Radioactivity and Half-Life

Beginner

Start from the picture of the atom you already know: a tiny nucleus in the middle, made of protons and neutrons, with electrons moving around it. The number of protons is the atom's identity card. Six protons always means carbon; 92 always means uranium. Chemical reactions only rearrange the outer electrons, so they can never change that number. Only a change inside the nucleus can.

Some nuclei are unstable: they carry too much energy, or an awkward mix of protons and neutrons. Sooner or later such a nucleus throws out a tiny particle or a flash of very energetic light, leaving a different nucleus, often of another element. That is radioactive decay.

When does it happen? By pure chance. Think of glow-in-the-dark stars on a bedroom ceiling: after the lamp goes off, their atoms give back stored light one at a time, at unpredictable moments, so the glow fades gradually. Radioactive nuclei behave the same way, with one surprise: they never age. A nucleus that has waited a million years is no closer to decaying than a brand-new one.

In the animation, each gold disc is a nucleus. They turn grey one by one, each spitting out a red particle, and you cannot guess which is next. Yet the crowd is predictable: each time the clock hand makes a full turn, half of the gold nuclei still left decay. One turn, 50 remain; two turns, 25; three turns, about 12. One turn of the clock is the half-life.

Check yourself: a sample starts with 800 radioactive nuclei. How many are left after three half-lives? Answer: $800\to 400\to 200\to 100$, so 100, not zero.

Half-lives range from a split second to billions of years. Carbon-14's, about 5,700 years, lets scientists date old wood and bone.

Intermediate

What decays, and what we count

Radioactive decay can turn one element into another, something no chemical reaction can do. Every nucleus is labelled by two whole numbers. The atomic number $Z$ counts its protons and fixes the element: $Z=6$ is carbon, $Z=92$ uranium. The mass number $A$ counts all its nucleons, protons plus neutrons, so carbon-14, ${}_6^{14}\mathrm{C}$, has 6 protons and 8 neutrons. Nuclei with the same $Z$ but different $A$ are isotopes of one element: carbon-12 is stable, carbon-14 radioactive.

In a sample we follow $N$, the number of nuclei that have not yet decayed, starting from $N_0$ at $t=0$. A detector registers decays, and the number of decays per second is the activity, measured in becquerels (1 Bq is one decay per second). The letter $A$ serves for both the mass number and the activity, so check the context; $N$ always counts undecayed nuclei.

Halving in equal times

A radioactive nucleus does not wear out: whatever its age, its chance of surviving the next half-life $T_{1/2}$ is exactly one half. In a large sample, half of the nuclei are therefore left after one half-life, half of those after the next, and so on. After $n=t/T_{1/2}$ half-lives, whether $n$ is a whole number or not,
$N=N_0{\left(\frac{1}{2}\right)}^{t/T_{1/2}}$

The chance of decaying per second is the decay constant $\lambda$, and the activity is that chance times the number of nuclei present:

• $\lambda =\frac{\ln 2}{T_{1/2}}$
• $A=\lambda N$

So the activity halves every half-life too. To turn a measured fraction into a time, count the halvings: $n={\log }_2(N_0/N)$, so a sample down to 1/16 of its start is four half-lives old. With $T_{1/2}$ in seconds, $\lambda$ comes out in ${\mathrm{s}}^{-1}$ and $A$ in becquerels.

Three ways to decay

The legend of the animation names three kinds of radiation:

• In $\alpha$ decay the nucleus emits a helium-4 nucleus of 2 protons and 2 neutrons, so $A$ falls by 4 and $Z$ by 2. A sheet of paper stops these heavy, doubly charged particles.
• In ${\beta }^{-}$ decay a neutron becomes a proton and the nucleus emits an electron and an antineutrino, so $Z$ rises by 1 at the same $A$; in ${\beta }^{+}$ decay a proton becomes a neutron and a positron is emitted, so $Z$ falls by 1. A few millimetres of aluminium stop $\beta$ particles.
• In $\gamma$ emission an excited nucleus, often one just formed by $\alpha$ or $\beta$ decay, releases a high-energy photon, and $A$ and $Z$ stay the same. Lead never stops $\gamma$ rays completely: a centimetre of it halves a beam of 1 MeV photons.

For $\alpha$ decay, counting nucleons and charges gives
$X(A,Z)\to Y(A-4,Z-2)+{}^4\mathrm{He}$
so uranium-238 ($Z=92$) becomes thorium-234 ($Z=90$). The animation's sample is an $\alpha$ emitter: each gold parent $X$ that turns grey has become a daughter $Y$, and the red dot that flies off is its $\alpha$ particle.

Worked example: the sample in the animation

The animation starts with $N_0=100$ nuclei and $T_{1/2}=2.5\mathrm{s}$, and the clock hand turns once per half-life. After one, two, three and four turns, at 2.5, 5, 7.5 and 10 s, the law gives 50, 25, 12.5 and 6.25 nuclei. The $N$ readout shows 50, 25, 12 and 6, since a count is a whole number; the $N/N_0$ readout shows the same numbers in per cent.

The formula works between the turns as well. At $t=4.0\mathrm{s}$,
$t/T_{1/2}=4.0/2.5=1.6,N=100\cdot 0.5^{1.6}=33.0$
and the readout shows 33. Run backwards, the law gives the time at which 10 nuclei are left: $n={\log }_{2}10=3.32$ half-lives, so $t=8.3\mathrm{s}$.

Fifty nuclei decay during the first turn, 25 during the second and 13 during the third: the decays thin out because fewer nuclei are left, not because the survivors have grown more stable. The animation times its decays to follow the law. A truly random sample of 100 nuclei would typically miss 50 by about 5 after one half-life, while a gram of material, with some $10^{21}$ nuclei, follows the law almost exactly.

Worked example: iodine-131 in medicine

Iodine-131, used to treat the thyroid gland, has a half-life of 8.0 days. A pharmacy prepares a capsule of 400 MBq, which the patient takes 3.0 days later. How much activity is left then, and when does a stored capsule fall below 1 % of its starting activity?

The activity is proportional to $N$, so it follows the same law:
$A=400\cdot 0.5^{3.0/8.0}=400\cdot 0.771=308\mathrm{MBq}$
The pharmacy must therefore prepare about 30 % more than the dose it means to deliver. For the 1 % mark, $n={\log }_{2}100=6.64$ half-lives, that is 53 days. After 24 days, three half-lives, $(1/2{)}^3=12.5$ % is left, and after 20 days, 2.5 half-lives, $0.5^{2.5}=17.7$ %.

With $\lambda =\ln 2/(8.0\cdot 86400\mathrm{s})=1.00\times 10^{-6}{\mathrm{s}}^{-1}$, the fresh capsule, $4.0\times 10^8\mathrm{Bq}$, holds $N=A/\lambda =4.0\times 10^{14}$ nuclei, less than a tenth of a microgram: far too little to weigh, yet easy to detect, which is why radioactive tracers are so useful.

Reading the decay curve

Plot the readouts, $N$ against $t$, and the curve starts steeply, flattens out and approaches zero without ever reaching it. A straight line from 100 nuclei at $t=0$ down to zero at $t=2T_{1/2}$ is the classic wrong picture; at that time the real curve still holds a quarter.

The steepness of the curve is the decay rate, so the activity can be read from its slope: for the animation, about 28 nuclei per second at the start, 14 at 2.5 s and 7 at 5 s. Real counts scatter around the smooth curve, relatively more when the numbers are small.

Common mistakes

• The half-life is not half the time needed for everything to decay. Each half-life removes half of what is left, so 6400 nuclei become 3200, 1600, 800 and 400 over four half-lives and never suddenly run out.
• Decay is not a fixed loss per hour. A sample that loses 10 % in its first hour loses 10 % of what is left in every hour, so $0.9^{10}=0.35$ of it, 35 %, remains after 10 hours.
• Old nuclei are not overdue. A nucleus that has survived ten half-lives still has exactly a 50 % chance of surviving the next one.
• Heat, pressure and chemistry do not change a half-life. Chemical bonds involve a few electronvolts and nuclear changes millions, so burning or dissolving a radioactive substance leaves its $T_{1/2}$ practically unchanged.

Where you meet it

Radiocarbon dating uses carbon-14, with a half-life of 5,730 years. Cosmic rays keep making it, and living things take it up until they die, so wood with a quarter of the carbon-14 of a living tree is about 11,500 years old. Uranium-238, with a half-life of 4.5 billion years, decays through a chain into lead-206 and dates the oldest rocks. Medicine chooses short half-lives: technetium-99m (6.0 hours) for imaging, fluorine-18 (110 minutes) for PET scans.

Smoke detectors contain americium-241, an $\alpha$ emitter with a half-life of 432 years, and radon-222 (3.8 days) gives most people their largest natural dose. The next lesson, Fission, Fusion and E = mc², explains where nuclear energy comes from, and the expert level here shows why $\alpha$ half-lives range from microseconds to billions of years.

Advanced

A constant chance per unit time

Radioactive decay rests on a single assumption: a nucleus has no memory. Whatever its age, the probability that it decays in the next short interval $dt$ is $\lambda dt$, where the decay constant $\lambda$ is a fixed probability per unit time for that kind of nucleus. A nucleus formed a moment ago and one that has waited a billion years face exactly the same odds. The animated sample starts with $N_0=100$ nuclei and has $\lambda =0.277{\mathrm{s}}^{-1}$, so each nucleus has a 0.28 % chance of decaying in the next hundredth of a second.

From this one rule follows the probability $P(t)$ that a nucleus survives a whole time $t$. Cut $t$ into $m$ equal steps of length $t/m$. In each step the nucleus decays with probability $\lambda t/m$, so it survives that step with probability $1-\lambda t/m$. The steps are independent, so the chances of surviving all of them multiply. Making the steps ever shorter gives
$P(t)={\lim }_{m\to \infty }{\left(1-\frac{\lambda t}{m}\right)}^m=e^{-\lambda t}$

The exponential is the only continuous survival law without memory. To check that it has none: a nucleus that has already lasted a time $s$ survives a further time $t$ with probability $P(s+t)/P(s)=e^{-\lambda (s+t)}/e^{-\lambda s}=e^{-\lambda t}=P(t)$, exactly as if it were new.

Keep two levels of description apart. The decay constant $\lambda$ belongs to a single nucleus and says only how likely it is to decay. The smooth decay curve belongs to a large crowd of nuclei, and it is smooth only because the crowd is large.

Solving the decay law

Now take a sample of $N$ identical nuclei. Each one decays in the next $dt$ with probability $\lambda dt$, so the expected number of decays is $\lambda Ndt$. Each decay removes one nucleus, so $N$ changes by $dN=-\lambda Ndt$, that is
$\frac{dN}{dt}=-\lambda N$

The rate of loss is proportional to what is left. To solve the equation, gather $N$ on one side: $dN/N=-\lambda dt$. Integrate the left side from the starting number $N_0$ up to $N$, and the right side from $0$ to $t$: $\ln (N/N_0)=-\lambda t$. Taking the exponential of both sides,
$N=N_0{e}^{-\lambda t}$

This is $N_0$ times the survival probability $P(t)$ of one nucleus, as it must be: on average a fraction $P(t)$ of the crowd survives. The exponential loses the same fraction in any two intervals of equal length, wherever they start, and that is what turns a radioactive sample into a clock.

Half-life and mean life

The half-life $T_{1/2}$ is the time in which $N$ falls to half its value. Setting $N=N_0/2$ gives $e^{-\lambda T_{1/2}}=1/2$; taking logarithms, $\lambda T_{1/2}=\ln 2$, so
$\lambda =\frac{\ln 2}{T_{1/2}}$

Substituting back, $e^{-\lambda t}=e^{-(\ln 2)t/T_{1/2}}=(1/2{)}^{t/T_{1/2}}$: the halving law and the exponential are one function written in two ways. For the animation $T_{1/2}=0.693/0.277=2.50\mathrm{s}$.

The mean life $\tau$ is the average lifetime of a nucleus. A nucleus decays between $t$ and $t+dt$ if it first survives to $t$, with probability $e^{-\lambda t}$, and then decays within $dt$, with probability $\lambda dt$. Weighting each lifetime $t$ by this probability and integrating by parts gives
$\tau ={\int }_0^{\infty }t\lambda e^{-\lambda t}dt=\frac{1}{\lambda }=\frac{T_{1/2}}{\ln 2}\approx 1.44T_{1/2}$

After one mean life a fraction $1/e$ of the nuclei remains. For the animation $\tau =3.61\mathrm{s}$, when $N_0/e=36.8$ nuclei should remain; the $N$ readout shows 37. Why is $\tau$ longer than $T_{1/2}$? The half-life is the median lifetime: half of the nuclei decay before it. The other half includes a long tail of late survivors, some lasting many half-lives, and they pull the average out to $\tau$.

Activity

A detector does not count the nuclei that remain; it counts decays. What it measures is the decay rate $-dN/dt$, called the activity $A$ and measured in becquerels (1 Bq is one decay per second). From the decay law,
$A=\lambda N$

Because $A$ is proportional to $N$, it falls with the same exponential, $A=A_0{e}^{-\lambda t}$ with $A_0=\lambda N_0$, and so it too is a clock. While a tree lives it keeps exchanging carbon with the air, so its share of carbon-14 stays steady; once it dies, its carbon-14 decays and is not replaced. Comparing the activity per gram of carbon in old wood with that in living wood gives $A/A_0$, and solving $(1/2{)}^{t/T_{1/2}}=A/A_0$ for $t$ gives the age of the wood. In the animation the $A$ readout of the expert layer starts at $A_0=27.7\mathrm{Bq}$ and shows 13.9 Bq at 2.5 s, one half-life later.

Activity depends on two things: how many nuclei there are, and how unstable each one is. A gram of radium-226 is $1/226$ of a mole, so it holds $N=6.02\times 10^{23}/226=2.66\times 10^{21}$ nuclei. With $T_{1/2}=1600$ years, $\lambda =\ln 2/T_{1/2}=1.37\times 10^{-11}{\mathrm{s}}^{-1}$ and $A=\lambda N=3.7\times 10^{10}\mathrm{Bq}$, the activity that once defined the curie. A gram of uranium-238 holds about as many nuclei, but its half-life of 4.5 billion years is about three million times longer, so it is about three million times less active.

Reading the decay curve

A graph of $N$ against $t$ shows the half-life, the mean life and the activity at a glance. Figure 1 draws it with time counted in half-lives, so the same picture serves every nuclide, from microseconds to billions of years.

Legend

$N$	Number of undecayed nuclei at time $t$.
$N_0$	Number of undecayed nuclei at $t=0$.
$N_0/2$	Half the starting number, reached after one half-life.
$N_0/4$	A quarter of the starting number, reached after two half-lives.
$t$	Time since the start of the count.	$\mathrm{s}$
$T_{1/2}$	Half-life: each step of this length halves $N$.	$\mathrm{s}$
$\tau$	Mean life $1/\lambda$, about $1.44T_{1/2}$: where the tangent at $t=0$ meets the time axis.	$\mathrm{s}$
$A_0$	Activity at $t=0$, $A_0=\lambda N_0$; the tangent there has slope $-A_0$.	$\mathrm{Bq}$

First look at the three dots. They sit at $T_{1/2}$, $2T_{1/2}$ and $3T_{1/2}$, equally spaced along the time axis, and their heights $N_0/2$, $N_0/4$ and $N_0/8$ fall by the same factor of two at each step. That is the halving law read off the graph: equal times, equal fractions.

Next look at the straight line marked $-A_0$, the tangent to the curve at $t=0$. Its slope is the starting rate of change, $dN/dt=-\lambda N_0=-A_0$. A line that starts at height $N_0$ and falls at the rate $A_0$ reaches the time axis after $N_0/A_0=1/\lambda =\tau$. So the tangent lands at the mean life $\tau$, a little beyond the first dot, because $\tau \approx 1.44T_{1/2}$. Put differently: if the sample kept decaying at its starting rate, it would be empty after one mean life.

The same construction works anywhere on the curve. At time $t$ the tangent has slope $-\lambda N=-A$ and meets the axis at $t+\tau$, one mean life later. In the animation the tangent at $t=0$ meets the axis at 3.6 s.

The expert graph of the animation shows the same features: the blue count follows the dashed curve $N_0{e}^{-\lambda t}$, with markers at $T_{1/2}$, $2T_{1/2}$, $3T_{1/2}$ and $4T_{1/2}$, equally spaced as in Figure 1. In the laboratory, half-lives are read from a different plot. Taking the logarithm of the decay law gives $\ln N=\ln N_0-\lambda t$, so a graph of $\ln N$ against $t$ (or of $\ln A$, which differs only by the constant $\ln \lambda$) is a straight line of slope $-\lambda$.

Fluctuations: a sharp law from chance

The decay law gives the expected number of survivors. A real sample is a collection of independent random events, so its count scatters around that value. How large is the scatter, and why does the law still look exact for a large sample?

Legend

$N$	Number of nuclei still undecayed in one sample.
$P$	Chance that a sample has exactly this many nuclei left.
$N_{0}p$	Mean of $N$: the starting number $N_0$ times the survival chance $p$ of one nucleus, here $p=1/2$, so 50.
$\sigma$	Standard deviation of $N$, $\sigma =\sqrt{N_{0}p(1-p)}$, here 5: the bracket is $2\sigma$ wide.

Each of the $N_0$ nuclei survives a time $t$ independently, with the same probability $p=e^{-\lambda t}$, like $N_0$ tosses of a biased coin. The number left is therefore binomially distributed, with mean $N_{0}p$ and standard deviation $\sigma =\sqrt{N_{0}p(1-p)}$. Figure 2 plots the exact probabilities for $N_0=100$ nuclei watched for one half-life, so that $p=1/2$. Each bar gives the probability $P$ that a sample has exactly $N$ nuclei left.

The bars are centred on the dashed line at the mean, $N_{0}p=50$. The bracket marked $2\sigma$ runs from 45 to 55, one standard deviation either side of the mean, because $\sigma =\sqrt{100\cdot \frac{1}{2}\cdot \frac{1}{2}}=5$. The green bars inside it add up to 73 %. So a real sample of 100 nuclei keeps $50\pm 5$ after one half-life, and about one sample in four ends up outside the range 45 to 55.

The animation shows the ideal, noise-free case instead: its $k$-th decay comes when the expected number of decays reaches $k-\frac{1}{2}$, so its staircase is the smooth law rounded to whole nuclei.

Counting experiments deal with decays rather than survivors. In a short interval only a small fraction $q$ of the nuclei decays. The number of counts is then binomial with $q\ll 1$, so its variance $N_{0}q(1-q)$ is almost equal to its mean $\bar{n}=N_{0}q$. This limit is the Poisson distribution, whose spread is $\sqrt{\bar{n}}$. The relative uncertainty is $\sqrt{\bar{n}}/\bar{n}=1/\sqrt{\bar{n}}$, so 400 counts are uncertain by 5 %, and 1 % needs 10,000 counts.

The same square root explains why the decay law looks exact. The relative scatter of the survivors falls as $1/\sqrt{N_0}$, and for $10^{21}$ nuclei it is about $3\times 10^{-11}$. A sharp law emerges from pure chance, just as the steady pressure of a gas emerges from random molecular impacts.

Conservation laws decide what can happen

Before asking how fast a nucleus decays, ask whether it can decay at all. Charge, nucleon number and lepton number must each be the same before and after. And the decay must release energy, paid for out of mass: the products together must have less mass than the parent. The energy released, called the Q-value, is
$Q=\left(m_{\mathrm{parent}}-\sum m_{\mathrm{products}}\right)c^2>0$

For the $\alpha$ decay ${}_{92}^{238}\mathrm{U}\to {}_{90}^{234}\mathrm{Th}+{}_2^4\mathrm{He}$, charge balances ($92=90+2$) and so does nucleon number ($238=234+4$). The atomic masses give $Q=4.27\mathrm{MeV}$, about a million times the energy of a chemical bond.

With only two products, momentum conservation fixes how $Q$ is shared. The parent was at rest, so the two products fly apart with momenta of equal size and opposite direction. Kinetic energy is the momentum squared divided by twice the mass, so for equal momenta the lighter particle takes the larger share, in inverse proportion to the masses. The $\alpha$ particle therefore takes $234/238$ of $Q$, or 4.20 MeV. Every $\alpha$ particle from this decay has that same energy, so $\alpha$ spectra are sharp lines.

In ${\beta }^{-}$ decay, $n\to p+e^{-}+{\bar{\nu }}_e$, a neutron becomes a proton. Charge balances ($0=+1-1+0$), and lepton number demands the antineutrino: the electron counts $+1$ and the antineutrino $-1$, so the total stays 0. Now three bodies share $Q$, and they can do so in endlessly many ways. The electrons therefore have a continuous spectrum, from zero up to $Q$ ($0.78\mathrm{MeV}$ for a free neutron). That continuum, which at first seemed to break energy conservation, led Pauli to propose the neutrino in 1930. Finally, $\gamma$ rays come from jumps between energy levels of the nucleus, so, like $\alpha$ particles, they form sharp lines.

Tunnelling decides how fast

Conservation laws only say whether a decay is allowed. How fast it goes depends on something else: whether the $\alpha$ particle can get out of the nucleus at all. Classically it cannot; quantum mechanically it leaks out, very slowly, and the rate of that leak sets the half-life.

Legend

$U$	Potential energy of the $\alpha$ particle at distance $r$ from the centre of the nucleus.	$\mathrm{J}$
$r$	Distance from the centre of the nucleus.	$\mathrm{m}$
$R$	Radius of the nucleus: the nuclear well lies inside it, the Coulomb barrier starts at it.	$\mathrm{m}$
$r_2$	Exit point $r_2=R{U}_{\max }/E$, where the barrier has fallen back to the energy $E$.	$\mathrm{m}$
$U_{\max }$	Height of the barrier at $r=R$.	$\mathrm{J}$
$E$	Energy of the $\alpha$ particle, which it carries away as kinetic energy once it is far outside.	$\mathrm{J}$
$\psi$	Wave function of the $\alpha$ particle along a radius: it oscillates inside, dies away across the barrier and reappears, smaller, beyond $r_2$.	${\mathrm{m}}^{-3/2}$

Figure 3 plots the potential energy $U$ of the $\alpha$ particle against its distance $r$ from the centre of the nucleus. Inside the nuclear radius $R$, the short-range nuclear attraction holds the particle in a deep well. Outside $R$ only the electric repulsion of the daughter nucleus acts, $U=C/r$ with $C$ a positive constant. So the barrier is highest at $r=R$, where it reaches $U_{\max }$, and falls away with distance.

Now look at the horizontal line at height $E$, the energy of the $\alpha$ particle. Energy is conserved, so $E$ is the same everywhere; far away, where $U$ has dropped to zero, all of it is kinetic energy. The line meets the barrier at the exit point $r_2$, where $C/r_2=E$. Since $C=R{U}_{\max }$, this gives $r_2=R{U}_{\max }/E$. Between $R$ and $r_2$ the kinetic energy $E-U$ would be negative, which no classical motion allows: a classical particle would rattle inside the well for ever.

The wave function $\psi$ does not stop at the wall. Follow it from left to right in Figure 3. It oscillates inside the well, dies away exponentially across the forbidden region between $R$ and $r_2$, and beyond $r_2$ oscillates again, with a smaller amplitude.

Across the barrier it falls off locally like $e^{-\kappa r}$, with $\kappa =\sqrt{2m_{\alpha }(U-E)}/\hslash$, where $m_{\alpha }$ is the mass of the $\alpha$ particle and $\hslash$ the reduced Planck constant: the higher and wider the barrier, the more it falls. The chance of escape on each attempt is the square of the ratio of the amplitude outside to the amplitude inside. In the figure the amplitude shrinks only by a modest factor, so that the ripples outside stay visible; in a real nucleus it shrinks enormously.

Here are the real numbers for uranium-238. The $\alpha$ particle has an energy $E$ of only about 4.2 MeV, far below the top of the Coulomb barrier, $U_{\max }\approx 28\mathrm{MeV}$, so classically it could never escape. The exit point lies $r_2/R=28/4.2\approx 6.7$ times further from the centre than the surface (Figure 3, not to scale, shows 4). Bouncing to and fro inside, the particle strikes the barrier about $10^{21}$ times per second. Yet the mean life of uranium-238 is $2.0\times 10^{17}\mathrm{s}$, so its decay constant is $\lambda =1/\tau =5\times 10^{-18}{\mathrm{s}}^{-1}$. Dividing by the number of attempts per second, each attempt succeeds with a probability of about $5\times 10^{-39}$, close to $e^{-88}$: the square of a wave amplitude that leaks through as $e^{-44}$.

Because the escape probability is an exponential, modest changes in the barrier make huge changes in the half-life. Working out the exponent shows that it grows roughly in proportion to $Z/\sqrt{E}$, where $Z$ is the charge number of the daughter nucleus: a larger charge makes the barrier higher, and a larger energy makes the forbidden region shorter. So polonium-212, emitting 8.8 MeV $\alpha$ particles, has a half-life of 0.3 µs, while thorium-232, at 4.0 MeV, has one of $1.4\times 10^{10}$ years: roughly halving the energy stretches the half-life by a factor of about $10^{24}$.

Decay chains

Often the daughter of a decay is itself radioactive, and a decay chain forms. The number of daughters then rises and falls, and a counter registers the decays of both. Figure 4 follows the simplest chain: parents $N_1$, with decay constant ${\lambda }_1$, turn into radioactive daughters $N_2$, with decay constant ${\lambda }_2$, which decay in turn into stable nuclei $N_3$.

Legend

$N$	Number of nuclei of each kind present at time $t$.
$N_1$	Parent nuclei still undecayed.
$N_2$	Radioactive daughter nuclei present.
$N_3$	Stable nuclei formed from the daughters, $N_3=N_0-N_1-N_2$.
$t$	Time since the start, when only parents were present.	$\mathrm{s}$
$t_m$	Time at which the daughters are most numerous.	$\mathrm{s}$

The daughters have a source and a sink. They are created by the parents' decays, at the rate ${\lambda }_1{N}_1$, and lost through their own decays, at the rate ${\lambda }_2{N}_2$:
$\frac{d{N}_2}{dt}={\lambda }_1{N}_1-{\lambda }_2{N}_2$

Read the curves of Figure 4 with this equation in mind. At $t=0$ there are no daughters, so the sink is zero and the daughters rise exactly as fast as the parents fall. As daughters accumulate, their own decay rate ${\lambda }_2{N}_2$ grows, while the supply ${\lambda }_1{N}_1$ shrinks with the parents. Where the two are equal, $d{N}_2/dt=0$: the curve of $N_2$ turns over at the time $t_m$, and then fades.

The stable nuclei $N_3$ start slowly, because at first there are hardly any daughters to decay, and climb until every nucleus has ended up there. No nucleus is lost along the way, so the three numbers always add up to the starting number $N_0$. The figure is drawn for the chain of the worked problem below: 100 parents, and a daughter that lives twice as long as its parent. At $t_m$ the curves of $N_1$ and $N_3$ cross at 25.

Worked problem: a radioactive daughter

Suppose the grey daughters of the animation decay into a stable nucleus with $T_{1/2}=5.0\mathrm{s}$, so that ${\lambda }_2=0.693/5.0=0.139{\mathrm{s}}^{-1}={\lambda }_1/2$. Starting from 100 parents and no daughters: when are the daughters most numerous, how many are there, and what does a counter that registers both decays read then?

Step 1: solve for $N_2$. Insert $N_1=N_0{e}^{-{\lambda }_{1}t}$ and move ${\lambda }_2{N}_2$ to the left: $d{N}_2/dt+{\lambda }_2{N}_2={\lambda }_1{N}_0{e}^{-{\lambda }_{1}t}$. Multiplying both sides by $e^{{\lambda }_{2}t}$ turns the left side into a single derivative, $d(N_2{e}^{{\lambda }_{2}t})/dt$, by the product rule. The right side becomes ${\lambda }_1{N}_0{e}^{({\lambda }_2-{\lambda }_1)t}$, which integrates directly. With $N_2(0)=0$,
$N_2=\frac{{\lambda }_1{N}_0}{{\lambda }_1-{\lambda }_2}\left(e^{-{\lambda }_{2}t}-e^{-{\lambda }_{1}t}\right)$

Step 2: use ${\lambda }_1=2{\lambda }_2$. The prefactor ${\lambda }_1/({\lambda }_1-{\lambda }_2)$ is then 2, and with $x=e^{-{\lambda }_{2}t}$ the second exponential is $e^{-{\lambda }_{1}t}=x^2$. So
$N_2=2N_0\left(x-x^2\right),x=e^{-{\lambda }_{2}t}$

Step 3: find the maximum. As $t$ runs from 0 to infinity, $x$ falls from 1 to 0, and $x-x^2$ is largest where its derivative $1-2x$ vanishes, at $x=\frac{1}{2}$. That happens after one daughter half-life, at $t_m=5.0\mathrm{s}$, when $N_2=2\cdot 100\cdot \frac{1}{4}=50$.

Step 4: the other two populations. The parents number $N_1=N_0{e}^{-{\lambda }_{1}t}=N_0{x}^2$, and the stable nuclei $N_3=N_0-N_1-N_2=N_0(1-x{)}^2$. The three curves of Figure 4 are therefore $x^2$, $2(x-x^2)$ and $(1-x{)}^2$ times $N_0$, and at $x=\frac{1}{2}$ they stand at 25, 50 and 25 nuclei. The $N$ readout of the animation, which counts the parents, shows 25 at that moment, so 25 stable granddaughters have formed.

Step 5: the counter. At the peak $d{N}_2/dt=0$, so the two activities are equal: ${\lambda }_1{N}_1=0.277\cdot 25$ and ${\lambda }_2{N}_2=0.139\cdot 50$ are both 6.93 Bq. A counter that registers both decays reads their sum, 13.9 Bq, twice the $A$ readout, which shows the parent alone.

Later the parent dies out ($x^2$ becomes negligible beside $x$), and $N_2\approx 2N_0{e}^{-{\lambda }_{2}t}$: the mixed sample fades with the daughter's 5.0 s half-life. The lesson for measurement is that the tail of a decay curve belongs to the longest-lived member of the chain.

Long-lived parents: secular equilibrium

When the parent lives far longer than the daughter, ${\lambda }_1\ll {\lambda }_2$, the parent hardly changes while the daughter builds up. After a few daughter half-lives the daughter's decay rate has caught up with its supply and then stays matched to it: $d{N}_2/dt\approx 0$, so ${\lambda }_2{N}_2={\lambda }_1{N}_1$. This is secular equilibrium: the daughter is exactly as active as the parent, however few of its nuclei there are. Radon-222 (half-life 3.8 days) growing from radium-226 (1600 years) numbers only $N_2/N_1={\lambda }_1/{\lambda }_2=6.5\times 10^{-6}$ of the radium atoms, yet is just as active.

Limits of validity

The constancy of $\lambda$ is extremely well tested, but it is not absolute. The exceptions all involve the atom's electrons, the only part of the atom that chemistry can touch. In electron capture the nucleus absorbs one of the atom's own inner electrons, so the decay rate depends slightly on how those electrons are arranged: the half-life of beryllium-7 differs by up to about 1 % between chemical environments. Stripping off every electron can transform a decay. Neutral rhenium-187 has a half-life of about $4\times 10^{10}$ years, the bare nucleus one of 33 years, because its $\beta$ electron can then be created directly in an empty inner orbit. Otherwise temperature, pressure and chemistry change nothing measurable.

Quantum mechanics predicts slight departures from the exponential law at extremely short and extremely long times, but none has been seen in a nuclear decay. Finally, $\lambda$ holds in the rest frame of the decaying particle. Seen from the laboratory, a fast particle's clock runs slow by the factor $\gamma$ of special relativity: muons, with a mean life of 2.2 µs at rest, live 22 µs in the laboratory at $\gamma =10$. That is how cosmic-ray muons born high in the atmosphere reach the ground.

Where this leads

The tunnelling that sets $\alpha$ half-lives is worked out in Schrödinger's Equation and Tunnelling, and the accounting $Q=\Delta m{c}^2$ drives Fission, Fusion and E = mc². The weak interaction behind $\beta$ decay belongs to Particles and Forces: the Standard Model, and the stretched lifetimes of fast particles to Special Relativity. Sharp $\gamma$ lines are the nuclear counterpart of the spectral lines in Atomic Models: the Bohr Atom. Constant fractional rates return in a capacitor discharging through a resistor, in Capacitors, and a sharp law built from random events is the core of Kinetic Theory of Gases.