UNDERSTANDING THE ORIGIN AND MEANING OF THE
RADIOACTIVE DECAY EQUATION
Stephen P. Huestis
Department of Earth and Planetary Sciences, University of New Mexico,
Albuquerque, New Mexico 87131 [email protected]
ABSTRACT
The radioactive decay equation can be derived, as an
exercise in calculus and probability, as a consequence of
two physical principles: a radioactive nucleus has no
memory, and decay times for any two nuclei of the same
isotope are governed by the same probability
distribution. The first principle implies that this
distribution has a continuous exponential probability
density function. Then, knowing the probability of
survival of a single nucleus to a specified time, the
second principle allows the number of a collection of
nuclei surviving to this time to be treated as a random
variable governed by the discrete binomial distribution.
The familiar radioactive decay equation does not give the
actual number remaining at this time, but rather the
expected value of this distribution. With this proper
interpretation of the radioactive decay equation, the
number of nuclei need not be a large. Algebraic and
numerical experiments illustrate, however, that as the
number of nuclei grows to the large values associated
with geochronological studies, the probability of
significant departure from the expected value becomes
negligibly small.
Keywords: Education - Geoscience; Geochronology.
INTRODUCTION
Radioactive decay of various naturally-occurring
isotopes provides a profoundly important means of
absolute age-dating of geological events. Geochronology
students are told that, for a given radioactive isotope, the
radioactive decay equation is
N (t) = N 0e − λt
(1)
where N0 is the number of nuclei present at time t = 0 and
λ is the decay constant specific to this isotope.
Commonly (e.g. Faure, 1986; Dickin, 1995), it is claimed
that the number of atoms decaying per unit time is
proportional to N(t), the number of atoms present at time
t, so that
dN (t)
= −λN (t).
dt
(2)
Then (1) is “derived” by integrating (2), an approach that
most students will find unilluminating because of the
lack of clear justification of (2). And while the
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probabilistic nature of radioactive decay is often
acknowledged (e.g. Dalrymple and Lanphere, 1969), the
term N(t) is usually described simply as the number of
nuclei remaining after time t > 0, implying an apparent
exactness which would seem to be incompatible with the
underlying randomness of the process.
This interpretation of N(t) must have some validity,
as evidenced by the success of radiometric dating
schemes. Yet, the discerning student might notice
another conceptual problem. Equation (1) expresses N(t)
as a continuous function of t: for almost all t it will predict
a non-physical situation of a non-integer value for N(t).
But, radioactive decays are discrete, instantaneous
events, each of which discontinuously reduces N(t) by 1,
at the time of decay. Presumably, (1) can at best be an
approximation. If we could observe N(t) with fine
enough resolution, we must see that it is actually a
monotone-decreasing
piecewise-constant
function
which assumes only integer values, with negative steps
of unit magnitude at times of decay of individual nuclei.
Furthermore, if we were to repeat the experiment
with the same N0, a different N(t) would result because of
the probabilistic nature of radioactive decay. For
example, in the extreme case of N0 = 1, N(t) will remain
constant at 1 until the single unpredictable time of decay,
whereafter it remains at 0. Such a result is far from the
form of Equation (1).
Some expositions (e.g. Nagy et al., 1998) will add to
(1) the proviso that N0 must be large. Presumably, for
large N 0, the actual N(t) is well-enough approximated by
(1) that differences are, if not indiscernible, at least
unimportant. But, how large must N0 be? More
specifically, can we study the approach of N(t) to the
form of Equation (1), as N0 increases?
Students who have completed a calculus-based
course in probability are in a position to understand the
true meaning of (1), not as the number of nuclei
remaining at t, or even an approximation to this number,
but instead as the very different concept of the mean, or
expected value, of a probability distribution. Under its
correct interpretation, the exponential form of (1) will be
seen to be an inevitable result of just two plausible
physical principles, here called axioms I and II, which
govern the probability distribution for decay time:
(I) A single radioactive nucleus has no “memory” of its
past history.
(II) Decay times for any two nuclei of a given isotope are
governed by exactly the same probability
distributions.
Journal of Geoscience Education, v. 50, n. 5, November, 2002, p. 524-527
In the next section we use axiom I to derive the
probability distribution for the decay time of a single
nucleus. Then, in the following section, the actual
meaning of the radioactive decay equation (1), for a
collection of N0 identical nuclei, is developed using
axiom II. It should be noted that it has long been known
that these axioms lead to the radioactive decay law
(Meyer and Schweidler, 1927). But most sources,
particularly those likely to be visited by geology
students, avoid this interpretation and hence obfuscate
the nature of (1) as a description of radioactive decay.
The different derivation presented here can remedy this
for adequately prepared students, and has the pedagogic
advantage of using a variety of textbook tools from
calculus and probability, in a non-artificial problem.
and, dividing by non-zero τ – t,
P (τ) − P (t) P (t)[P (τ − t) − 1 ]
=
τ −t
τ −t
Now take the limit as τ→ t:
dP (τ)
dτ
τ=1
=
dP (t)
 P (τ − t) − 1 
= P (t) lim
τ→t 
dt
τ −t 
Because P(0) = 1, this limit is an indeterminate form, 0/0;
an application of L’Hospital’s rule gives
dP (t)
dP (t)
= P (t)
dt
dt
t =0
(4)
dP (t)
is some constant , say –λ, whose value is
dt t= 0
not fixed by the argument. Instead, it is a free parameter
DECAY PROBABILITY FOR A SINGLE
whose value governs the decay rate of the isotope under
NUCLEUS
consideration.
Equation (4) becomes
First consider the experiment of beginning to observe a
single unstable nucleus at a time t = 0, and asking for the
probability, P(t), that this nucleus will survive without dP (t) = −λP (t),
decaying to a time t > 0 or beyond. Obviously, P(0) = 1 dt
and P(t) is a decreasing function of t.
The student must now be reminded of the notion of a simple first-order differential equation for P(t), whose
conditional probability: the probability of occurrence of solution is
Now,
two (not necessarily mutually exclusive) events A and B P (t) = e −λt .
(5)
is given by the conditional probability of A occurring,
given that B has occurred, multiplied by the probability This is the probability asked for and is what is needed for
the argument below, but it is not yet the continuous
of occurrence of B. We write this as
probability density function governing the decay time
P(A ∩ B) = P(A| B) P(B).
random variable. In fact, denoting this probability
density function by f(t), we have found
Let t > 0 be some arbitrary time and consider any τ > t. Let
∞
(6)
P (t) = e − λt = ∫t f (τ)dτ
A = P(τ) and B = P(t), so
P(τ ∩ t) = P(τ | t) P(t).
(3)
Differentiating both sides of (6) with respect to t gives
Clearly, P(τ ∩ t) = P(τ) because they represent the same f (t) = λe − λt ,
event: if the nucleus has survived to τ > t, it also has
survived to τ and t.
the familiar exponential distribution, which in fact
Now invoke axiom I. If the nucleus has no memory
always arises for processes without memory.
of past history, survival to t resets its clock in the sense
that
RADIOACTIVE DECAY OF MULTIPLE
P(τ | t) = P (τ - t).
Thus, (3) becomes
P(τ) = P(τ - t) P(t) for all τ > t > 0.
Then
P(τ) – P(t) = P(t) [P(τ-t) – 1],
Huestis - Understanding the Radioactive Decay Equation
NUCLEI
While (1) and (5) have the same functional form, they
represent very different concepts. Equation (5) refers
strictly to a single nucleus and symbolizes the
probability of the particular event of survival to t, related
to the underlying continuous probability density
function by (6). On the other hand, (1) is not a probability
at all; nor, however, does (1) give the exact value of nuclei
525
remaining at t, when interpreted correctly. Instead as
shown below, N(t) in (1) must be interpreted as the
expected value (mean) of a particular discrete binomial
distribution. It is this true meaning of N(t) which is
usually overlooked in description of the radioactive
decay equation. And, although some discussions state
that N0 must be large (without specifying how large) for
(1) to be valid, this is in fact not the case.
Recall first the relevant discrete probability
distribution, the general binomial distribution. A
binomial experiment comprises n repeated trials, each of
which is characterized by an outcome whose (constant)
probability of success is p. The probability of x successes
in n trials is
With n = N0 and p given by (5), (9) becomes
µ(t) = N0 e-λt.
(10)
This is just the radioactive decay equation (1) if we
identify N(t) not with the number of nuclei remaining at
t, but rather the expected value of the binomial survival
random variable, a very different concept. This is the
meaning of the radioactive decay law that will be used
throughout the remainder of the paper. As an expected
value, µ(t) given by (10) is exact for all t, even at times for
which it has a non-integer value: the expected value of a
probability distribution does not necessarily correspond
to a possible outcome of the random variable. And,
reiterating an important concept, under this
interpretation of (10), N0 need not be a large integer.
n 
b (x ; n , p ) =  p x (1 − p ) n − x x= 0, 1, ..., n
(7)
To further explore the decay law (8), and µ(t) as its
x 
expected value, it is convenient to work at the half-life, t½.
While usually so defined, t ½ is in fact not the time it takes
where the binomial coefficient
N0/2 nuclei to decay; the time required for half to decay
n!
is a random variable whose exact value cannot be
n 
 =
predicted, and would vary as the experiment is repeated.
x  x !(n − x ) !
Rather t ½ is that time for which the expected value of the
Now, for the isotope under consideration, let us start number, x, of surviving nuclei is µ(t1/2)=N0/2. From (10)
with N0 nuclei at t = 0. By axiom II, the probability of
λ t1/2 = ln 2
survival past fixed time t > 0 is the same for all nuclei,
given by (5). We can therefore model the decay of N0 and from (8)
nuclei as a binomial experiment with the fate of each
N0 
nucleus as a single trial, and “success” corresponding to
(11)
b (x ; N 0 , P (t 1/2 )) = b (x ; N 0 , .5) =   (.5) N0
 x 
survival to t.
Then, for fixed t, the probability of x nuclei
remaining is
regardless of the particular value of the decay constant,
λ.
N
N 0 


x
N −x
0
b (x ; N 0 , P (t)) =  [P (t)] [1 − P (t)]
=  e − λt (1 − e − λt ) N − x
(8)
x
x
0


0


Notice that (8) represents a different binomial
distribution (different p) for each choice of t.
Furthermore, although N0 is a positive integer, equation
(8) imposes no further requirements on its size.
In terms of giving probabilities, (8) is the true and
complete radioactive decay law: for any initial N0 and
any integer x between 0 and N0, we can evaluate the
probability that exactly x nuclei remain after t. For large
N0, however, calculation of a single such probability is of
little interest: any specific outcome x, even that with the
greatest probability, will be very unlikely. More useful
for large N0 is the expected value of the probability
distribution. For the general binomial distribution (7),
the expected value, µ, is
µ = n p.
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(9)
EXPLORATION OF THE APPROACH TO
LARGE N0
Because (11) is valid for any N0 > 0, it is fruitful for the
student to explore probabilities for small N0, in order to
clarify the distinction between the number of particles
remaining and the expected value of this number. For
example, let N0 = 10. At the half-life, (11) is
 10 
b (x ; 10 , .5) =   (.5) 10 .
x 
The expected value of surviving nuclei is 5, but the actual
probability of 5 surviving is only 0.246. The probabilities
of 4 or 6 nuclei surviving are each not much less than this,
at 0.205. And, even none or all nuclei surviving to the
half-life could occur, although with extremely small
probabilities of 9.77 x 10-4 each. Of course, such
calculations can also be performed at times other than the
half-life, using equations (8) and (10).
Journal of Geoscience Education, v. 50, n. 5, November, 2002, p. 524-527
N0
200
500
1000
2000
5000
10000
20000
50000
100000
P(.495 N0 ≤ x ≤ .505 N0)
.168
.212
.272
.361
.529
.688
.845
.975
.998
b(x; N0, .5)
5.63 x 10-2
3.57 x 10-2
2.52 x 10-2
1.78 x 10-2
1.13 x 10-2
7.98 x 10-3
5.64 x 10-3
3.57 x 10-3
2.52 x 10-3
Table 1. Probabilities associated with the number of surviving nuclei at the half life.
Students then should investigate the effect of
increasing N0 to larger values, specifically with the
goal of understanding why the usual definition of
N(t) as the number of nuclei remaining at time t
becomes a harmless misstatement when applied to
age-dating. To this end, they might initially consider
the standard deviation of the binomial distribution
as a measure of its width around the expected value.
For the general binomial distribution, (7), the
standard deviation is
σ = np (1 − p )
which, for (8), becomes
P (x 1 ≤ x ≤ x 2 ) =
x2
∑ b (x ; N
x = x1
0
, P (t))
(12)
For t = t1/2, µ = N0/2 and we choose x1 and x2 as the
endpoints of a small interval centered on N0/2. For
example, let x1 = .495 N0 and x2 = .505 N0. For
increasing value of N0, Table 1 shows the result of
(12) as well as the probability that x assumes the
value of µ exactly, from (11). As N0 increases, x =
N0/2 becomes an increasing unlikely outcome, but
the probability that x falls in the range [.5 ± .005]N0
grows, reaching near certainty by N0 = 100,000.
COMMENTS
Clearly, as N0 reaches the very large values
associated with age-dating, it becomes virtually
guaranteed that x will not equal N0/2, contradicting
Again, appealing to the convenience of the half-life: the casual statement that, after one half-life, (exactly)
N0/2 nuclei remain. Yet, we have also seen that it
σ(t 1/2 ) = N 0 / 2 ,
becomes almost certain that the actual number
remaining will be negligibly different from N0/2.
so that the spread of the distribution grows with N0, (Students should understand, however, that for any
but less rapidly than µ(t1/2). We can define an ad hoc finite N0, there is a non-zero probability that the
number of surviving nuclei might be very different
fractional spread, σ’(t) = σ(t)/µ(t), so that
from N0/2). Herein lies the justification for treating
N(t) in (1) as “exactly” equal to the number of nuclei
σ' (t 1/2 ) = 1 / N 0
remaining at time t, instead of its true meaning as the
expected value of this number.
which approaches zero as N0 grows: the radioactive
decay distribution becomes relatively more REFERENCES
concentrated around the mean.
Calculation of some actual probabilities can also Dalrymple, G.B. and M.A. Lanphere, 1969.
Potassium-Argon Dating, San Francisco, W.H.
illuminate the increased narrowing of the spread of
Freeman, 258 p.
(8) about µ(t), as N0 increases. It is particularly Dickin, A., 1995, Radiogenic Isotope Geology,
interesting to compare the probability that the
Cambridge, Cambridge Univ. Press, 490 p.
outcome, x, equals µ with the probability that x falls Faure, G., 1986, Principles of Isotope Geology, 2nd
ed., New York, John Wiley and Sons, 589 p.
in some narrow percentage range around µ. To
Meyer, S. and E. Schweidler, 1927, Radioaktivität,
compute the latter, recall that the probability that x in
Leipzig, B.G. Teubner, 721 p.
(8) assumes a value between x1 and x2 at time t, is
Nagy, S., K. Süvegh, and A. Vértes, 1998, Basics of
nuclear science, in Nuclear Methods in
Mineralogy and Geology, edited by A. Vértes, S.
Nagy, and K. Süvegh, New York, Plenum, 555 p.
σ(t) = N 0 e − λt (1 − e − λt ).
Huestis - Understanding the Radioactive Decay Equation
527