient, µ. It gives the probability per unit length of a certain process occurring. When photons are incident on an absorber the intensity, I, varies exponentially with distance such that I(x) = I(0) exp(–µx). Attenuation coefficients are additive, so µtotal = µCompton + µphotoelectric. For Compton scattering, µCompton = NZσ, where N is the number of atoms or molecules per unit volume, Z is the atomic number of the material and σ is the total Compton cross section per electron. A useful resource for finding information relating to X-­‐ray and gamma-­‐ray cross sections is the collection of NIST databases. In particular, the XCOM database [3] gives you the facility to lookup attenuation coefficients for elements and compounds specifying up to 75 user-­‐defined energies in each search. 3
Exercise 2. Sodium iodide (NaI) crystal has a nominal density of 3.67 g/cm . Look up the Compton and photoelectric attenuation coefficients for 662 keV gamma rays in NaI. (a) Work out the interaction probability for 662 keV photons in 1 inch (2.54 cm) of NaI at normal incidence. What is the fraction of photons that interact via Compton scattering, i.e. the ratio Compton/total? (b) Use the Compton attenuation coefficient at 662 keV in NaI to verify your calculation of the total Compton cross section per electron in Exercise 1. (Hint: the number density of a material is given by N = ρNA/A, where ρ is the density of the material, NA is Avogadro’s number and A is the atomic weight (g/mol), which is roughly given by the mass number.) Why might the measured ratio of events in the Compton continuum to those in the photopeak be smaller than the ratio calculated in b)? [12 marks] 4. Design Considerations The back projection shown in Figure 1 is perfect; it is what you would expect from a system free of any detector resolution effects. There are two types of resolution that concern us in this project: position resolution in the scattering plane and energy resolution in the absorption plane. You might argue that both effects are present in both detectors and you would be right. However, in certain situations it might be possible to argue that one effect is more important that the other. For example, let us assume that it is possible to build a scatter detector with excellent position, but poor energy resolution and an absorption detector with excellent energy resolution, but poor position resolution. If the final photon energy is well Page 2 of 4 measured then the angle of scatter is known to high accuracy, assuming that the initial photon energy is known. Combining this information with a known point in the scatter detector gives us all the information we need. Redundancy is a good thing in general, but it may not be cost effective to devise a system where both elements have both high energy resolution and high position resolution. It may be possible to design a system where the scatter detector has high energy resolution and the absorber has high position resolution. It may also be possible to conceive a camera system that has both high position resolution and high energy resolution in one detector (e.g. the absorber) and a passive scattering plane. However, most examples of Compton cameras that you can find in the literature seem to combine good position resolution in the scatter detector and good energy resolution in the absorber. You need to decide if this geometry the best choice. There are several considerations for the scatter detector. It should have a large Compton/total ratio in the energy range of interest. It should probably be a pixelated detector of some kind, which will determine the limiting resolution in two dimensions perpendicular to the incoming photons. It should probably also be a thin detector to give good resolution in the third dimension, parallel to the incoming photons. The issue is how to get sufficient thickness (interaction probability) without compromising efficiency, or requiring too many channels (increased cost)? As well as pixelated silicon detectors, there are also strip detectors and planar drift detectors. There are also other materials to consider, such as Germanium, which has a higher energy resolution but needs to be operated at liquid nitrogen temperature. There are also some interesting developments in scintillating fibre trackers that might be worth considering as an alternative to pixelated silicon detectors. Scintillating fibres can be made small giving good position resolution in two dimensions and can be readout via photomultiplier tubes (PMTs) connected to each end of the fibre. The time difference between the two ends potentially enables the position along the fibre to be located to high accuracy. Bundles of fibres can readout using multi-­‐anode photomultiplier tubes (MAPMTs), or silicon photomultipliers (SiPMs). It might be possible to procure a couple of MAPMTs and some scintillating fibres to do some timing tests during this Group Study. Gas detectors, such as proportional chambers, might also be considered for the scatter detector. 5. Resolution For all detectors, fluctuations in measured energy stem from the stochastic nature of the processes by which the well-­‐defined energy given to an atomic electron via the photoelectric effect or Compton scattering is shared amongst the information carriers. In scintillators, the information carriers are scintillation photons; in semiconductors, they are electron-­‐hole pairs; in gas detectors, they are electron-­‐ion pairs. Since a large number of information carriers, N, are typically produced on average, in any given event the observed number is expected to fluctuate by ±sqrt(N) about the mean in accordance with Poisson statistics. This translates into an uncertainty in the energy in the absorption detector, which, in turn, translates into an uncertainty in the scattering angle. The energy resolution is defined as R = FWHM(E)/E, where FWHM(E) is the full width at half maximum measured in units of energy and E is the energy. Since, N is proportional to E and the FWHM is simply related to the standard deviation of N, R is expected to vary as 1/sqrt(E). Exercise 3. The energy resolution of a certain 2 inch sodium iodide detector is found to be 7% at 662 keV. The detector is placed 150 cm from a thin scattering rod, which is illuminated by a 662 keV collimated source. For photons scattered at an angle of 30 degrees into the centre of the detector, what is the uncertainty in the scattering angle from the measurement of the scattered photon energy? How does this uncertainty compare to the angular resolution determined by the geometry of this configuration? Comment on your findings. [9 marks] The situation for semiconductor detectors is similar. However, in addition to the effects of energy resolution we also have to be concerned with the issue of noise. Noise becomes increasingly problematic the smaller the signal you are trying to measure, which is often the case in the scattering detector. This problem is exacerbated if we consider building a Compton camera for X-­‐ray imaging when the incoming energy is already rather small. Overall, energy resolution will depend on the quadrature sum of the noise contribution and the contribution from counting statistics. In silicon imaging sensors there are various sources contributing to the noise. The main sources are readout noise, thermal noise (dark current) and noise induced by light falling on the detector. Noise is usually quoted in terms of electrons. Readout noise is irreducible and reflects that all Page 3 of 4 real devices have some inherent variability in their response. For the Medipix2 (Timepix) chip, the total noise is –
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quoted to be around 100e RMS per channel. Exercise 4. The mean energy for electron-­‐hole pair production in silicon for electrons is approximately 3.7 eV. Calculate the energy resolution in terms of electrons for a 662 keV photon that scatters through an angle of 30 –
degrees in a silicon detector with a noise value of 100e RMS. Compare the noise contribution to the counting 2
statistics in the detector. Given a pixel pitch of 55 µm and using the NIST ESTAR database [4] to find the CSDA range of the recoil electron in silicon, comment on whether it would be possible to track the recoil electron. Repeat the calculation for a 140 keV photon. You might also like to calculate the probability that a 662 keV and 140 keV photon will Compton scatter in 300 µm of silicon. Based upon your findings, comment on the suitability of the Timepix chip as the scatter detector of a Compton Camera. (Suggestion: you might find it convenient to work your solution in EXCEL so that it can be generalised later for other scattering angles.) [15 marks] References [1] Kenneth S. Krane, Introductory Nuclear Physics, Wiley, 1988. [2] http://en.wikipedia.org/wiki/Klein-­‐Nishina_formula [3] http://physics.nist.gov/PhysRefData/Xcom/html/xcom1.html [4] http://physics.nist.gov/PhysRefData/Star/Text/ESTAR.html 1
RMS = Root Mean Square. For a Gaussian distribution RMS is equivalent to one standard deviation. 2
CSDA = Continuous Slowing Down Approximation Page 4 of 4