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22.01 Fall 2015, Problem Set 4 Solutions (Analytical Version)
Due: October 21, 11:59PM on Stellar
October 31, 2015
Complete all the assigned problems, and do make sure to show your intermediate work. Please upload your full problem set in PDF form on the Stellar site. Make sure to upload your work at least 15 minutes early, to account for computer/network issues.
Other Worldly Decay
Suppose a parallel universe exists, where for an ensemble of radioactive particles in a box, the probability of one particle decaying in a fixed amount of time Δt is proportional by a constant η to the number of unstable particles remaining in the box. Derive a new expression for the half life of a given other-worldly isotope, as a function of the number of particles in a box (N 0 ) at time t 0. Graph this expression compared to the normal half-life equation, and comment on the reasons for the differences in the shapes of the curves. This time, instead of the probability of an ensemble of particles decaying just being a proportionality constant λ, giving the following equation: dN = −λN (1) dt
The probability of decay is instead proportional to the number of atoms left over:
dN = (−ηN ) N (2) dt
Separating variables, we get: dN = −ηdt (3) N 2 − 1 1 = −ηt + C; N = (4) N ηt + C
Now to find the integration constant: we assume that N = N 0 @t = 0 ;
1 1 1 N = ; C = ; N = (5) ηt + C N 0 ηt + (^) N^10
Graphed, the functions are as follows:
2 Activity and Half Lives
Back to earth for now... suppose a given smoke detector is rated at an activity of 0. 8 μCi of 241 Am upon manufacture. It is powered by a betavoltaic power module, which encases a tritium source 3 H rated at an activity of 2Ci upon its manufacture.
- Which component of the smoke detector will fail to provide exactly 1nA of current first, and after how long? Assume that the equation for allowable current through an ionization chamber is as follows: Em 3. 7 · 1010 Bq Imax = ec S (t) (6) Ei Ci where Imax is the saturation current through the ionization chamber, ec is the charge of an electron in Coulombs, Em is the energy of the ionizing particle (you will have to look this up), Ei is the ion formation pair energy in air (14eV ), and S (t) is the activity of the 241 Am source. Here, we want to compare when each component will reach a current of 1 nA. For the 241 Am source, we use the formula in Equation 6, while for the tritium isotope we just equate its activity to the current it can provide directly by beta decay:
9 19 5.^638 ·^106 eV^3.^7 ·^1010 Bq −λ (^241) Amt I (^241) Am = 10 −^ A = 1. 6 · 10 −^ C 8 · 10 −^7 Ci e (7) 14 eV Ci
- 7 · 1010 Bq I 3 9 −λ^3 H^ t H =^10 − (^) A = 1. 6 · 10 − (^19) C (2 Ci) e (8) Ci ln (^2) − 1 ln (^2) − 1 λ (^3) H = = 1. 93 · 10 −^9 s ; λ (^241) Am = = 5. 35 · 10 −^11 s (12 yrs) 3 · 107 secyr (432. 2 yrs) 3 · 107 secyr (9) tH, 1 nA = 42. 7 yrs; tAm, 1 nA = 402. 3 yrs (10)
Clearly the tritium source will run out sooner.
- Suppose this first component were to be inexhaustible. How long would it take for the second com ponent to drop to a supplied/allowable current of exactly 1nA? For this question, make sure to take the uncertainty due to significant figures into account, and give a maximum, minimum, and expected value. Here we must account for the fact that the activity of the 241 Am source was specified as 0.8 μCi, which carries an uncertainty of ± 0. 05 μCi. The expected value is given in part 2.1 above. The minimum and maximum values for the time to Americium failure: tAm,min = 362. 1 yrs; tAm, 1 nA = 440. 1 yrs (11)
(a) What is the source of these radioactive daughter products? Write as complete of a decay chain as you can for the creation of radioactive substances found in tobacco. The source for these decay products is ultimately from uranium, but more recently from radium:
This decay chain summarizes it nicely. It would give a total of four alpha and four beta particles per atom of radium atom ingested if everything were retained in the body. However, since radon is a gas and many isotopes afterwards have very short half lives, one can assume that an atom of radon that decays within settling distance of a tobacco leaf will have decayed to Pb-210 by the time the leaves are picked. This leads to a set of two sequential radioactive decay equations, assuming that the half life of bismuth is very, very short compared to both Pb-210 and Po-210. Therefore, each atom of Pb-210 ingested imparts two beta particles and one alpha particle worth of energy to the smoker.
(b) Many of these radioactive products are alpha emitters, which are particularly damaging to tissue. Estimate the additional radiation activity to someone’s lungs if they smoke one pack of cigarettes a day. Let’s say that the absorption rate of Pb-210 on tobacco leaves is S atoms, and that the leaves m^2 leaf don’t live that long until they are picked. One can then estimate the total Pb-210 in one pack as: atoms m^2 leaves cigarettes NP b−per−pack = S 16 (19) m^2 leaf cigarette pack Assuming that each pack was made the day it was smoked, or something like that, we can calculate the total time-dependent activity imparted to a user from the day they smoke each pack until the day they die, directly from the Bateman equations on Yip, p. 101: dNP b− 210 = −λP b− 210 NP b− 210 (20) dt dNP o− 210 = λP b− 210 NP b− 210 − λP o− 210 NP o− 210 (21) dt dNP b− 206 = λP o− 210 NP o− 210 (22) dt The equation for the total activity from one pack is the sum of the first two equations, with one small change to account for the nearly instantaneous decay of Bi-210:
Aone−pack (t) = 2 (λP b− 210 NP b− 210 ) + (λP o− 210 NP o− 210 ) (23)
We also know the equations for the numbers of these two isotopes, directly from the Bateman equations (6.15-6.16 in Yip, p. 101):
NP b− 210 (t) = NP b−per−packe^ − λP b−^210 t (24)
Courtesy of National Academies Press. Used with permission. Source: National Research Council. Health Effects of Exposure to Radon: BEIR VI. The National Academies Press, 1999. doi:10.17226/5499.
λP b− 210 NP o− 210 (t) = NP b−per−pack e −λP^ b−^210 t^ − e −λP^ o−^210 t (25) λP o− 210 − λP b− 210 From these equations, one can define the total number of radioactive disintegrations absorbed by the smoker, assuming (justifiably) that all the metals involved are heavy metals, and bio accumulate: tdeath Disintegrations − per − pack = t (26)
Aone−pack ( ) dt t Now we just have to define a couple of fixed times, tstart−smoking and tdeath , as fixed times in days. Then we can construct an equation accounting for all the disintegrations that will happen during
t
the smoker’s lifetime: t (^) death−t (^) start−smoking⎡ ⎤ 4 86 , 400 secday 86 , (^400) ˆtdeath Lif etime Disintegrations = Aone−pack (t) dt i=0 (^86) , 400 i
(c) How long would you have to keep these cigarettes in storage for this radioactivity to reach 1% of
its original level, from the day the tobacco was picked? You will need to consider simultaneous production and decay of radioactive isotopes in your answer. Using the equations above, one can solve for t 1 % using the following equation:
t´1% Aone−pack (t) dt 0 ∞´ Aone−pack (t) dt 0
Statistics and Certainty
For these problems, consider that you have just measured the activity of a source stamped at 1 μCi to be
70 μCi, by measuring a count rate of 15 counts per second.
Taking uncertainty into account, how long would you have to count to ensure 95% confidence in your measurement? What would the standard deviation of your measurement be in this case? In this case, the standard deviation of this count rate is as follows:
σ = (29)
15 CP S
tcounting
If we measured the specimen to have an activity of 0. 70 μCi, then that means the uncertainty on this measurement due to significant figures alone is 0. 005 μCi, or roughly 0.7%. If we want to be 95% confident in this measurement, then we construct our equation as follows:
15 CP S 2 σ = 2 = 0 .007; tcounting = 1. 22 · 107 sec (30) tcounting
This means that we would have to count for almost five months to ensure this confidence in our mea surement!!!
- Taking uncertainty into account, when could this source have been calibrated at 1 μCi? A stated calibration activity of 1 μCi means that the actual activity could range from 0. 5 − 1. 5 μCi. In the case that the starting activity was 1. 5 μCi, then the calibration date in days would have been as follows, relative to today:
(ttoday − tcal) = −ln
- 70 μCi
- 5 μCi (λ)
1 day 86 , 400 sec = 8. 8 · 10 −^6 λ days ago (31)
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22.01 Introduction to Nuclear Engineering and Ionizing Radiation
Fall 2015
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