Approximations to Convolutions of Exponential Functions
Approximations for Radioactive Decay Formulas - Old and New
Frank Massey and Jeffrey Prentis
1. Introduction. A radioactive decay chain
(1.1) …
is a series of radioactive nuclei where each one decays into the next. An example is the first four decays of the 238U series
(1.2) 238U92 234Th90 234Pa91 234U92
The rate of decay of a nucleus Xn is proportional to the amount. The proportionality constant λn is the decay constant, it's reciprocal, 1/λn, is the average lifetime and τn = ln(2)/λn is the half-life. For (1.2) one has (Adloff [1, pp. 125-128])
|k |λk |1/λk ( 1.44τk |τk = ln(2)/λk ( 0.69/λk |
|1 |1.54 ( 10-10 year-1 |6.49 ( 109 years |4.5 ( 109 years |
|2 |0.0288 day-1 = 10.5 |34.8 days = 0.0952 years |24.1 days |
| |year-1 | |= 0.066 years |
|3 |0.592 min-1 = 312000 |1.69 min = 3.21 ( 10-6 |1.17 min = 2.22 ( 10-6 years |
| |year-1 |years | |
|4 |2.83 ( 10-6 year-1 |353,000 years |245,000 years |
Let
(1.3) Nn(t) = the amount of Xn present at time t
The Nn(t) satisfy the radioactive decay equations:
(1.4) N = - λ1N1
(1.5) N = λn-1Nn-1 - λnNn for n ( 2
It is often more convenient to work with
(1.6) An(t) = λnNn(t) = the rate of decay of Xn at time t
rather than Nn(t) itself. The An(t) satisfy equations similar to (1.4) and (1.5), namely
(1.7) A = - λ1A1
(1.8) A = λnAn-1 - λnAn for n ( 2
For simplicity we shall assume N1(0) = 1 and Nn(0) = 0 for n ( 2. Then A1(0) = λ1 and An(0) = 0 for n ( 2. Using these initial condition, equations (1.7) and (1.8) give
(1.9) A1(t) = λ1e-λ1t
(1.10) An(t) = λne-λnt * An-1(t)
where * denotes convolution, i.e.
(1.11)
Formula (1.10) implies
(1.12) An(t) = λ1e-λ1t * … * λne-λnt
Parentheses are omitted on the right since convolution is commutative and associative. Integration gives
(1.13) λ1e-λ1t * λ2e-λ2t = e-λ1t + e-λ2t
Using induction one obtains
(1.14) An(t) = An(t;λ1,...,λn) = c1e-λ1t + ( + cne-λnt
(1.15) ck = ck(λ1,...,λn) =
Unless stated otherwise, when we write An(t) we shall assume the decay constants are λ1,...,λn, i.e. An(t) = An(t;λ1,...,λn). In the 238U series (1.2) one has (assuming t is in years)
A4(t) = - 4.72 ( 10-26 e-312000t + 4.15 ( 10-17 e-10.5t – 1.54 ( 10-10 e-2.83 ( 10-6 t
+ 1.54 ( 10-10 e-1.54 ( 10-10 t
Here is a graph of A4(t) using a range of t on the order of the largest half-life. It looks like λ1e-λ1t.
Here is a graph of A4(t) using a range of t on the order of the second largest half-life. It looks like λ1(1 - e-λ4t).
Here is a graph of A4(t) using a range of t on the order of the second smallest half-life.
Here is a graph of A4(t) using a range of t on the order of the smallest half-life. The scale on the vertical axis is in units of 10-25.
In order to get a more complete picture of A4(t) over a range of t running from the smallest half-life to the largest, we make a plot of log[ A4(t) ] vs log t. Notice that there are portions where the graph appears to be a straight line segment. In these portions A4(t) is approximately proportional to a power of t, i.e. A4(t) ( Ctm for some C and m. More generally, we shall show the following. Let μ1 ................
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