You already know the shape of radioactive decay: wait one half-life and half of what is left decays; wait another and half of the remainder goes. That halving is the fingerprint of an exponential. In this lesson we stop counting in whole half-lives and do the real, continuous mathematics — the calculus that turns "it halves every so often" into a single, exact equation you can evaluate at any instant, for any isotope.

The whole thing rests on one physical idea. A single nucleus decays completely at random — no schedule, no warning. But in a real sample there are so many nuclei (a microgram holds of order 10^{16} of them) that the average behaviour is rock-solid: in each second, a fixed fraction of the survivors decays. That single sentence is enough to derive everything that follows.

Activity: the rate is proportional to how many are left

Let N be the number of undecayed nuclei in a sample at time t. The activity A is the rate at which nuclei are decaying — the number of decays per second — which is exactly how fast N is falling:

The minus sign is there because N is decreasing, so \mathrm{d}N/\mathrm{d}t is negative — while activity, a count of decays, is a positive number. Activity is measured in becquerels (Bq), where 1\ \text{Bq} is one decay per second.

Now the physics. Because a fixed fraction of the survivors decays each second, twice as many undecayed nuclei means twice as many decays per second. In other words the activity is proportional to N. Writing the constant of proportionality as \lambda:

That constant \lambda is the decay constant: the probability that any given nucleus decays per unit time. It is a fixed property of the isotope, measured in per second (\text{s}^{-1}), or per year, per hour — whatever unit of time you use. A big \lambda means each nucleus is very likely to go soon (a frantic, short-lived isotope); a tiny \lambda means each nucleus is very reluctant (a long-lived one).

Solving it: the exponential decay law

The relation \dfrac{\mathrm{d}N}{\mathrm{d}t} = -\lambda N is a differential equation: the rate of change of N is set by N itself. We can solve it exactly by separating the variables.

Step 1 — separate. Gather all the N's on one side and the t's on the other:

Step 2 — integrate both sides.

Step 3 — fix the constant. At the start, t = 0, the sample holds N_0 nuclei, so \ln N_0 = c. Substituting back and using \ln N - \ln N_0 = \ln\!\left(\tfrac{N}{N_0}\right):

Step 4 — undo the logarithm by raising e to each side. This is the radioactive decay law:

And because activity is just \lambda times the number of nuclei (A = \lambda N), multiplying through by \lambda shows the activity obeys the very same law — as does the count rate C a detector reads, and the mass m of the isotope:

All four quantities fall along identical exponential curves, because each is simply proportional to N. That is enormously convenient: you can work with whichever one you can actually measure — usually the count rate on a Geiger–Müller tube — and the maths is unchanged.

The link to half-life

The half-life T_{1/2} is the time for the number of nuclei (or the activity) to fall to half. We can pin down exactly how it relates to \lambda by putting N = \tfrac{1}{2}N_0 into the decay law at t = T_{1/2}:

Take natural logs of both sides. Since \ln\tfrac{1}{2} = -\ln 2:

Notice there is no N_0 left in the result: the half-life does not depend on how much material you started with, only on the isotope's \lambda. The two are inversely related through the fixed number \ln 2 \approx 0.693 — double the decay constant and you halve the half-life. A handy sanity check: after one half-life \lambda t = \lambda \cdot \tfrac{\ln 2}{\lambda} = \ln 2, so e^{-\ln 2} = \tfrac{1}{2}, exactly as it must.

See it decay

The curve below is N = N_0\,e^{-\lambda t} with N_0 = 100. Drag the decay constant \lambda slider and watch two markers move with it. The first sits at the half-life T_{1/2} = \ln 2/\lambda, where the curve has dropped to 50 — turn \lambda up and it slides left (shorter half-life), exactly as T_{1/2} = \ln 2/\lambda demands.

The second marker sits at t = 1/\lambda, the mean lifetime \tau, where the curve has fallen to N_0/e \approx 37\% of the start. Notice it always lies a little to the right of the half-life: the average nucleus lives slightly longer than one half-life, by that factor 1/\ln 2 \approx 1.44. Whatever \lambda you choose, the curve keeps the identical shape — only the timescale stretches or squeezes.

Worked examples

Example 1 — how many are left? A source has decay constant \lambda = 0.20\ \text{s}^{-1} and starts with N_0 = 5.0\times10^{8} undecayed nuclei. How many remain after t = 10\ \text{s}?

First the exponent: \lambda t = 0.20 \times 10 = 2.0. Then

About 68 million survive — a little under \tfrac{1}{7} of the original, as you would expect after roughly three half-lives (T_{1/2} = \ln 2 / 0.20 \approx 3.5\ \text{s}).

Example 2 — decay constant from half-life (and back). Cobalt-60, a gamma source used in radiotherapy, has a half-life of 5.27 years. Find its decay constant.

To get it in \text{s}^{-1}, divide by the number of seconds in a year (1\ \text{yr} \approx 3.16\times10^{7}\ \text{s}): \lambda = 0.132 / (3.16\times10^{7}) = 4.16\times10^{-9}\ \text{s}^{-1}. Going the other way is just as easy: an isotope with \lambda = 0.0289\ \text{s}^{-1} has T_{1/2} = 0.693/0.0289 = 24\ \text{s}.

Example 3 — how long to reach a given activity? A source of activity A_0 = 6.0\times10^{5}\ \text{Bq} has \lambda = 0.030\ \text{s}^{-1}. How long until its activity falls to A = 1.0\times10^{5}\ \text{Bq}?

This is where the logarithm earns its keep. Start from A = A_0 e^{-\lambda t}, divide, and take natural logs:

So it takes about a minute. Rearranging the decay law with a log is the single most useful exam skill on this topic — reach for it whenever the unknown is the time.

Dating the past: radiocarbon

The decay law is a clock. Living things absorb carbon from the air, a fixed fraction of which is radioactive carbon-14. When the organism dies the topping-up stops and the carbon-14 decays with a half-life of about 5730 years, i.e. \lambda = \ln 2 / 5730 = 1.21\times10^{-4}\ \text{yr}^{-1}. Measure how much is left and you can read off how long ago it died.

Worked example. A wooden beam from an ancient tomb contains only 25\% of the carbon-14 of fresh wood. How old is it?

Here N/N_0 = 0.25, so from N = N_0 e^{-\lambda t}:

About 11{,}500 years — which makes sense, because 25\% = \left(\tfrac{1}{2}\right)^{2} is exactly two half-lives, and 2 \times 5730 = 11{,}460 years. The calculus and the "count the half-lives" picture agree perfectly, as they must; the decay law simply lets you handle the in-between cases (37%, 12%, anything at all) that don't land on a whole number of half-lives.