Approximations for Radioactive Decay Formulas - Old and New
Frank Massey and Jeffrey Prentis
1. Introduction. A radioactive decay chain
1
X1
(1.1)

2
X2


n-1
 
n
Xn

is a series of radioactive nuclei where each one decays into the next. An example is the first
four decays of the 238U series
238U92
(1.2)
1
2
3
4
 234Th90  234Pa91  234U92 
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. 125-128])
k
k
1/k  1.44τk
τk = ln(2)/k  0.69/k
1
1.54  10-10 year-1
6.49  109 years
4.5  109 years
2
0.0288 day-1
34.8 days
24.1 days
= 10.5 year-1
= 0.0952 years
= 0.066 years
0.592 min-1
1.69 min
1.17 min
= 312000 year-1
= 3.21  10-6 years
= 2.22  10-6 years
2.83  10-6 year-1
353,000 years
245,000 years
3
4
Let
(1.3)
Nn(t) = the amount of Xn present at time t
The Nn(t) satisfy the radioactive decay equations:
(1.4)
N1 = - 1N1
(1.5)
Nn = n-1Nn-1 - 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)
A1 = - 1A1
(1.8)
An = nAn-1 - 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.
t
(1.11)
f(t) * g(t) = 
f(s) g(t-s) ds
0
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 =  1- 2 e-1t +  1- 2 e-2t
2
1
1
2
Using induction one obtains
(1.14)
An(t) = An(t;1,...,n) = c1e-1t +  + cne-nt
(1.15)
ck = ck(1,...,n) =
1n
(1-k)(k-1-k)(k+1-k)(n-k)
2
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  10-26 e-312000t + 4.15  10-17 e-10.5t – 1.54  10-10 e-2.83  10
+ 1.54  10-10 e-1.54  10
Here is a graph of A4(t)
A t
using a range of t on the order
-6 t
-10 t
in units of 10
10
1.5
1.25
of the largest half-life. It looks
1
like 1e
-1t
.
0.75
0.5
0.25
5 10
A t
Here is a graph of A4(t) using a
9
1 10
in units of 10
10
1.5
10
10
2 10
10
2.5
10
t
10
3 10
10
10
1.5
1.25
range of t on the order of the second
1
largest half-life. It looks like
1(1 - e-4t).
0.75
0.5
0.25
500000
3
1 10
6
1.5
10
6
2 10
6
2.5
10
6
3 10
6
t
Here is a graph of A4(t)
A t
in units of 10
using a range of t on the order
17
1.5
1.25
of the second smallest half-life.
1
0.75
0.5
0.25
0.02
0.04
-6
0.00001
0.06
0.08
0.1
t
Here is a graph of A4(t)
6
using a range of t on the order of
5
the smallest half-life. The scale on
4
the vertical axis is in units of 10-25.
3
2
1
5 10
0.000015
0.00002
In order to get a more
-30
complete picture of A4(t) over a range
of t running from the smallest half-
-40
life to the largest, we make a plot of
-50
log[ A4(t) ] vs log t. Notice that there
are portions where the graph appears
-10
10
20
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 < μ2 < … < μn be the values 1, ..., n arranged in increasing order. Then
4
(1.16)
An(t)  μ1μmt m-1/(m-1)!
for 1/μm+1 << t << 1/μm and 1  m  n. ((1.16) holds for t << 1/μn if m = n.) For simplicity,
we shall assume in the remainder that μ1 << μ2 << … << μn. Then the relative error in the
m
2(m-1)
n
approximation (1.16) is less than  for    t < 2 when m = 1, 2, 3, …, n-1 and for t < 
m+1
m
n
when m = n; see Theorems 3 and 5. Here a  t means b < t for some b  a. In (1.2) one has
A4(t)  14t for 9.6 years < t < 7,000 years and A4(t)  124t2/2 for 5.7 hours < t < 1 day
and A4(t)  1234t3/6 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)   1- 2 e-1t  1e-1t
2
1
1
1
The left hand approximation holds with relative error less than  for t   ln(  ); see
2
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 t >> 1/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)
2

with relative error is less than  for    t   ; see Theorem 5(b) and (c). In (1.2) one has
3
1
A4(t)  1(1 – e-4t) with an error less than 2% for 10 years < t < 1.3  108 years. (1.19) can
also be generalized; see Theorem 6.
5
2. Properties of An(t). Let
Fn(t) = Fn(t;1,...,n) = e-1t *  * e-nt = An(t)/(1n)
(2.1)
Qn(t) = Qn(t;1,...,n) =
n
n
j=1
j=1
 Nj(t) =  Aj(t;1,...,j)/j
(1,...,n; ) = 1n / [ (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) Qn (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) = e-αtFn(t;1-,...,n-) and
An(t;1,...,n) = e-αt(1,...,n;)An(t;1-,...,n-) for any .
t
 f(t) dt for
(d) Fn(t;0,...,0) = 1**1 = t n-1/(n-1)! and Fn(t;,...,) = t n-1e-t/(n-1)!. f(t) * 1 = 
0
any function f(t) and f(t) * t n-1/(n-1)! is obtained by integrating f(t) from 0 to t a total of n
-t(#)
times. (e) Fn(t) = tn-1 
d = tn-1e- t/(n-1)! where the integral is an (n-1)-fold
e

multiple integral over  = {(1,…,n-1): 0  1  1, 0  2  1,…, 0  n-1  n-2} with
# = 1+1(2-1)+…+n-1(n-n-1) and d = d1…dn-1 and 1    n.
(f) Fn(t)  t n-1e-1t/(n-1)! for t  0. (g) If the j are non-negative then t Fn (t)  (n-1)Fn(t) for
t ≥ 0.
6
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) * t n-1/(n-1)! follows from this and
the formula for 1**1. (e) Simonsen [2] showed that (-1)n-1Fn(t) is the (n-1)st divided
difference with respect to  of the function e-t. The (n-1)st divided difference of an arbitrary
function g() is given by
(2.2)
(n-1)
(n-1)
g[1,...,n] = 
g (#) d = g ()/(n-1)!

where g (k)() = d kg/dk; see Issacson [1, pp. 249-250]. The result follows from this.
(f) Since e-jt  e-1t it follows that Fn(t)  Fn(t;1,...,1) = t n-1e-1t/(n-1)!. (g) It follows from
-t(#)
(e) that tFn (t) = (n-1)Fn(t) - tn
d. Since # ≥ 0 for all , one has
(#)e

-t(#)

d ≥ 0 and the result follows. //
(#)e

Theorem 2. Assume 0 < 1 < ... < n and f(t)  0 for t  0. (a) If f(t) is
non-decreasing for t  0 then f(t) * An(t)  f(t) for t  0. (b) Suppose there is an  such that
 < 1 and eαtf(t) is non-decreasing for t  0. Then f(t) * An(t)  (1,...,n;)f(t) for t  0.
(c) If 1 < ... < m and 1 < 1, then eα1tFm(t; 1,...,m) is increasing for t  0 and
Fm(t; 1,...,m) * An(t)  (1,...,n;1)Fm(t; 1,...,m) for t > 0.
7
(d) An(t;1,...,n)  (m+1,...,n;1)Am(t;1,...,m) for t > 0. (e) Qn(t)  (2,...,n;1)e-1t for
t > 0.
t
t
Proof. (a) One has f(t) * An(t) = 
f(t-s) An(s) ds  f(t) 
 An(s) ds. The result follows
0
0
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
f(t) * An(t) = (1,...,n;)e-t [(etf(t)) * An(t;1-,...,n-)]. By part a one has
(etf(t)) * An(t;1-,...,n-) ≤ etf(t) from which the result follows. (c) Theorem 1(c) gives
t
m(t; 1,...,m) = Fm(t;0,2-1,...,m-1) which equals 
 Fm-1(s;2-1,...,m-1) ds. The
e1tF
0
integrand is positive for t > 0, so e1tFm(t; 1,...,m) is increasing. The inequality then follows
from part b. (d) One has An(t) = 1m Fm(t;1,...,m) * An-m(t;m+1,...,n), so the inequality
follows from part c. (e) By part (d) one has Aj(t)/j  (2,...,j;1)1e-1t/j for 2  j  n.
n
Therefore Qn(t)  ce-1t where c = 1 + 1 2j-1/[(2-1)(j-1)]. It is not hard to show
j=2
that c = (2,...,n;1), so the result follows. //
8
3. Approximation theorems.
_
Theorem 3 (Small t). Let 1,…,n be non-negative, m be the mean of 1,…,m and
f(t)  0 and 1  m  n - 1. Then for t  0 one has
(3.1)
(3.2)
(3.3)
1n t n-1
(n-1)!
_
 n t n-1
(1 - nt)  An(t)  1(n-1)!
_
1n
n-1
[
f(t)
*
t
]
(1

f(t) * t n-1
nt)  f(t) * An(t) 
(n-1)!
(n-1)!
1n
_
1m
m-1
[
A
(t;

,...,

)
*
t
]
(1

A (t;m+1,...,n) * t m-1
n-m
m+1
n
mt)  An(t) 
(m-1)!
(m-1)! n-m
1m
Proof. The right inequality in (2.1) follows from Theorem 1f. To prove the left
inequality we need to show
(3.4)
tn-1
(n-1)!
-
(1 +  + n) tn
n!
 Fn(t)
If n = 1, this is the well known inequality 1 - t  e-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)
tn-1
* e-n+1t
(n-1)!
-
(1 +  + n) tn
n!
* e-n+1t  Fn+1(t)
Since 1 - n+1t  e-n+1t it follows that
(3.6)
tn-1
*
(n-1)!
tn-1
tn-1
1 - n+1 (n-1)! * t  (n-1)! * e-n+1t
9
tn-1
tn
tn-1
tn+1
tn
tn+1
tn-1
Since (n-1)! * 1 = n! and (n-1)! * t = (n+1)! it follows that n! - n+1 (n+1)! * t  (n-1)! * e-n+1t.
Since e-n+1t  1 it follows that
(1 +  + n) tn
n!
* e-n+1t 
(1 +  + n) tn
n!
*1 =
(1 +  + n) tn+1
(n+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)
tn-1
*
(n-1)!
_
tn-1
tn-1
f(t) - nt [(n-1)! * f(t)]  Fn(t) * f(t)  (n-1)! * f(t)
To prove this, note that from (3.4) it follows that
(3.7)
tn-1
*
(n-1)!
_ tn-1
tn-1
f(t) - n (n-1)! * f(t)  Fn(t) * f(t)  (n-1)! * f(t)
One can estimate the second term on the left by
t
tn

* f(t) = 
t
sn
 s n-1 f(t-s) ds = t [tn-1 * f(t)]
f(t-s) ds  t 
0
0
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
 and using the fact that Am(t) * An-m(t;m+1,...,n) = An(t). //
m(t; m+1,..., n)
10
^
Theorem 4 (Large t). Assume 0 < 1 < ... < n. Let # and m be defined by
n
1
#
=

n
1
and
j
j=1
1
^
m
=
   . Then the following are true. (a) Suppose f(t)  0 and f(t) is
1
j- 1
j=m
non-decreasing for t  0 and there is an  such that  < 1 and etf '(t) is non-decreasing for
t  0. Then
f(t) - f(0)Qn(t) -
(1,...,n;) f  (t)
 f(t) * An(t)  f(t)
#
for t > 0. (b) If q  1 then [1 - q/(#t)] t q  t q * An(t)  t q for t  0. (c) If 0  1 < ... < m
and 1 < 1. Let  = 1. Then
[1 -
q(1,...,n;)
]
#t
Fq+1(t;0,1,...,q)  Fq+1(t;0,1,...,q) * An(t)  Fq+1(t;0,1,...,q)
for t  0. (d)
[1 - (3-1,...,n-1;2-1)e-(2 1 t] (2,...,n;1)1e-1t  An(t)  (2,...,n;1)1e-1t
for t  0. (e)
[1 -
(m-1)(m+1-1,...,n-1;2-1)
] (m+1,...,n;1)Am(t)  An(t)  (m+1,...,n;1)Am(t)
^
m+1t
for t  0.
Proof. (a) The right inequality is Theorem 2(a). If one integrates by parts and uses
the fact that Qn = - An (by Theorem 1(a)) one obtains f(t)*An(t) = f(t) - f(0)Qn(t) - f (t)*Qn(t).
Theorem 2(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).
Then f  (t) = Fq(t;1,...,q), so it follows from Theorem 2(c) that etf  (t) is increasing. So f(t)
11
and  satisfy the hypotheses of part (a) and f(0) = 0. By Theorem 1(i) one has f  (t)/f(t)  q/t.
So (c) follows from (a). (d) The fact that An(t) ≤ (2,...,n;1)1e-1t follows from Theorem
2(d). Using Theorem 1(b) and (c) one can write
An(t) = (2,...,n;1)1e-1t[1*An-1(t;2-1,...,n-1)]. By Theorem 1(a) one has
1*An-1(t;2-1,...,n-1) = 1 - Qn-1(t;2-1,...,n-1). Using Theorem 2(e) one obtains
1 - (3 1,...,n-1;2-1)e-(2 1 t ≤ 1 * An-1(t;2-1,...,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)1m e-1t[Fm(t)*An-m(t)], where
Fm(t) = Fm(t;0,2-1,...,m-1) and An-m(t) = An-m(t;m+1-1,...,n-1)]. By part (c) one has
(1 -
(m-1)(m+1-1,...,n-1;2-1)
) Fm(t) ≤ Fm(t)*An-m(t).
^ t
m+1
When combined with the previous, this proves the left inequality in part (e). //
Theorem 5 (Intermediate t.). Suppose 0 < 1 < 2 <  < n and let #m be defined
n
by
1
=
#
m
  . Then for t > 0 and 2  m  n-1 one has
1
j
j=m
[1
(m-1)
-^
m+1t
_
-  mt ]
1m tm-1
(m-1)!
 An(t) 
1m tm-1
(m-1)!
Proof. The right inequality is included in Theorem 3. Combining Theorem 3 and
(m-1)
_
Theorem 5b gives [ 1 - ^
] [ 1 -  mt ]
m+1t
1m tm-1
(m-1)!
 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. //
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Theorem 6 (Intermediate t.). Let 0 < 1 < 2 <  < n and 2  m < m+r  n-1. Let
b = 1 if m > 1 and b = (r+2n;2) if m = 1. Let
tm-1
Hn,m(t) = Hn,m(t;1,...,n) = (m-1)! * An(t;1,...,n)
Then for t > 0 one has
(m+r-1)b
[1- ^
m+r+1t
_
- mt ] 1mHr,m(t;m+1,…,m+r) < An(t) < 1mHr,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
_
1 - mt  An(t)/g(t)  1 where g(t) = An-m-r(t;m+r+1,...,n) * U(t) with
U(t) = 1m+r Fm+r(t;0,…0,m+1,...,m+r). Apply Theorem 4(c) with Fq+1(t;0,1,...,q) = U(t)
^
to obtain 1 - (m+r-1)/(
m+r+1t) < g(t)/U(t) < 1. Combining this with the previous inequality
and using the fact that U(t) = 1mHr,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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