Radioactivity & Nuclear Decay

A lump of uranium sitting in a drawer has been there, as far as it is concerned, for four and a half billion years. Nothing in the room tells it what time it is. It has no clock, no memory, no stored-up fatigue. And yet if you had a tonne of it and waited, you could predict to several digits how many atoms would break apart this afternoon.

That is the strange thing worth getting straight before any formula. A single nucleus is completely unpredictable: you cannot say when this one will go. A mountain of them is completely predictable. The whole of radioactivity comes from taking that one fact seriously.

The last lesson gave every particle a wavelength and left the nucleus as a small hard thing at the centre of the atom. Now look inside it.

One rule, and the law that follows from it

Here is the entire physical input. Each nucleus has a fixed probability per second of decaying, and that probability never changes. It does not depend on how long the nucleus has already survived, on temperature, on pressure, or on what chemical the atom is bound into. Call that probability per second $\lambda$, the decay constant, with units of one over time.

Now count. If you have $N$ nuclei and each has chance $\lambda$ per second of going, the expected number of decays in a short time $dt$ is $\lambda N\,dt$. Every one of those is a nucleus removed from the pile, so

Read that in words: the rate at which the pile shrinks is proportional to how much pile is left. Twice the sample, twice the decays per second. The minus sign is just bookkeeping: $N$ is falling.

That equation says the rate of change of something is proportional to the thing itself, and the only function that does that is the exponential. Separate and integrate:

so

$N_0$ is the number you started with at $t = 0$. Nothing here is special to nuclei; the same equation governs a draining capacitor. What is special is that $\lambda$ is a fixed property of the isotope, the same in a star and in a freezer.

Half-life and mean life

$\lambda$ is the honest quantity but nobody quotes it. What gets quoted is the half-life $T_{1/2}$: the time for half the sample to go. Put $N = N_0/2$ into the law:

Because the rule has no memory, the half-life applies from any starting moment, not just from $t=0$. A sample that has already lost 90% of its nuclei still loses half of what remains in the next $T_{1/2}$. That is why powers of one half are so useful:

Three half-lives leave $(1/2)^3 = 1/8$. Ten leave about a thousandth. If a problem gives you a tidy multiple of the half-life, use this form and skip the logarithms entirely.

There is a third timescale in circulation. The mean lifetime $\tau = 1/\lambda$ is the average survival time of a nucleus, and

A muon with a mean lifetime of 2.20 μs has a half-life of $0.693 \times 2.20 = 1.52$ μs, which is shorter. Check which one a problem has handed you; the two differ by 44% and that is enough to wreck an answer.

Activity: what a detector actually measures

You never count nuclei in a sample. You count the decays coming out of it. That rate is the activity,

measured in becquerels (Bq), one decay per second. The older unit is the curie: $1\ \text{Ci} = 3.70\times10^{10}$ Bq, originally the activity of a gram of radium.

Since $A$ is just $\lambda$ times $N$, and $\lambda$ is constant, activity obeys exactly the same exponential:

That is the most useful sentence on this page. You can do every dating and counting problem with measured activities and never find out how many nuclei were there.

Notice what $A = \lambda N$ implies about long-lived isotopes. Uranium-238 has an enormous $N$ in any ore sample but a tiny $\lambda$, so a lump of it is barely radioactive. A medical tracer with a six-hour half-life has a $\lambda$ about $10^{13}$ times larger, so a speck of it lights up a detector. Long-lived means feeble; short-lived means fierce but brief.

Worked example 1: half-life from two readings

A Geiger counter reads 800 Bq from a fresh sample and 100 Bq exactly 30.0 minutes later. Find the half-life.

Write down the ratio first, because the ratio is what the exponential controls:

One eighth is $(1/2)^3$, so exactly three half-lives have passed. Then

No logarithms needed. When the ratio is not a power of two, take logs of $A/A_0 = e^{-\lambda t}$:

Check it against the numbers above: $\ln 8 = 2.079$, so $T_{1/2} = 30.0 \times 0.693/2.079 = 10.0$ min. Same answer, and this version works for any ratio.

Three ways a nucleus sheds instability

A nucleus is protons and neutrons held together by the strong force against the electrical repulsion of the protons. Write it as $^A_Z X$: $A$ nucleons in total, $Z$ of them protons, so $A - Z$ neutrons. When the mix is wrong, the nucleus rearranges along one of three routes.

Alpha decay throws out a helium nucleus, two protons and two neutrons bound together:

Heavy nuclei do this because the strong force only reaches its nearest neighbours while the Coulomb repulsion reaches across the whole nucleus. Past about $Z = 83$ the repulsion wins.

Beta-minus decay turns a neutron into a proton inside the nucleus, spitting out an electron and an antineutrino:

$A$ is unchanged, $Z$ goes up by one. Neutron-rich nuclei do this. Beta-plus decay runs the other way (a proton becomes a neutron plus a positron and a neutrino, so $Z$ drops by one) and electron capture achieves the same thing by swallowing an inner atomic electron instead of emitting a positron.

Gamma decay emits a photon from a nucleus left in an excited state. Neither $A$ nor $Z$ changes; only the energy does. It nearly always follows an alpha or beta decay that dropped the daughter into an excited level.

The bookkeeping rule covers most identification questions: $A$ and $Z$ must each balance across the arrow. Uranium-238 emitting an alpha leaves $A = 238 - 4 = 234$. Carbon-14 with $Z = 6$ beta-minus decaying leaves $Z = 7$, nitrogen. And a nucleus that emits one alpha and then two beta-minus particles has $Z$ changed by $-2 + 1 + 1 = 0$, back where it started, with $A$ down 4.

Reading the clock backwards: dating

If you know $\lambda$ and you can measure how far along the decay a sample is, you can solve for $t$. That is all dating is. The work is in figuring out what plays the role of $A_0$.

Worked example 2: radiocarbon

Charcoal from an old campfire gives 0.150 Bq per gram of carbon. Living wood gives 0.250 Bq per gram. Carbon-14 has $T_{1/2} = 5730$ yr. How old is the charcoal?

Living wood keeps exchanging carbon with the atmosphere, so its carbon-14 stays topped up at the atmospheric level. The moment the tree died the topping-up stopped. So living wood today is the $A_0$ for a sample that died at any time in the past.

First the decay constant:

Then

The ratio is $1.667$ and $\ln 1.667 = 0.5108$, so

About 4220 years old.

Worked example 3: uranium-lead

A zircon crystal contains 0.300 atoms of lead-206 for every atom of uranium-238, and all of that lead came from uranium decay. Uranium-238 has $T_{1/2} = 4.468\times10^9$ yr. How old is the crystal?

Here nobody tells you $N_0$. You have to build it. Every lead atom present is a uranium atom that decayed, so

That single line is the whole idea; the rest is algebra. Divide through by $N_{\text{U}}$:

And $N_0/N = e^{\lambda t}$, so $\lambda t = \ln 1.300 = 0.2624$. With $\lambda = 0.6931/(4.468\times10^9) = 1.5513\times10^{-10}\ \text{yr}^{-1}$,

Roughly 1.7 billion years. The same trick (reconstruct $N_0$ from what is left plus what it turned into) handles any parent-daughter dating problem, and it handles the two-isotope version too. If two isotopes were made in equal numbers and you know their ratio now, write $N_1/N_2 = e^{(\lambda_2 - \lambda_1)t}$ and solve for $t$.

Parents, daughters and equilibrium

Real decays rarely stop at one step. A parent decays to a daughter which is itself unstable. Now two things happen to the daughter at once: it is being created at the rate the parent decays, and destroyed at its own rate. So

Solving that gives the Bateman result, which problems will usually hand you:

for a daughter that starts at zero. Two limits matter more than the formula.

Long after a short-lived daughter settles, the $e^{-\lambda_d t}$ term has died away and the daughter simply tracks the parent, decaying with the parent's half-life at a fixed activity ratio:

This is transient equilibrium. For molybdenum-99 ($T_{1/2} = 66.0$ h) feeding technetium-99m ($T_{1/2} = 6.01$ h), $\lambda_d/\lambda_p = 66.0/6.01 = 10.98$, so the ratio is $10.98/(10.98 - 1) = 1.10$.

If the parent is enormously longer-lived ($\lambda_p \ll \lambda_d$) the ratio goes to 1: the daughter's activity equals the parent's. This is secular equilibrium, and it is why radium-226 sits in old uranium ore at a tiny but fixed abundance. Setting $\lambda_{\text{Ra}}N_{\text{Ra}} = \lambda_{\text{U}}N_{\text{U}}$ gives $N_{\text{Ra}}/N_{\text{U}} = T_{\text{Ra}}/T_{\text{U}} = 1600/(4.468\times10^9) = 3.58\times10^{-7}$.

Worked example 4: when is the daughter strongest?

A freshly purified sample of molybdenum-99 ($T_{1/2} = 66.0$ h) decays to technetium-99m ($T_{1/2} = 6.01$ h), initially absent. The daughter activity peaks at $t = \ln(\lambda_d/\lambda_p)/(\lambda_d - \lambda_p)$. When?

Get both decay constants in the same units, hours:

Their ratio is $0.11533/1.0502\times10^{-2} = 10.981$, and $\ln 10.981 = 2.3962$. Their difference is $0.11533 - 0.01050 = 0.10483\ \text{h}^{-1}$. So

Just under a day, which is why hospitals milk these generators daily. The result makes sense as a rough check: it is a few daughter half-lives, long enough for the daughter to build up, short enough that the parent has hardly faded.

Rates that add, and rates that are fed in

Three more situations show up constantly, and each is one line.

Two competing ways to lose the same nuclei. Probabilities per second add, so decay constants add. If a tracer decays physically with $\lambda_p$ and is also flushed out of the body biologically with $\lambda_b$, the effective constant is $\lambda_{\text{eff}} = \lambda_p + \lambda_b$, which in half-lives is

Iodine-131 with $T_p = 8.02$ d and $T_b = 10.0$ d has $1/T_{\text{eff}} = 0.1247 + 0.1000 = 0.2247\ \text{d}^{-1}$, so $T_{\text{eff}} = 4.45$ days. Branching ratios are the same idea split the other way: potassium-40 at 1000 Bq with 10.7% electron capture gives 107 captures per second.

Two isotopes mixed in one vial. Activities add, but each falls with its own $\lambda$. Decay them separately and sum at the end: never average the half-lives.

Steady production. A cyclotron making nuclei at a constant rate $R$ while they decay gives $dN/dt = R - \lambda N$, and starting from nothing,

The activity climbs towards $R$ and saturates there, because at saturation you are destroying them exactly as fast as you make them. Fluorine-18 ($T_{1/2} = 110$ min) made at $5.00\times10^{10}$ per second for 120 min has $\lambda t = 0.693 \times 120/110 = 0.756$, $e^{-0.756} = 0.4695$, so $A = 5.00\times10^{10}\times 0.5305 = 2.65\times10^{10}$ Bq. Waiting forever would only get you to $5.00\times10^{10}$ Bq, so there is no point irradiating for more than about three half-lives.

Finally, two conversions that turn counts into physics. A detector that records a fraction $\varepsilon$ of decays reads $A_{\text{measured}} = \varepsilon A$, so divide by $\varepsilon$ to get the source activity. And absorbed dose is energy deposited per kilogram, in grays: $1\ \text{Gy} = 1\ \text{J/kg}$. Multiply the number of particles by the energy each, convert eV to joules with $1\ \text{eV} = 1.602\times10^{-19}$ J, and divide by the mass.

How to recognise these problems

Almost everything in this topic announces itself.

A half-life, a mean life, or a decay constant appears. Convert whichever you are given to $\lambda$ immediately. If the elapsed time is a whole number of half-lives, use $(1/2)^n$; otherwise use $e^{-\lambda t}$ and take logs.

Two measurements of the same sample at different times. You are being asked for $T_{1/2}$ or for the elapsed time. Form the ratio $A/A_0$ first: the absolute numbers almost never matter.

A daughter product is mentioned and you are asked for an age. This is a dating problem. Write $N_0 = N_{\text{now}} + N_{\text{daughter}}$ and carry on.

The words equilibrium, generator, or ore. Parent-daughter. If the parent is vastly longer-lived, use $\lambda_p N_p = \lambda_d N_d$, meaning equal activities. Otherwise use the transient ratio $\lambda_d/(\lambda_d - \lambda_p)$.

Something is being produced while it decays. Use $A = R(1 - e^{-\lambda t})$ and check whether the question is really asking about saturation.

Two loss mechanisms, or two isotopes. Add the $\lambda$s if both act on the same nuclei; add the activities if they are separate populations.

Only $A$ and $Z$ appear, with no times. Pure bookkeeping. Balance the mass number and the charge across the arrow and you are done.

What none of this touches is why a particular decay releases the energy it does, or how much. That comes from the masses: the daughter plus the emitted particle weigh slightly less than the parent, and the difference is the energy budget $Q$. Next up: nuclear binding energy and reactions.