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CHAPTER 3A
Radioactive decay equations
3.1 The radioactive decay law As discussed in Chapter 2, the decay rate of a radioactive substance was found to decrease exponentially with time. Radioactive decay was found to follow a statistical law, that is, it is impossible to predict when any given atom will disintegrate. The rate at which a radioactive substance N 1 decays is given by
1 1
1 N
dt
dN
where N is the number of radioactive atoms and λ 1 is called the decay constant , which
represents the probability per unit time for one atom to decay. The fundamental assumption in the statistical law for radioactive decay is that this probability for decay is constant, that is, it does not depend on time or on the number of radioactive atoms present. Eq. 3.1 can be integrated with the boundary condition that the initial number of
radioactive atoms is N. This leads to the exponential law of radioactive
decay :
1 o^ = N 1 (^0 )
N No^ e^1^ t 1 1 = −^ λ (3.2)
Denoting the daughter product of N 1 as N 2 , and for the case in which N 2 is a stable daughter product, the accumulation of N 2 is simply given by
1
1
2 2 1
2 2 1 = + −
o t
o o t
N N N e
N N N e λ
λ (3.3)
where No 2 is the initial amount of N 2 prior to the accumulation of radioactive decay
products of N 1. For measurement purposes, it is often convenient to express Eq. 3.2 in terms of the activity A of a radioactive substance, which is defined to be the number of decays per
unit time or simply A= λ 1 N 1. If we multiply both sides of Eq. 3.2 by λ 1 , we find that
N No^ e^ t dt
dN (^) 1 1 1 1 1
− 1 =λ =λ −^ λ (3.4)
It follows that
A = Ao^ e −^ λ t 1 1 (3.5) where is the initial activity of the radioactive substance. The unit most commonly
used in activity measurements is the curie (Ci), which is defined as 3.7×
A^ o 1 (^10) decays/s. The half-life of a radioactive substance is defined as the elapsed time over which the radioactive substance has decreased to half its original abundance. Thus, by letting
N (^) 1 = N 1^ o / 2 in Eq. 3.2, the half-life is given by
ln 2 0. 69315 t 0 (^). 5 = = (3.6)
The mean life, t , or the average amount of time a radioactive element lives, is given by
(^01) 0 0 1 1 1
= − ∫ = ∫ λ =λ∫ λ^ =
∞ ∞ ∞ (^) − t Ndt te dt N
tdN N
t t o o
Thus, the mean life is simply the inverse of the decay constant. During the mean life, the abundance of a radioactive substance decreases by a factor of 1/e.
3.2 Branched decay 3.2.1 Basic equations In some cases, the parent N 1 decays to two different products, a process termed branched decay. Calling these two decay modes a and b , we define the probability of each decay mode as
1
1
1
1
N
dN dt
N
dN dt
b b
a a
λ
λ (3.8)
The total decay rate of N 1 is
( a b ) total
total a b
N N
dt
dN dt
dN dt
dN 1 1 , 1 , 1 1 ,
where ( λ 1 , a + λ 1 , b ) = λ 1 , total is the total decay constant. It follows that N 1 decays according
to the following law
N No^ e^1 total^ t 1 1 = −^ λ (3.10)
Thus, if we were to measure only the abundance or activity of the radioactive nuclide, N 1 , we would have no way of distinguishing between the two decay modes because we can only measure the total decay constant. However, if we were to monitor the evolution of the abundance or activity of the two daughter products of N 1 as a function of time, we would be able to distinguish
between the two decay modes. For example, we know that a fraction λ 1 , a / λ 1 , total of N 1
decays by mode a and a fraction λ 1 , b / λ 1 , total decays by mode b. Denoting the two types
of daughter products as N2,a and N2,b , the formulas describing the accumulation of each of the daughter products are
1
1 ,
2 , 2 , 1 , 1
2 , 2 , 1 , 1 o t b total
o b b
o t a total
o a a total
total
N N N e
N N N e λ
λ
−
−
= + −
where N (^) 2 o , a and N (^) 2^ o , b are the initial amounts of daughter product N2,a and N2,b present
before the decay of N 1.
3.2.2 Example: decay of 40 K to 40 Ar and 40 Ca An example of a branched decay is the decay of naturally occurring 40 K to stable (^40) Ar and 40 Ca, the former by electron capture and the latter by β- emission (negatron
emission). Electron capture decay to 40 Ar is favored by 11.2 % of 40 K atoms and beta decay to 40 Ca is favored by 88.8 %. Accordingly, the decay constants for each branching
decay, λ K,Ar and λ K,Ca , are
10 1 ,
10 1 ,
- 962 10
− −
− −
= ×
= ×
yr
yr
KCa
KAr
A better understanding of Eq. 3.17 can be had if we look at some examples. For
example, let us assume that λ 1 << λ 2
− λ 1 t
, that is, N 1 decays much slower than N 2. It follows
that for long times the quantity e is much larger than e , and therefore Eq. 3. simplifies approximately to
Secular equilibrium in a decay chain
A
B
N 1 N 2
N 1
N 1 N 2
N 2
λ 2 N 2
λ 1 N (^1)
A
t
N 1 N 2 N 3 N 4
Figure 1. Secular Equilibrium.
− λ 2 t
1 1 1 2
1 2
N ~ Noe^1 t λ N
or in other words
λ 1 N 1 = λ 2 N 2 (3.19)
which essentially means that the activities (or decay rate) of N 1 and N 2 are equal. This asymptotic state is called secular equilibrium and is illustrated schematically in Figure X. Secular equilibrium is another way of saying a system has reached steady-state, that is, there is no longer a change in the decay rates of each nuclide. For example, let N 2 represent the number of people in a restaurant. At 5 PM, the restaurant doors open. Initially, there are no people in the restaurant, but immediately after 5 PM, there is a steady stream of people entering the restaurant. Assuming each person finishes eating and leaves the restaurant in about 45 minutes, we would see that initially the number of people in the restaurant would increase, but after a while, a maximum number will be reached as people start to exit. At this point, the amount of people leaving the restaurant equals the amount entering. This situation is called steady state and is a form of secular equilibrium. At secular equilibrium , it can be shown that the stable daughter nuclide (in this case, N 3 ) increases as
N (^) 3 = N 1 o^ ( 1 − e −^ λ^1 t ) (3.20)
from which it can be seen that the growth of N 3 depends only on the decay constant of N 1. We leave the proof of Eq. 3.20 as an exercise for the student. Another scenario of Eq. 3.16 is when N 1 decays only slightly slower than N 2 , e.g.
λ 1 < λ 2. In this case, we can combine Eq. 3.2 and Eq. 3.17 for the case in which there is
no initial nuclide of N 2. We thus have
( 1 ( ) ) 2 1
2 1 1
(^2 2) e 2 1 t N
N λ λ
For long times, the exponential term goes to zero such that the ratio of the activities of N 2 and N 1 approaches a limiting constant, that is,
2 1
2 1 1
2 2 1
2
N
N
A
A
This situation is known as transient equilibrium.
Another situation is when λ 1 > λ 2 , where N 1 decays faster than N 2. For long times,
the contribution from N 1 becomes insignificant and the abundance of N 2 rises to a maximum and decays with its characteristic decay constant. For the case in which there is no initial N 2 , Eq. 3.17 simplifies to
N No^ e^2^ t 1 1 2
1 2 ~
λ
3.3.2 Generalized decay series We now treat the more general case where a decay series is composed of n-chains. For each nuclide i in a decay series, the following general equation must hold
i i i i
i (^) N N dt
dN
The general solution for the case of N 1 o atoms of N 1 with no other nuclides initially
present is given by
o ( t t n t )
n
i
t i
o n n n
n
i
N ce ce ce
A N N ce
λ λ λ
− − −
=
−
1 1 1 2 2 K
1
1 (3.25)
where
( (^) m )( (^) m ) ( (^) n m )
n
n
i
i m
n
i
i cm
=
=
L
L
1 2
1 2
1
1 ' ( ) (3.26)
where the prime on the lower product indicates that the i=m term is omitted. Secular
equilibrium in an n-series decay chain is given by λ 1 N 1 = λ n Nn. Some examples of long
decay series include 238 U, 235 U, and 232 Th, which decay to 206 Pb, 207 Pb, and 208 Pb, respectively, through a series of intermediary decay products. For example, Fig. 2 shows
A 1 = λ 1 N 1 = R (3.31)
Eq. 3.31 essentially states that for long times, the production rate equals the decay rate, and therefore the activity of N 1 becomes constant. The production of radioactivity occurs in a number of situations. A man-made example is the production of radioactive nuclides by neutron bombardment of target nuclei in a nuclear reactor or in an accelerator. In such a case, the production rate of a radionuclide is
R = No σ I (3.32)
where No is the number of target nuclei, I is the neutron flux (number of incident neutrons per unit area) and σ is called the cross-section (cm^2 ), which represents the probability of neutron capture by the target nuclei. Because the number of target nuclei No must decrease for every radionuclide N 1 produced, the production rate R must decrease. However, because the probability of neutron capture is so small R can generally be considered constant. An example from nature is the production of radioactive nuclides by the bombardment of stable nuclides by secondary cosmic ray neutrons. In the Earth’s atmosphere, cosmic ray bombardment of oxygen, nitrogen, and argon (“spallation reactions”) can lead to the formation of such radionuclides as 3 H, 10 Be, 14 C, 26 Al, 32 Si, (^36) Cl, 29 Ar and 81 Kr. The example of 14 C is discussed below.
3.4.1 Application: Cosmogenic 14 C dating An example of a cosmogenic radionuclide is 14 C. Carbon-14 is produced by the reaction of cosmic-ray neutrons with the nucleus of stable 14 N
o^1 n +^147 N →^146 C + 11^ H (3.32)
The neutrons are created by reactions between cosmic rays (dominated by protons) and atmospheric gases. At the same time as 14 C is being produced it is also decaying by emission of a negative beta particle (t0.5 = 5730±40 y):
146 C → 147 N + β −+ v + 0. 156 MeV (3.33)
An equilibrium abundance of 14 C (in the form of 14 CO 2 ) is therefore attained in the atmosphere due to its simultaneous production and decay. The search for measurable amounts of 14 C in the atmosphere and in living matter was pioneered by W. F. Libby at the University of Chicago. Libby demonstrated that the activity of 14 C in the atmosphere was nearly constant throughout all latitudes. He and co- workers then showed that 14 C could be used to date the time of death of organic matter, and for this pioneering work, Libby received the Noble Prize for chemistry in 1960. The use of 14 C to date organic material is known as “radiocarbon dating” and is based on the following assumptions. Because of the rapid mixing times of the atmosphere, the 14 C/^12 C concentration in the atmosphere is nearly homogeneous. The activity of 14 C in living plants is maintained at a constant level because plants are constantly exchanging CO 2 with the atmosphere, which itself has attained steady state levels of 14 CO 2. However, when a plant dies, it ceases to exchange with the atmosphere. As a result, the activity of (^14) C in a dead plant begins to decay because no new 14 C is being replenished in the dead
plant. The age of decayed matter, that is, the elapsed time since such matter was alive, is determined by the exponential decay law t A Ao e = −^ λ (3.34)
where A is the present activity of 14 C and Ao was the original 14 C activity when the plant was alive. A very important assumption in Eq. 3.34 is that the current activity of 14 C in living matter has been constant through time and does not depend on geography or elevation. While short atmospheric circulation times render the second assumption mostly valid, the assumption that the 14 C activity of the atmosphere has not changed in the past is not generally valid. This is because it is likely that the cosmic ray flux to Earth has changed due to changes in the source of cosmic rays and to changes in the magnetic field of the Earth, which ultimately guide the incoming charged particles in cosmic rays. While the source of cosmic rays is not known exactly, it is likely that a large component originates from the sun because variations in the cosmic ray flux appear to correlate with solar flares. In order to account for these variations, geologists have attempted to calibrate the 14 C timescale with tree-ring data (which takes us back 6000 years) and with U-Th series data on corals (which takes us back 10,000 years). Radiocarbon ages therefore need to be corrected to account for changes in the 14 C content of the atmosphere. We will discuss the assumptions of 14 C-dating in more detail later.
3.5 Geochronology From the above discussions, it is easy to see how Eq. 3.2 or 3.3 can be rearranged so that one can obtain the elapsed time over which a nuclide has decayed. Provided one knows the amount of initial radionuclide, the present amount of radionuclide and the decay constant, calculating the elapsed time is trivial. This concept is illustrated qualitatively by using an hour glass as an example. Assuming we know the rate at which sand falls through the neck of an hour glass, we could always determine the time at which the hourglass clock started ticking if we know the initial amount of sand in the upper vessel and the current amount remaining in the upper vessel. Similarly, if we knew the initial amount of sand in the lower vessel and the current amount in the upper and lower vessels, we could also derive the elapsed time. However, if we want to apply this to the general case of dating geologic materials, knowing the initial amount of radionuclide or daughter product is not trivial. To see how we can apply Eq. 3.3 to dating rocks, let us explore the Sm-Nd isotopic system (Sm = Samarium; Nd = Neodymium) in which 147 Sm alpha decays to (^143) Nd with a half life of 106 Gy (λ=6.54 x 10-12 (^) / yr). Eq. 3.3 is expressed as follows
(^147) Sm ( e λ t (^) − 1 )+ (^143) Ndo = (^143) Nd (3.35)
Eq. 3.35 is not particularly useful to geologists as most geologists measure isotope ratios on mass spectrometers rather than the absolute number of atoms. For this reason, we will modify Eq. 3.35 so that it can be expressed in terms of isotope ratios. Let us use another isotope 144 Nd, which is stable and has no radioactive parent (meaning that its total abundance in the solar system has remained constant), to normalize Eq. 3.35. This yields
Nd
Nd Nd
Nd e Nd
Sm o
t o 144
143 144
143 144
147 ( λ^ − 1 )+ = (3.36)
granite therefore represents the time at which it passed through its closure temperature, that is, the time at which the granitic pluton crystallized. It is at this point that the “clock” starts ticking. More generally, the age of a rock refers to the time when the constituent minerals were last homogenized. For sedimentary rocks, this could represent the time of diagenesis. We now return to Eq. 3.36 and apply it to the granite example. Let us pick mineral phases with different parent-daughter ratios, e.g. different [^147 Sm/^144 Nd] ratios. Depending on the degree to which Sm and Nd are partitioned into a particular mineral, two different minerals are likely to have different parent daughter ratios. Those with high parent-daughter element ratios will generate, by radioactive decay, more daughter products for a given interval of time than those minerals with low parent-daughter ratios. Those without any parent element will retain the initial daughter isotopic composition indefinitely. Thus, if we assume that all the minerals in a system began with the same initial daughter isotopic composition, e.g. [^143 Nd/^144 Nd]o, we will find that with time, every different mineral in a plot of [^143 Nd/^144 Nd] versus [^147 Sm^44 Nd] will fall on a straight line having a slope of
m = ( e^ λ t^ − 1 ) (3.37)
The y-intercept (e.g., where [^147 Sm^44 Nd]=0 ) would correspond to the initial Nd isotopic composition [^143 Nd/^144 Nd]o of the system and all the minerals. The structures of Eqs. 3.36 and 3.37 show that the tieline connecting all of the mineral points in such a plot will rotate counter-clockwise with age. At zero age, i.e. the time at which the “clock” started ticking, the tieline has a slope of zero. At any other time, the age of the rock can be trivially obtained from the slope of the line
ln( 1 )
t = m +
where m is the slope.
Problem Set
- Calculate the binding energy per nucleon for 6 Li, 31 P, 108 Pd, 195 Pt, and 238 U from a) the masses given in Appendix Table 1, and b) from the semi-empirical binding- energy equation.
- Define alpha, beta (negatron and positron) electron capture decay. Without consulting the chart of nuclides, specify Z , N , and A for the daughters of the following radionuclides which decay by positron or electron capture decay: 2211 and. Do the same for 14 and which decay by negatron emission, and for 147 and which decay by alpha emission.
Na^8839 Y
6 C 3787 Rb 62^ Sm (^23492) U
- What is fission decay? For 238 U, estimate the energy available for a) alpha decay and b) spontaneous fission into equal fragments.
- Osmium is made up of 7 isotopes (184, 186, 187, 188, 189, 190, and 192). Based on mass spectrometry measurements, a sample of Os is found to have the following isotopic ratios 186 Os/^188 Os = 0.11996, 187 Os/^188 Os = 0.17444, 189 Os/^188 Os = 1.21966, (^190) Os/ (^188) Os = 1.98389, and 192 Os/ (^188) Os = 3.08261 (assume 184 Os/ (^188) Os is almost negligible). Calculate the atomic weight of Os. Note that 187 Os is the radiogenic daughter product of 187 Re by β- decay and 186 Os is the daughter product of 190 Pt by α decay. The half-life of 187 Re is ~43 Gy and therefore the relative abundance of 187 Os varies in nature. The half-life of 190 Pt is so long (650 Gy) that only under certain circumstances can variations in 186 Os be measured.
- (Faure 11.1) Given that the atomic abundances of Pb isotopes relative to 206 Pb are stated as 204 Pb = 0.143, 206 Pb = 1.00, 207 Pb = 12.95, 208 Pb = 21.96, recalculate their abundances in terms of atom percent and determine the atomic weight of this sample of Pb. The masses of the isotopes in amu are 204 Pb = 203.9730, 206 Pb = 205.9744, (^207) Pb = 206.9759, and 208 Pb = 207.9766. Answer (207.55).
- The rate at which a radioactive nuclide decays is linearly proportional to the amount of radioactive nuclides present, e.g.
1 1
1 N
dt
dN
This equation is trivially derived from the following equation
dt N
dN 1 1
Explain in words what Eq. 2 implies. Do so by first explaining what the left-hand side of Eq. 2 means. Then explain what the decay constant means. Explain why the decay constant does not vary with the number of radioactive nuclides present. Can you think of any other situation in physics, biology, ecology, etc., which can be described by a first-order differential equation.
