Page 1 of 5 Queen s University Faculty of Applied Science; Faculty of Arts and Science Department of Physics, Engineering Physics and Astronomy PHYS 352 Measurement, Instrumentation and Experiment Design Final Examination April 2010 Instructor: M. Chen Time: 2 hours Please put your student number on the front of all answer booklets. Approved calculators are permitted. No outside notes or pages are allowed. Complete the short-answer questions section and then answer two out of the remaining three longer questions. Please indicate clearly which two answers you want marked (otherwise your first two answers will be marked). Please note: Proctors are unable to respond to queries about the interpretation of exam questions. Do your best to answer questions as written. Please note: the candidate is urged to submit with the answer paper a clear statement of any assumptions made if doubt exists as to the interpretation of any question that requires a written answer. This material is copyrighted and is for the sole use of students registered in PHYS 352 and writing this exam. This material shall not be distributed or disseminated. Failure to abide by these conditions is a breach of copyright and may also constitute a breach of academic integrity under the University Senate's Academic Integrity Policy Statement.
Page 2 of 5 Short-Answer Questions (24 points) Each answer need not be more than one to a few sentences in length (point form is acceptable too if you write enough to convey your understanding). (a) Explain how a photoconductive sensor made of CdSe measures light intensity. (b) A thermistor bolometer is used to measure IR radiation intensity. You could also detect IR with a photoconductive PbS sensor. What s an advantage of the PbS detector (with explanation)? What s an advantage of the thermistor bolometer over the PbS detector? (c) Name two types of photon detectors that have sensitivity to single photons. (d) State two reasons why it is a good idea to operate a photodiode under reverse bias. (e) When an electron beam accelerated to several 10 s of kev energy is incident on a metal target two different types of X-rays are produced. Very briefly describe the two types of X-rays produced. (f) Nuclei that have too many protons might undergo radioactive decay. Write down two decay reactions that can occur for these nuclei. (g) Explain why heavy charged particles incident on biological tissue have a depth-dose curve that peaks near the maximum penetration depth of the radiation (i.e. at the end of the charged particle s path). (h) Explain two physics reasons why energy loss by electrons is quite different from energy loss by heavy charge particles (how they are different from the modelbuilding perspective). (i) Gamma ray and X-ray photons undergo different interactions in matter. What interaction is the predominant one at very high energies? What interaction is the predominant one for lower energies (like 10 s of kev)? (j) A gas radiation detector can operate in proportional mode. Describe briefly how it works in this operating region. (k) What are two reasons why the energy resolution in scintillation detectors is not as good as than the energy resolution in a semiconductor or gas radiation detector? (l) Why is germanium chosen (over silicon) when making semiconductor gamma-ray detectors? For the gamma-ray detection efficiency to be as large as possible in a Ge detector what are the other main considerations (in designing/building and/or operating)?
Page 3 of 5 Question #1 (24 points) [Choice: Answer 2 out of 3 Questions, this is #1 of the 3] X-ray Detection, Photon Interactions, Photodiodes, Light Detectors A side-view diagram of an X-ray panel is shown below. material µ/ρ [cm 2 /g] @ 30 kev µ/ρ [cm 2 /g] @ 80 kev density [g/cm 3 ] CsI 9.045 3.677 4.51 Si 1.436 0.223 2.33 Bone 1.331 0.223 1.92 Soft Tissue 0.379 0.182 1.06 relative permittivity of silicon ε r =11.68; permittivity of the vacuum ε 0 =8.854 pf/m (a) The panel produces scintillation light when X-rays interact in the CsI layer. The energy of the X-rays incident on the CsI layer is 30 kev. What fraction of these X- rays (i.e. the efficiency) will produce scintillation light? (b) If it were just 30 kev X-rays hitting a silicon detector array, how thick would the active region in the silicon detector have to be to achieve 45% efficiency? (c) A graph of µ/ρ versus photon energy in CsI is shown above. What is the dominant interaction that takes place for 30 kev X-rays in the CsI? What would happen if the X-ray energy were increased to 40 kev? Would this be good/bad/not important for an imaging X-ray panel (with explanation, please)? (d) Prof. Chen had X-ray images of his foot taken prior to surgery. Why is 30 kev a better choice than 80 kev for the energy of those X-rays? (e) The photodiode array has 100 µm pixel size. The depletion region thickness in a photodiode determines the junction capacitance. The junction capacitance per area is a function of the reverse bias voltage: C [nf/cm 2 ] = 1 / 0.26V + 0.20, where V is in volts. Calculate the depletion region thickness if the reverse bias voltage is 50 V. (f) Would an array of photomultiplier tubes with the typical bi-alkali photocathode be better for detecting these scintillation photons than a Si photodiode array? Explain why or why not. Facts about CsI: 56,000 photons/mev, 550 nm emission peak.
Page 4 of 5 Question #2 (24 Points) [Choice: Answer 2 out of 3 Questions, this is #2 of the 3] Charged Particle Energy Loss, Resolution, Scintillator, Electron Energy Loss A cosmic ray muon is a heavy charged particle (same charge as an electron). The rest mass energy of a muon is 105.66 MeV. The electron rest mass energy is 0.511 MeV. A plastic scintillator panel is used to detect muons that pass through. 1 de ρ dx = [4πr 2 e m e c 2 N A ] z2 β 2 4π N A r e 2 m e c 2 = 0.307 MeVcm 2 /g; E = γ mc 2 ; γ = ln I eff = a i Z i ln I i ; Z eff = a i Z i ; A eff = a i A i Z eff Z A B(v); B(v) = ln(2m ev 2 ) ln(1 β 2 ) β 2 I a i is the # of atoms of 'i' in the molecule mean excitation energy for hydrogen: I = 19.2 ev mean excitation energy for carbon: I = 78.0 ev 1 1 β 2 ; β = v c (a) A muon with total relativistic energy 1 GeV (i.e. 1000 MeV) goes through 5 cm of plastic scintillator. The composition of plastic scintillator is CH 2. The density of plastic scintillator is 1.032 g/cm 3. Calculate how much energy (the typical, average amount) the muon deposits in the plastic scintillator. (b) You must have made an assumption when calculating the energy loss along the path of the particle in (a). Justify that assumption (lengthy calculation not required though a number or two would help your explanation). (c) The light output from plastic scintillator is 13,000 photons/mev. Scintillation light from the plastic scintillator propagates down the length of the panel via total internal reflection to a photomultiplier tube, which collects about 10% of the light produced by the muon passing through. The average quantum efficiency of a photomultiplier tube over the emission spectrum of the plastic scintillator is 25%. Calculate the expected energy resolution of this measurement. [Note: if you don t have an answer for (a) assume that 10 MeV energy was deposited in the scintillator.] (d) One way to measure the energy of a muon is to measure its time of flight the time interval between a muon traversing two detectors separated by a distance. Why would a plastic scintillator be a good choice for making this measurement? (e) If this were a 1 GeV electron passing through instead of a muon, explain the dominant energy loss mechanism. Facts about electrons in plastic: the critical energy is E c = 100 MeV, radiation length in plastic L rad = 42.9 cm. How much energy does the electron lose going through 5 cm of plastic scintillator (on average)?
Page 5 of 5 Question #3 (27 points) [Choice: Answer 2 out of 3 Questions, this is #3 of the 3] Geiger-Müller Counters, Compton Scattering, Electron Energy Loss A Geiger-Müller counter is used to detect gamma rays. The cylindrical counter has dimensions shown in the figure below. It is filled with 0.1 atm Ar gas (and a small amount of quench gas which you can ignore in subsequent calculations). hν i hν f = 1+ hν i (1 cosθ) 2 m e c N Av = 6.022 10 23 material total Compton cross Z A density section per electron argon 0.26 10 24 cm 2 18 39.95 1.784 g/l at STP steel (iron) 0.26 10 24 cm 2 26 55.85 7.85 g/cm 3 (a) Calculate the interaction probability for 662 kev gamma rays crossing 5 cm of gas (ignore the anode wire). Don t forget the gas is at 0.1 atm pressure. [Note: no formula given since you derive it in your answer.] (b) Calculate the interaction probability for 662 kev gamma rays crossing 1 mm of steel. Where do gamma rays interact in a Geiger counter? (c) What is the maximum kinetic energy of a Compton scattered electron produced by a 662 kev gamma ray that interacted in the steel wall? (d) Will the Compton scattered electrons emerge from the wall into the gas? A short, approximate calculation is sufficient given the following facts for electron energy loss: at the energy being considered the electron s energy loss from collisions is 1.434 MeV cm 2 /g and the radiative energy loss is 0.024 MeV cm 2 /g. If you don t have an answer for (c) then assume Compton scattered electrons have 500 kev kinetic energy. (e) Can a Geiger-Müller counter distinguish between Compton scattering and photoelectric effect interactions of gamma rays? (f) Explain, in detail, how a gas radiation counter works in the Geiger-Müller region, making sure your explanation describes the reason behind your answer to (e).