22.01 Fall 2016, Problem Set 4 Solutions

22.01 Fall 2016, Problem Set 4 Solutions October 30, 2016 Complete all the assigned problems, and do make sure to show your intermediate work. Part I Skill-Building Problems (50 points) 1 Successive Decay Chains The molten salt reactor (MSR) could be used not u just p for power, but u to pproduce two incredibly valuable gases from its 7 Li-enriched salt coolant: Tritium 3 H and helium-3 3 He . Tritium is extremely useful as fusion reactor fuel, while helium-3 is the most effective gaseous neutron detector available. The latter is valuable enough that NASA has considered mining it on the Moon. 1. Starting with the capture of a neutron by 7 Li, write a complete set of nuclear reactions that describe the production and destruction of 3 H and 3 He. Consider that both gases can also capture neutrons to be destroyed, with microscopic cross sections σH,n and σHe,n . Rxn − 1 : 7 3 Li + 10 n → 31 H + 42 He + 10 n 3 1H Rxn − 2 : 3 2 He Rxn − 3 : Rxn − 4 : 3 1H + → 1 0n → 3 2 He − +β +ν (2) + 10 n → 42 He 4 1H → 4 2 He (1) (3) − +β +ν (4) 2. Develop a set of differential equations, similar to the Bateman equations, describing the production and destruction of 3 H and 3 He. First, we set up the equations for the concentration of each isotope, assuming we have some amount of 73 Li to begin with: dLi = −NLi σc,7−Li Φreactor dt (Rxn − 1) dH = NLi σc,7−Li Φreactor − NH σc,H Φreactor − λH NH (Rxns − 1, 2, 4) dt dHe − 3 = λH NH − NHe−3 σc,He−3 Φreactor (Rxns − 2, 3) dt dHe − 4 = NHe−3 σc,He−3 Φreactor + NH σc,H Φreactor (Rxns − 3, 4) dt 1 (5) (6) (7) (8) Next, we make an assumption that because σc,H−3 is so small, that it gives a negligible change in concentrations. This reduces the equations to the following: dLi = −NLi σc,7−Li Φreactor dt (9) dH = NLi σc,7−Li Φreactor − λH NH dt dHe − 3 = λH NH − NHe−3 σc,He−3 Φreactor dt dHe − 4 = NHe−3 σc,He−3 Φreactor dt (10) (11) (12) 3. Graph the solutions to this set of differential equations, showing the relative levels of 7 Li, 3 H, and 3 He in the reactor. Use the following data to make your graphs: σc,7−Li = 10−4 b σc,3−H = 10−10 b σc,3−He = 10−5 b λ3 H = 1.8·10−9 1 s Φ = 1·1014 n cm2 − s Without numerically solving the equations, we do know the following things: (a) 7 Li proceeds by simple exponential decay, at a pretty slow rate because of the small cross section. (b) Because 3 H decays so much faster than it is produced, its level remains very low. For short times, the slope of the 3 H level will be the negative value of 7 Li. (c) 3 He is burned even more slowly than 7 Li, so its slope will be proportional to the value of 3 H at all times. The maximum production rate of 3 He occurs at the maximum value of 3 H. (d) 4 He just builds up, very slowly, from both the burning of 3 He and 7 Li. Using this information, the following graph (showing exaggerated burn/production rates) was generated, which is quite accurate. If I were to draw it a second time, I would “stretch out” the 3 H line, and adust the slopes of 3 He and 4 He accordingly: 2 4. Now assume that 3 H costs $30,000/g and 3 He costs $53,000/g. Graph the value of the gases in the MSR as a function of time, assuming an initial 7 Li number density of 1022 atoms cm3 . Because we know from (b) above that there is almost no 3 H in the reactor at any given time, the price of the gases in the reactor is simply 53,000*3 He for most times. Therefore, except for a short blip up during peak 3 H production, it will proportionally follow the 3 He inventory. 2 Statistics and Certainty Since we know that smoking is a major source of radioactivity, smoke shops should also be major sources of radioactivity. How long would you have to count in a smoke shop to be 95% sure that you can distinguish your count from the background? Assume a new background activity of ASmoke−Shop inside the store. For this problem, we must use the relation for uncertainty in quadrature, because both the new background counting experiment and the counting experiment will have their own un­ certainties associated with their respective√count rates. Using Poisson statistics, the standard deviation of a number of counts n is σ = n. A count rate c is defined as nt , where t is the √ √ counting time. Therefore, the standard deviaton of a count rate is σc = tn = tct = ct . The uncertainties for the background count rate cb and the gross count rate in the smoke shop cg added together in quadrature become: σnet = σb2 + σg2 = cb tb 2 + cg tg 2 = cb cg + tb tg (13) Assuming we can control the background counting time tb , and we measure a known back­ ground count rate cb , then it’s up to us to find an acceptable uncertainty rate that we can determine with 95% confidence. The 95% confidence interval for our acceptable uncertainty (error) means that the following equation must be satisfied: 2σnet = (% Error) cnet (14) We substitute the expression for σnet , and the relation cnet = cg − cb , to relate the count rate in the smoke shop to the required counting time inside: 2  cb cg + = (% Error) (cg − cb ) tg tb 2  2 cb cg % Error + = (cg − cb ) tg 2 tb (15) (16) 2 cb cg (% Error) 2 + = (cg − cb ) tg 4 tb (17) 2 p cb cg (% Error) u 2 + = cg − 2cg cb + c2b tb tg 4 (18) 2 p cb cg (% Error) u 2 = cg − 2cg cb + cb2 − 4 tb tg tg 1 = (% Error)2 u p cg 2 cg − 2cg cb + cb2 − ctbb 4 cg tg = (% Error)2 u p c2g − 2cg cb + c2b − ctbb 4 (19) (20) (21) We now have an equation relating the required counting time in the smoke shop tg to all the other parameters in the experiment, including the background counting rate and time, the measured counting rate cg , and the percent error that we are willing to accept. Finally, we recognize that: 3 • Our detector will have some efficiency ηdetector (a number between zero and one) in count­ ing particles that enter the detector, • Each of the radon decay products may have a number of associated radioactive decays npart/dis associated with each disintegration, and • There will be some geometric factor relating the distributed source in the smoke shop, much like an effective solid angle Ω. Therefore, we can relate the net activity to a net count rate as follows: cn = An Ωef f ective ηdetector npart/dis 4π Substituting into our final expression yields:  tg = ηdetector (% Error)2 4  An Ωef f ective 4π  npart/dis An ηdetector (22) Ωef f ective 4π − cb 2 npart/dis (23) − cb tb Now we have a direct relation between the activity of the smoke shop, our background counting experiment, known quantities like detector efficiency, and our acceptable percentage error that can be reported with 95% confidence. Let’s throw in a few example parameters: Suppose we are willing to accept a 25% error (0.25) with 95% confidence, and we took a normal one-hour background reading (60 minutes) yielding a background count rate of 25 CPM (counts per minute). Our detector has an efficiency of 1% (0.01), we’ll assume that each disintegration yields two detectable gamma rays on average, and our effective solid angle being surrounded by our source is π Sr (steradians). This yields a much simpler expression: (200An − 25) (24) tg (minutes) = 25 625A2n − 60 A theoretical estimate of the activity of a smoke shop would therefore give you a good guess as to how much time you would have to count for. As an example, let’s say you measure a gross count rate (cg) of 35 CPM inside. Equation 21 yields a required counting time of only 30.6 minutes to be 95% confident in your result. This makes it a potentially awkward, but not impossible, task. 1. Bonus Question (25 points): Go do this. Tell us how radioactive a local smoke shop is, state/calculate your uncertainty, and how long you had to count. You should get the shop owner’s permission before doing this, to avoid arousing suspicion. 3 Radioactive Dating with Confidence For this problem, consider the methods used to radioactively date the Shroud of Turin, thought to be the burial cloth of Jesus of Nazareth. 1. Why did the investigators use carbon as the dating isotope? Consider what other isotopes could have been present, and give at least three reasons that carbon was chosen. The investigators used carbon-14 because its half life is 5,730 years, and if the shroud were indeed the burial cloth of Jesus, then a sizeable chunk of radioactivity would be measurable. If the half life were too short or too long, then very little counts would be acquired, greatly increasing the uncertainty of the count rate. The goal here would be to maximize the activity of the isotope being counted. Too short a half life, and it would have all decayed away. Too long, and very few atoms would decay at any given time at all. 4 2. How did the investigators use statistics to prove beyond a reasonable doubt that the Shroud of Turin was not the burial cloth? What is a χ2 test, and how did they arrive at the χ2 values in the paper? Samples of the shroud were sent to different laboratories, for their own dating procedures and uncertainty to be measured. Each laboratory used their own procedures, showing that each sample (1 - the shroud, and 2-4, the “control” samples known to be from different time periods) were able to be measured within a certain range of ages. The chi-squared test here is applied to measure the “goodness of fit” of each sample’s ages, and tells us the confidence of a random measurement of that specimen falling within the uncertainty bounds of the expected, or “theoretical,” measured age. The fact that many laboratories were only able to measure the shroud sample with 5% confidence within the specified limits does not mean that they were wrong, as their dating methods showed great confidence for the other, control samples. 3. Why did the investigators send so many unknown control samples to so many laboratories, and why did they use different cleaning procedures? This was done to avoid any chance of introducing bias into the results. Many, many people encounter significant quantities of feelings when dealing in matters of faith and religion, therefore those who would “want” the shroud to be or not to be the real burial cloth could have tampered with the evidence if they knew which piece they had. Keep­ ing everyone guessing completely removes the incentive to tamper with the specimens. Different cleaning procedures were also used, to ensure that if one cleaning procedure artificially reduced the carbon-14 content of a specimen, then it would be discovered, and that method could be eliminated from altering the actual amount of carbon-14 present. 4 Generating Cobalt-60 and Profit How many days should one irradiate a 100g source of 59 Co in order to maximize profit from the reactor? Assume a fully homogeneous, thermal reactor, with the following parameters: n 60 Reactor Cost = $1000/day Co = $100/µCi cm2 s Start by writing down what is physically happening (the nuclear reactions), model them using a system of differential equations, solve the system of equations for the amount of 60 Co as a function of time, and construct & solve an equation to maximize the profit of the reactor. Look up any nuclear data that you need from the JANIS cross section database and the KAERI Table of Nuclides. Let’s say we have the following nuclear reactions (like on the lecture on October 6th), where we are producing 60 Co from the neutron bombardment of 59 Co, and 60 Co has its own decay constant λ where it decays by β − decay into 60 Ni: Φ = 1014 59 27 Co N1 n λ 60 −→ 60 27 Co −→ 28 N i N2 (25) N3 We also have to account for the fact that both 59 Co and 60 Co are “burned” in the reactor by capturing neutrons. The first reaction produces 60 Co, while the second one depletes it. We therefore define a couple of neutron capture cross sections: σc59Co = σ59 = 20 b σc560 = σ60 = 2 b 1 b = 10−24 cm2 (26) where we have looked up the values of the cross sections from the JANIS database, using the ENDF VII library for incident neutron data, and we’ve chosen the values at 0.025eV (the kinetic energy of thermal neutrons). Let’s just pretend that our reactor has only thermal neutrons in it, a “one group” approximation. Oh, let’s also define the neutron flux of our reactor, or the number of neutrons zipping through every square centimeter per second: Φ = 1014 5 n cm2 s (27) We next look up the half life of 60 Co, and use Equation ?? to get its decay constant: t1/2 = 1925.4 days = 1.66 · 108 s λ= ln (2) = 4.17 · 10−9 s−1 1.66 · 108 s (28) Finally, we have to define some initial amount of 59 Co that we put into our reactor. Let’s just say it was 100g of 59 Co, or 1.69 moles, or 1024 atoms: N10 = 1024 atoms (29) Finally we are ready to construct our differential equations to physically model this real system. First, we recognize from the KAERI Table of Nuclides that 59 Co is stable, so it has no production term, and the only way for it to be destroyed is to be “burned” by neutron capture: dN1 = −σ59 ΦN1 (30) dt Next, the production rate of 60 Co is equal to the destruction rate of destroyed both by natural radioactive decay and artificial “burning:” dN2 = σ59 ΦN1 − σ60 ΦN2 − λN2 dt Finally, the only way to produce 60 Co does NOT produce 60 Ni: 60 59 Co, while 60 dN2 = σ59 ΦN1 − (λ + σ60 Φ) N2 dt Ni is by radioactive decay of 60 60 (31) Co. Note that “burning” dN3 = σ60 ΦN2 dt Here we are ignoring the “burning” of we don’t care about it: Co can be (32) Ni. In fact, let’s just ignore 60 Ni altogether, because dN3  = σ60 ΦN2 (33)  dt Now we start with the easy equation for N1 (t). Note that the quantity σ59 Φ has the same units as λ, so the equation takes the same form:  ˆ ˆ  dN1 dN1 = (−σ59 Φ) N1 = (−σ59 Φ) dt N1 (t = 0) = N10 N1 (t) = N10 e−σ59 Φt (34) dt N1 Now we take this expression for N1 and substitute it into Equation 31: dN2 = σ59 ΦN10 e−σ59 Φt − (λ + σ60 Φ) N2 dt (35) Next we rearrange terms so that all the stuff is on one side of the equation: dN2 + (λ + σ60 Φ) N2 − σ59 ΦN10 e−σ59 Φt = 0 dt (36) Next we introduce our integrating factor , µ: ´ µ=e (λ+σ60 Φ)dt = e(λ+σ60 Φ)t (37) and we multiply every term in Equation 36 by µ:   u p dN2 e(λ+σ60 Φ)t + ((λ + σ60 Φ) N2 ) e(λ+σ60 Φ)t − σ59 ΦN10 e−σ59 Φt e(λ+σ60 Φ)t = dt (38) We recognize that our first two terms look eerily similar to the end result of the Product Rule: d (a (t) b (t)) db (t) da (t) = a (t) + b (t) dt dt dt equivalently 6 ' (ab) = ab' + a' b (39) We then smoosh the first two terms of Equation 38 together using the Product Rule in reverse, and combine the exponential parts of the third term:
d N2 e(λ+σ60 Φ)t = σ59 ΦN10 e((λ+σ60 Φ)−σ59 Φ)t (40) dt We then integrate both sides: ˆ d N2 e(λ+σ60 Φ)t dt
! dt = ˆ   σ59 ΦN10 e((λ+σ60 Φ)−σ59 Φ)t dt (41) This just kills the derivative on the left hand side, puts the exponential term in the denominator on the right hand side, and introduces a constant of integration: N2 e(λ+σ60 Φ)t = σ59 ΦN10 e((λ+σ60 Φ)−σ59 Φ)t + C ((λ + σ60 Φ) − σ59 Φ) (42) We now use the initial condition that we had no 60 Co when we first started producing it at t = 0: 1 0 0 σ59 ΦN10 (λ+σ60 Φ)
t ((λ+σ60 Φ)−σ59 Φ)
t   >  e +C (43) N2 (t = 0) = 0 N2 e =  ((λ + σ60 Φ) − σ59 Φ) :0  =  Φ)(1) (λ+σ 60 (0) e  0= :1   σ59 ΦN10  ((λ+σ60 Φ)−σ 59 Φ)(0)  e  +C ((λ + σ60 Φ) − σ59 Φ) (44) −σ59 ΦN10 ((λ + σ60 Φ) − σ59 Φ) (45) σ59 ΦN10 +C ((λ + σ60 Φ) − σ59 Φ) C= We finally plug this integration constant back into Equation 42 and do a bit of rearranging: N2 e(λ+σ60 Φ)t = σ59 ΦN10 σ59 ΦN10 e((λ+σ60 Φ)−σ59 Φ)t − ((λ + σ60 Φ) − σ59 Φ) ((λ + σ60 Φ) − σ59 Φ) h i σ59 ΦN10 e((λ+σ60 Φ)−σ59 Φ)t − 1 ((λ + σ60 Φ) − σ59 Φ) " #    (λ+σ Φ)−σ59 Φ)t 60 60 Φ)t  σ59 ΦN10 1 N2 e(λ+σ e( − (λ+σ Φ)t  =    60 60 Φ)t 60 Φ)t ((λ + σ60 Φ) − σ59 Φ) e   e(λ+σ e(λ+σ     σ59 ΦN10 1 N2 (t) = e(−σ59 Φ)t − (λ+σ Φ)t 60 ((λ + σ60 Φ) − σ59 Φ) e N2 e(λ+σ60 Φ)t = N2 (t) = h i σ59 ΦN10 e−(σ59 Φ)t − e−(λ+σ60 Φ)t ((λ + σ60 Φ) − σ59 Φ) (46) (47) (48) (49) (50) Not surprisingly, this looks exactly like Equation 4.40 from the Turner book: N2 (t) =  λ1 N10  −λ1 t e − e−λ2 t (λ2 − λ1 ) (51) where we have defined λ1 = σ59 Φ and λ2 = λ + σ60 Φ. Now let’s start plugging in some of the values:    −24  2 14 n λ1 = σ59 Φ = 20 · 10  cm 10 = 2 · 10−9 s−1 (52) 2 cm s       1 n 2 λ2 = λ + σ60 Φ = 4.17 · 10−9 cm 1014  = 4.37 · 10−9 s−1 (53) + 2 · 10−24 2 s cm s  7 How we plug these values into Equation 51: i 
1024 atoms
h 2 · 10−9 s−1 −(2·10−9 s−1 )t −(4.37·10−9 s−1 )t
N2 (t) = e − e  − 2 · 10−9  s−1 4.37 · 10−9 s−1
i 2 · 1015 atoms h −(2·10−9 s−1 )t −(4.37·10−9 s−1 )t − N2 (t) = e e (2.37 · 10−9 s−1 ) (54) (55) We can then graph this situation using Desmos: Note that the x-axis is in seconds, and the y-axis is in atoms. This shows you that it takes about 3.204 · 108 seconds, or 10 years, to reach a maximum inventory of 60 Co in the reactor. A very real example, using actual numbers, from a very theoretical derivation! Finally, we constuct an equation to calculate the profit that we can extract from the reactor, remembering that the activity in Bq is written as A = λN: !  �     �   60Co  $100 N2 (t) $1, 000 1 day 1 µCi    $ (t) = − t (s) (56)    3.7 · 104 Bq 60 Co λ + σ60 Φ 86, 400 s 1 day  1 µCi 60Co   $ (t) =   $100  atoms  >60 Co 3.7 · 104  Bq  $ (t) =      N2 (t) $1, 000 − t (
s) 1 86, 400
s 4.37 · 10−9 s− (57)       N2 (t) ( atoms) $1, 000  − t (
s)   1 atoms  86, 400
s 4.37 · 10−9 s− 60 4 >  3.7 · 10  Bq Co   $100N2 (t) $ (t) = − 0.011574t (3.7 · 104 ) (4.37 · 10−9 ) $100 (58) (59) Finding the maximum of this equation graphically yields the same time as the maximum in Co inventory, or 3.204 · 108 seconds. Clearly this reactor’s profit model isn’t quite accurate... we haven’t accounted for the isotopic isolation costs of extracting cobalt. 60 8 Part II Take-Home Lab: Estimating the Radioactivity of One Banana (50 points) Using the banana ashes accumulated from last year’s 22.01 class, estimate the radioactivity of one banana. Make the following assumptions: 1. The ashes were created from 50 pounds of peeled, fresh, ripe bananas. 2. No other sources of contamination are present in the bananas Use the high purity Germanium detector (HPGe) in the Nuclear Reactor Laboratory (NRL) to collect background and banana spectra. Devise a way to determine the total radioactivity of your sample based on any of the available features of the spectrum (peak height, number of counts, area under peak, etc.) of the most appropriate peak of the most appropriate isotope. Compare your answer to any reputable source from the literature, and make sure to cite your source: • Journal articles need the authors, title, journal, volume, pages, and year. • Books need the authors, editors, title, publisher, pages used, year, and ISBN number. • Other articles (like those online) need the author, URL, date that you accessed it, and date of publi­ cation. These should be used as a last resort. Solution: First, we look at the spectra from counting our bananas (left) and from our background (right): 40 We know that bananas are good sources of potassium, and that 0.011% of all potassium is radioactive K. The complete decay diagram for 40 K is as follows: Courtesy of Korea Atomic Energy Research Institute. Used with permission. These branching ratios have been slightly rounded from their true values to one significant digit, we will use these. This diagram shows us that 10% of the time, we expect 40 K to decay by electron capture, releasing a 1.461 MeV gamma ray. Looking at our spectra above, both of these have a strong peak at about 1.460 MeV. This is the peak of interest for us, and it is the photoelectric effect peak for 40 K. Because we have a large HPGe detector, there are very few single- and double-escape peaks from pair production, so we can assume that this peak contains both the photoelectron interactions and most of the pair production 9 interactions. The clear Compton edge and bowl shape tells us that we will be missing some of the photons from the 1.460 MeV peak to Compton scattering. We will deal with those later, though you don’t have to account for them to get full credit. We now look at the values of the 40 K photopeak in both spectra, and subtract the background peak area from the real spectrum peak area: Figure 1: Software-identified peaks from the banana counting spectrum Figure 2: Software-identified peaks from the background spectrum Subtracting the background count rate (0.0238 CPS, or 2,060 counts in 86,266 secnds) from the gross count rate (0.182 CPS, or 75,600 counts in 420,611 seconds), we get a net count rate of 0.158 CPS. Noting that the detector efficiency at 1400keV (closest to the 40 K photopeak at 1460keV) is 8.26953 · 10−4 , and that only 10% of the 40 K decays would result in a countable gamma ray, we divide the net count rate of 0.158 CPS by 8.26953 · 10−5 to get a true disintegration rate of 1,910 Bq in our sample of bananas. Now assuming that we had 50 pounds of bananas, and that each banana weighs a third of a pound, that means we had 150 bananas, meaning that each banana had a 40 K activity of 12.7 Bq. This is remarkably close to the true value of 3,520 pCi/kg, or 130.24 Bq/kg. Assuming that each banana weighs a third of a pound, or 150g, that would lead to a theoretical activity of 19.5 Bq. The missing activity could be due to (1) Slightly different true efficiencies in the detector, or (2) not counting the Compton scattered gamma rays in the photopeak activity. References [1] Brodsky, A. “Handbook of Radiation Measurement and Protection/” CRC Press 1978 and Environmental Radioactivity from Natural, Industrial and Military Sources, Eisenbud, M and Gesell T. Academic Press, Inc. 1997. Found at http://www.physics.isu.edu/radinf/natural.htm on 2016-10-25. 10