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)238U92234Th90234Pa91234U92
The rate of decay of a nucleus Xn is proportional to the amount. The proportionality constant n is the decay constant, it's reciprocal, 1/n, is the average lifetime and τn = ln(2)/n is the half-life. For (1.2) one has (Adloff [1,pp.125128])
k / k / 1/k 1.44τk / τk = ln(2)/k 0.69/k1 / 1.54 10-10 year-1 / 6.49 109 years / 4.5 109 years
2 / 0.0288 day-1 =10.5 year-1 / 34.8 days =0.0952years / 24.1 days
=0.066years
3 / 0.592 min-1 =312000 year-1 / 1.69 min =3.2110-6 years / 1.17 min =2.22106years
4 / 2.83106year-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 = -1N1
(1.5)N = n1Nn1 - nNn for n 2
It is often more convenient to work with
(1.6)An(t) = nNn(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 = -1A1
(1.8)A = nAn1 - nAn for n 2
For simplicity we shall assume N1(0) = 1 and Nn(0) = 0 for n 2. Then A1(0) = 1 and An(0)= 0 for n 2. Using these initial condition, equations (1.7) and (1.8) give
(1.9)A1(t) = 1e-1t
(1.10)An(t) = ne-nt * An-1(t)
where * denotes convolution, i.e.
(1.11)
Formula (1.10) implies
(1.12)An(t) = 1e-1t* … * ne-nt
Parentheses are omitted on the right since convolution is commutative and associative. Integration gives
(1.13)1e-1t* 2e-2t = e-1t + e-2t
Using induction one obtains
(1.14)An(t) = An(t;1,...,n) = c1e-1t+ +cne-nt
(1.15)ck = ck(1,...,n) =
Unless stated otherwise, when we write An(t) we shall assume the decay constants are 1,...,n, i.e.An(t)=An(t;1,...,n). In the 238U series (1.2) one has (assuming t is in years)
A4(t) = - 4.72 10-26e-312000t + 4.15 10-17e-10.5t – 1.54 10-10e-2.83 10-6t
+ 1.54 10-10e-1.54 10-10t
Here is a graph of A4(t) using a range of t on the order of the largest half-life. It looks like 1e-1t.
Here is a graph of A4(t) using a range of t on the order of the second largest half-life. It looks like 1(1e4t).
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 μ1μ2…μn be the values 1, ..., n arranged in increasing order. Then
(1.16)An(t) μ1μmtm-1/(m-1)!
for 1/μm+1t < 1/μm and 1 mn. ((1.16) holds for t< 1/μn if m = n.) For simplicity, we shall assume in the remainder that μ1μ2…μn. Then the relative error in the approximation (1.16) is less than for t whenm = 1, 2, 3, …, n-1 and for t when m = n; see Theorems 3 and 5. Here at means bt for some ba. In (1.2) one has A4(t)14t for 9.6 yearst7,000 years and A4(t)124t2/2 for 5.7 hourst1 day and A4(t)1234t3/6 for t < 4 sec, all with an error less than 2%
For (1.2) we noted above that A4(t) 1e-1t for large t. In general
(1.17)An(t) e-1t 1e-1t
The left hand approximation holds with relative error less than for t; see Theorem 4(d). In the case of (1.2) the error is less than 2% for t > 1.4 106 years. (1.17) can be generalized to An(t) Am(t;1,...,m) for t1/μm+1; see Theorem 4(e).
For (1.2) we noted above that A4(t) 1(1 – e-4t) when viewed on a time scale of the second largest half-life. In general
(1.19)An(t) 1(1 – e-2t)
with relative error is less than for t; see Theorem 5(b) and (c). In (1.2) one has A4(t)1(1 – e-4t) with an error less than 2% for 10 yearst1.3108years. (1.19) can also be generalized; see Theorem 6.
2. Properties of An(t). Let
Fn(t)=Fn(t;1,...,n)=e1t**ent = An(t)/(1n)
(2.1)Qn(t)=Qn(t;1,...,n) = =
(1,...,n; )=1n / [ (1)(n) ]
Qn(t) is the total amount of the nuclei X1,X2,,Xn at time t if at t = 0 there is one unit of X1 present and none of the other Xk. When we write Fn(t) or Qn(t) the decay constants are assumed to be 1,...,n.
Theorem 1. If 1...n. then the following are true. (a) Q(t) = - An(t). (b)[eαtf1(t)]**[eαtfn(t)] = eαt[f1(t)**fn(t)] for any and f1(t), , fn(t). (c)Fn(t;1,...,n)=eαtFn(t;1,...,n) and An(t;1,...,n)=eαt(1,...,n;)An(t;1,...,n) for any . (d)Fn(t;0,...,0)=1**1=tn1/(n-1)! and Fn(t;,...,)= tn-1e-t/(n-1)!. f(t) * 1 = for any function f(t) andf(t) * tn-1/(n-1)! is obtained by integrating f(t) from 0 to t a total of n times. where the integral is an (n-1)-fold multiple integral over with #=1+1(21)+…+n-1(n-n-1) and d=d1…dn-1 and 1n. (f)Fn(t)tn1e1t/(n1)! for t 0. (g)If the j are non-negative then for t≥ 0.
Proof. (a) This follows from (1.4) and (1.5). (b) The case n = 2 is proved by a direct computation, while the case n > 2 is proved by induction on n. (c) follows from (b). (d) The formula for Fn(t;0,...,0) is a straight forward computation and the formula for Fn(t;,...,) follows from the formula for Fn(t;0,...,0) and (c). The formula for f(t) * 1 follows from the definition of convolution and the formula for f(t) * tn-1/(n-1)! follows from this and the formula for 1**1. (e)Simonsen [2] showed that (1)n1Fn(t) is the (n1)st divided difference with respect to of the function et. The (n1)st divided difference of an arbitrary function g() is given by
(2.2)g[1,...,n] = = g(n-1)()/(n-1)!
where g(k)() = dkg/dk; see Issacson [1,pp.249-250]. The result follows from this. (f)Since ejte1t it follows that Fn(t) Fn(t;1,...,1) = tn1e1t/(n1)!. (g)It follows from (e) that Since # ≥ 0 for all , one has ≥ 0 and the result follows.//
Theorem 2. Assume 0 < 1...nand f(t)0 for t0. (a)If f(t) is nondecreasing for t0 then f(t)*An(t) f(t) for t0. (b)Suppose there is an such that 1 and eαtf(t) is nondecreasing for t0. Then f(t)*An(t)(1,...,n;)f(t) for t0. (c)If 1...m and 11, then eα1tFm(t;1,...,m) is increasing for t0 and Fm(t;1,...,m)*An(t)(1,...,n;1)Fm(t;1,...,m) for t> 0. (d)An(t;1,...,n)(m+1,...,n;1)Am(t;1,...,m) for t > 0. (e)Qn(t)(2,...,n;1)e1t for t> 0.
Proof. (a) One has f(t)*An(t)=f(t) . The result follows since the integral on the right is less than 1 which is because An(t) is a probability density function. (It is the probability density function of the sum of n independent exponential random variables with densities je-jt.) (b)Theorem 1(b) and (c) give . By part a one has (etf(t))*An(t;1-,...,n-) ≤ etf(t) from which the result follows. (c) Theorem 1(c) gives e1tFm(t;1,...,m)=Fm(t;0,21,...,m1) which equals . The integrand is positive for t > 0, so e1tFm(t;1,...,m) is increasing. The inequality then follows from part b. (d) One has An(t)=1mFm(t;1,...,m)*Anm(t;m+1,...,n), so the inequality follows from part c. (e)By part (d) one has Aj(t)/j(2,...,j;1)1e1t/j for 2 jn. Therefore Qn(t)ce1t where c=1+. It is not hard to show that c=(2,...,n;1), so the result follows. //
3. Approximation theorems.
Theorem 3 (Small t). Let 1,…,n be non-negative, be the mean of 1,…,m and f(t) 0 and 1 mn - 1. Then for t 0 one has
(3.1) (1 - nt) An(t)
(3.2) [ f(t)*tn-1 ] (1 - nt) f(t)*An(t) f(t)*tn-1
(3.3) [ An-m(t;m+1,...,n)*tm-1 ] (1 - mt) An(t) An-m(t;m+1,...,n)*tm-1
Proof. The right inequality in (2.1) follows from Theorem 1f. To prove the left inequality we need to show
(3.4) - Fn(t)
If n = 1, this is the well known inequality 1te-t. Assume it is true for n and write Fn+1(t)=Fn(t)*e-n+1t. Convoluting the left inequality in (3.4) with e-n+1t gives
(3.5) * e-n+1t - * e-n+1t Fn+1(t)
Since 1n+1te-n+1t it follows that
(3.6) * 1 - n+1 * t * e-n+1t
Since * 1 = and * t = it follows that - n+1 * t * e-n+1t. Since e-n+1t 1 it follows that
* e-n+1t * 1 =
Combining this and (3.5) with (3.6) gives tn/n! - (1 + + n+1) tn+1/(n+1)! Fn+1(t), which proves (3.4) for n+1.
In order to show (3.2) we need to show
(3.6) * f(t) - nt [ * f(t)] Fn(t) * f(t) * f(t)
To prove this, note that from (3.4) it follows that
(3.7) * f(t) - n * f(t) Fn(t) * f(t) * f(t)
One can estimate the second term on the left by
tn * f(t) = t = t [tn-1 * f(t)]
Combining this with (3.7) gives (3.6).
Finally, note that (3.3) follows from (3.2) by taking n = m, An(t) = Am(t), and f(t)=An-m(t;m+1,...,n) and using the fact that Am(t) * An-m(t;m+1,...,n) = An(t). //
Theorem 4 (Large t). Assume 0 < 1...n. Let # and m be defined by and . Then the following are true. (a)Suppose f(t) 0 and f(t) is nondecreasing for t0 and there is an such that 1 and etf'(t) is non-decreasing for t0. Then
f(t)f(0)Qn(t) f(t)*An(t) f(t)
for t0. (b) If q 1 then [1q/(#t)] tqtq*An(t)tq for t0. (c) If 0 1...m and 11. Let = 1. Then
[1]Fq+1(t;0,1,...,q) Fq+1(t;0,1,...,q) *An(t) Fq+1(t;0,1,...,q)
for t0. (d)
[1 - (3-1,...,n-1;2-1)e(21t] (2,...,n;1)1e1t An(t) (2,...,n;1)1e1t
for t 0. (e)
[1 - ] (m+1,...,n;1)Am(t) An(t) (m+1,...,n;1)Am(t)
for t0.
Proof. (a) The right inequality is Theorem 2(a). If one integrates by parts and uses the fact that Q = - An (by Theorem 1(a)) one obtains f(t)*An(t) = f(t) - f(0)Qn(t) - f(t)*Qn(t). Theorem2(b) gives f(t)*Aj(t)(1,...,j;)f(t)(1,...,n;)f(t) which implies f(t)*Qn(t)(1,...,n;)f(t)/#. Combining with the previous gives the left inequality in(a). (b) follows from (a) by taking f(t) = tq and = 0. (c) Let f(t) = Fq+1(t;0,1,...,q). Thenf(t)=Fq(t;1,...,q), so it follows from Theorem 2(c) that etf(t) is increasing. So f(t) and satisfy the hypotheses of part (a) and f(0) = 0. By Theorem1(i) one has f(t)/f(t)q/t. So (c) follows from (a). (d) The fact that An(t)≤(2,...,n;1)1e1tfollows from Theorem 2(d). Using Theorem 1(b) and (c) one can write An(t)=(2,...,n;1)1e1t[1*An1(t;21,...,n1)]. By Theorem 1(a) one has 1*An1(t;21,...,n1)=1Qn1(t;21,...,n1). Using Theorem 2(e) one obtains 1(31,...,n-1;2-1)e(21t≤1*An1(t;21,...,n-1) which proves the left inequality in (d). (e) From Theorem 2(d) we get An(t)≤(m+1,...,n;1)Am(t). Using Theorem 1(b) and (e) one can write An(t)=(m+1,...,n;1)1me1t[Fm(t)*Anm(t)], where Fm(t)=Fm(t;0,21,...,m1) and Anm(t) = Anm(t;m+11,...,n1)]. By part (c) one has
(1) Fm(t) ≤ Fm(t)*Anm(t).
When combined with the previous, this proves the left inequality in part (e). //
Theorem 5 (Intermediate t.). Suppose 012n and let be defined by . Then for t > 0 and 2 mn1 one has
[ 1 - - mt] An(t)
Proof. The right inequality is included in Theorem 3. Combining Theorem 3 and Theorem 5b gives [ 1 - ] [ 1 - mt] An(t). The left inequality follows from this and the fact that (1-a)(1-b) 1 – a – b if a 0 and b 0. //
Theorem 6 (Intermediate t.). Let 012n and 2mm+rn-1. Let b = 1 if m > 1 and b = (r+2n;2) if m = 1. Let
Hn,m(t) = Hn,m(t;1,...,n) = * An(t;1,...,n)
Then for t > 0 one has
[ 1 - - mt]1mHr,m(t;m+1,…,m+r) <An(t)1mHr,m(t;m+1,…,m+r)
Proof. We prove the case m > 1. The proof for m = 1 is similar. By (3.3) one has where g(t)=Anmr(t;m+r+1,...,n)*U(t) with Apply Theorem 4(c) with Fq+1(t;0,1,...,q) = U(t) to obtain 1(m+r1)/(m+r+1t) < g(t)/U(t) < 1. Combining this with the previous inequality and using the fact that U(t)=1mHr,m(t;m+1,…,m+r) proves the inequalities. //
References
[1]E. Issacson, H.B. Keller, Analysis of Numerical Methods, Wiley, New York, 1966.
[2]W. Simonsen, On divided differences and osculatory interpolation, Skandinavisk Aktuarietidsskrift 31 (1948) 157.
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