In short

A single unstable nucleus has no schedule. You cannot say whether the uranium-238 atom in your fingernail will decay in the next second, the next year, or the next billion years. But you can say, with high precision, what fraction of a large collection of such nuclei will have decayed in a given interval — because each nucleus decays independently with a fixed probability per unit time, the decay constant \lambda.

That single assumption forces the population to obey the radioactive decay law:

N(t) = N_0\, e^{-\lambda t},

where N_0 is the number of undecayed nuclei at t = 0 and N(t) the number still surviving at time t.

From this one equation fall three characteristic times:

  • Half-life t_{1/2} = (\ln 2)/\lambda \approx 0.693/\lambda — the time for half the sample to decay.
  • Mean life \tau = 1/\lambda = t_{1/2}/\ln 2 — the average lifetime of a single nucleus.
  • e-folding time \tau — the time for the population to drop to 1/e \approx 36.8\% of its starting value (same number as the mean life, non-coincidentally).

The activity A(t) = \lambda N(t) — the number of decays per second — obeys the same exponential law: A(t) = A_0\, e^{-\lambda t}. The SI unit is the becquerel (1 Bq = 1 decay per second). The older unit, the curie (1 Ci = 3.7 \times 10^{10} Bq), is still common on medical-isotope labels.

Short half-lives mean "fast clocks" for dating young material; long half-lives mean "slow clocks" for dating ancient rocks. Carbon-14 (t_{1/2} = 5730 y) dates charcoal from Harappan sites at Dholavira. Uranium-238 (t_{1/2} = 4.47 billion years) dates the solar system itself.

Put a Geiger counter next to a sealed source of caesium-137 on a laboratory bench. Leave it there for an hour. Listen. Every time a caesium nucleus somewhere inside the source decays, the counter clicks — click, click, click-click, click, click-click-click, click. The clicks do not come on a metronome; the gaps between them are random, some longer, some shorter. But count the clicks for ten minutes and you will get a number very close to the number you got in the previous ten minutes, and the one before that. The rate is steady. The individual events are not. The most ordered thing in physics and the most disordered thing in physics, happening in the same box.

Now come back in thirty years and listen again. You will still hear clicks, but the rate will be noticeably lower — closer to half. Come back after another thirty years, the rate halves again. Caesium-137 has a half-life of 30.2 years, and the sample obeys this number with the patience of a geological clock. The decay is pure chance for any single nucleus, inevitable statistical law for the many.

The goal of this article is the one-line assumption that turns the chance into the law, and the law's three characteristic times.

The one assumption that forces the decay law

Imagine N_0 = 10^{20} undecayed atoms of a single radioisotope in a sample at time t = 0. At any moment in time some of them have decayed and some have not. Let N(t) be the number of undecayed nuclei at time t. You want to know how N(t) changes as t runs forward.

Here is the core physical assumption, short enough to fit on one line:

Every undecayed nucleus has the same constant probability \lambda\,dt of decaying in the next infinitesimal interval dt, independently of every other nucleus and of how long it has already been sitting around.

The constant \lambda is the decay constant. It has units of 1/\text{time}. A big \lambda means "fast decay"; a tiny \lambda means "slow decay."

That a nucleus has no memory of its age is a strong claim — an atom of uranium-238 that has been waiting in a South Indian granite since the formation of Earth has the same chance of decaying in the next second as a uranium-238 atom that has just been made in a reactor this morning. This property — memorylessness — is a consequence of quantum tunnelling, the mechanism by which an alpha particle escapes the nuclear potential well. A tunnelling process does not accumulate "readiness"; at each instant it either happens or it does not, with a fixed probability set by the wavefunction geometry. Read Radioactivity — Types of Decay for the mechanism; here, accept the assumption and see what it forces.

From the assumption to the differential equation

Consider the interval [t, t + dt]. Each of the N(t) undecayed nuclei has probability \lambda\,dt of decaying in this window. The expected number of decays in the window is therefore

(\text{decays in } dt) \;=\; N(t)\cdot \lambda\,dt. \tag{1}

Why: "probability per nucleus × number of nuclei" is just the standard way to turn a per-unit rate into a total rate. It is the same logic as "number of cars per minute on a highway equals the number of cars passing a point times the rate at which each passes."

Each decay removes one nucleus from the undecayed population. So the change in N over the interval is the negative of the number of decays:

dN \;=\; -\lambda\, N(t)\, dt. \tag{2}

Why: the sign is bookkeeping. Decays reduce the undecayed population, so dN is negative when N > 0. The minus sign is what encodes that.

Divide both sides by dt:

\frac{dN}{dt} \;=\; -\lambda\, N. \tag{3}

Why: this is now a clean first-order linear ODE for N(t). The rate of change of the population is proportional to the population itself — the hallmark of exponential processes (bank interest, bacterial growth, RC circuits, and this).

Solving the ODE

Separate variables in equation (3):

\frac{dN}{N} \;=\; -\lambda\, dt.

Integrate both sides. The left integrates to \ln N; the right to -\lambda t, plus a constant:

\ln N \;=\; -\lambda t + C. \tag{4}

Why: \int dN/N = \ln N is the standard antiderivative that makes exponential decay fall out. Keep the constant of integration C until the end — it will be fixed by the initial condition.

Apply the initial condition: at t = 0, N = N_0. Equation (4) gives \ln N_0 = C. Substitute back:

\ln N - \ln N_0 \;=\; -\lambda t \quad\Longrightarrow\quad \ln\!\left(\frac{N}{N_0}\right) = -\lambda t.

Exponentiate both sides:

\boxed{\; N(t) \;=\; N_0\, e^{-\lambda t}. \;} \tag{5}

Why: this is the radioactive decay law. It is the same functional form as every other exponential decay in physics — the charge on a discharging capacitor, the amplitude of a lightly-damped oscillator, the temperature of a cooling cup of kulhad chai. The proportionality dN/dt \propto -N always produces e^{-\lambda t}.

The decay law is the entire physics of the situation. Everything that follows — half-life, mean life, activity — is a re-expression of the same one formula in more convenient units.

Half-life — the clock you can read

The decay law is beautiful but not how people talk. A chemist does not tell you "caesium-137 has a decay constant of 7.28 \times 10^{-10} per second." They say "caesium-137 has a half-life of 30.2 years." Why? Because the half-life is the time you can actually watch elapse — the interval over which a sample drops to half its starting amount.

Define the half-life t_{1/2} as the time at which N(t)/N_0 = 1/2. Substitute into equation (5):

\frac{1}{2} \;=\; e^{-\lambda t_{1/2}}.

Take the natural log of both sides:

\ln\!\left(\tfrac{1}{2}\right) \;=\; -\lambda t_{1/2} \quad\Longrightarrow\quad -\ln 2 \;=\; -\lambda t_{1/2}.

Why: \ln(1/2) = -\ln 2 by the log-of-a-reciprocal identity. Rearranging gives a clean positive formula for the half-life.

\boxed{\; t_{1/2} \;=\; \frac{\ln 2}{\lambda} \;\approx\; \frac{0.693}{\lambda}. \;} \tag{6}

One half-life is the same for every part of the decay curve. If you wait one half-life, half of what you had is gone. If you wait another half-life, half of what was left is gone — so a quarter of the original remains. Three half-lives: one-eighth. Ten half-lives: a thousandth. Twenty half-lives: a millionth. The useful pattern:

N(n\, t_{1/2}) \;=\; \frac{N_0}{2^n}.

You can show this directly. Put t = n\, t_{1/2} into (5):

N \;=\; N_0\, e^{-\lambda n t_{1/2}} \;=\; N_0\, e^{-n \ln 2} \;=\; N_0\, (e^{\ln 2})^{-n} \;=\; N_0 \cdot 2^{-n}.

Why: \lambda t_{1/2} = \ln 2 by (6), and e^{\ln 2} = 2 by definition of the natural logarithm. The exponential law in units of half-lives becomes a pure halving rule.

Watch a sample halve

The simulation below plays out equation (5) in a concrete case: N_0 = 1000 nuclei at t = 0, half-life 2 seconds (chosen so you can watch it happen). The red curve is N(t) = 1000\, e^{-(\ln 2/2)\, t}; the dashed horizontals mark N/N_0 = 1/2, 1/4, 1/8. Watch the moving dot trace the curve and the horizontal-line crossings.

Animated: the population of undecayed nuclei falls exponentially A curve plotting N of t equals one thousand times e to the minus ln two over two t, plotted from t equals zero to ten seconds. Dashed horizontal lines at N equals five hundred, two hundred fifty, one hundred twenty-five mark one, two, and three half-lives. A moving red dot traces the curve over the duration of the animation. t (s) N(t) 500 250 125 1000 0 2 4 6 8 10 ½ N₀ ¼ N₀ ⅛ N₀ t₁⁄₂
$N(t) = 1000\, e^{-\lambda t}$ with $\lambda = (\ln 2)/2 \approx 0.347~\text{s}^{-1}$. The red dot traces the decay. Dashed horizontal lines mark half, quarter, and eighth of $N_0$, and the ghost markers at $t = 2,\,4,\,6$ s sit exactly on those lines — each additional half-life halves the remaining population. Click replay to watch again.

Mean life — the average age of a nucleus at decay

A closely related quantity is the mean life (or "average lifetime") \tau, the expected lifetime of a single nucleus drawn from the population.

To compute it, note that the fraction of nuclei that decay between t and t + dt is |dN|/N_0 = \lambda\, e^{-\lambda t}\, dt (differentiate (5), flip the sign). Each such nucleus lived for a time t before decaying. The mean lifetime is the probability-weighted integral:

\tau \;=\; \int_0^\infty t \cdot \lambda\, e^{-\lambda t}\, dt.

Integrate by parts with u = t, dv = \lambda e^{-\lambda t}\, dt (so du = dt, v = -e^{-\lambda t}):

\tau \;=\; \Big[-t\, e^{-\lambda t}\Big]_0^\infty + \int_0^\infty e^{-\lambda t}\, dt.

The boundary term is zero at both ends (t = 0 kills it directly; t \to \infty gives t\cdot e^{-\lambda t} \to 0 because the exponential beats any polynomial). The remaining integral is 1/\lambda:

\boxed{\; \tau \;=\; \frac{1}{\lambda} \;=\; \frac{t_{1/2}}{\ln 2} \;\approx\; 1.44\, t_{1/2}. \;} \tag{7}

Why: the mean life is always longer than the half-life by a factor of 1/\ln 2 \approx 1.44. The reason is that the distribution is skewed — most nuclei decay early, but the long tail of late-deciding nuclei pulls the average up past the median. The half-life is the median lifetime; the mean life is the mean.

Equivalently, \tau is the time it takes the sample to drop to 1/e \approx 0.368 of its starting value — substitute t = \tau = 1/\lambda into (5) and you get N = N_0\, e^{-1}. The mean life and the e-folding time are the same number.

Activity — decays per second

The quantity that your Geiger counter measures is not N(t) (the number of remaining nuclei — you cannot count them directly). It is the rate of decays: how many clicks per second. This rate is called the activity A(t), and it is what appears on every medical-isotope label.

Start from the definition: A(t) is the number of decays per unit time, which is the magnitude of dN/dt:

A(t) \;=\; \left|\frac{dN}{dt}\right| \;=\; \lambda\, N(t). \tag{8}

Why: equation (3) says dN/dt = -\lambda N, so the rate at which nuclei disappear is \lambda N per second. That is exactly the rate at which decays occur — one-for-one.

Substitute the decay law (5):

\boxed{\; A(t) \;=\; \lambda\, N_0\, e^{-\lambda t} \;=\; A_0\, e^{-\lambda t}, \qquad A_0 \;=\; \lambda N_0. \;} \tag{9}

The activity decays with the same \lambda as the population. If the population halves in one half-life, the activity halves too. This is why you can measure the half-life of a short-lived isotope with a Geiger counter without ever knowing how many atoms are actually in the sample — the count rate gives you \lambda directly.

Units

Specific activity

A gram of caesium-137 is more active than a gram of uranium-238 by a factor of (4.47 \times 10^9 \text{ y})/(30.2\ \text{y}) \times (238/137) — a few hundred million to one. The reason: short-lived isotopes decay fast, so their \lambda is big, so their activity per gram is big. The specific activity A/m (Bq per gram) is inversely proportional to the half-life. A nuclear medicine isotope like technetium-99m has a half-life of 6 hours precisely so that its specific activity is large enough to image a patient but short enough that no significant dose remains 24 hours after the scan.

Explore — move the half-life, watch the curve reshape

Interactive: N/N₀ versus time for varying half-life A curve showing N over N zero equals two to the negative t over t half, for values of t half between one and thirty seconds. The curve reshapes as t half changes. Readouts display the current half-life, decay constant lambda, mean life tau, and fraction remaining at a fixed probe time t equals ten seconds. t (s) N/N₀ 0 0.5 1 10 20 30 40 drag red dot along axis to change t₁⁄₂ (probe at t = 10 s)
Drag the red point horizontally along the time axis to change the half-life $t_{1/2}$ between 1 and 30 seconds. The red curve shows $N/N_0 = 2^{-t/t_{1/2}}$, equivalent to $e^{-\lambda t}$ with $\lambda = (\ln 2)/t_{1/2}$. The red dot sits on the curve at $N/N_0 = 0.5$ — always at $t = t_{1/2}$ by construction. The dashed vertical probe at $t = 10$ s lets you read off the fraction remaining at a fixed time as the half-life changes.

Two things to notice. First, long half-lives give nearly-flat curves on this time scale — a thirty-second half-life has barely touched half when the axis ends at forty seconds. Second, the red dot that marks (t_{1/2}, 0.5) stays on the curve at exactly height 0.5 whatever you do with the slider. This is the algebraic identity 2^{-t_{1/2}/t_{1/2}} = 2^{-1} = 1/2, made into a visual.

Worked examples

Example 1: Carbon-14 dating of charcoal from Dholavira

A piece of charcoal is recovered from a fire pit at the Harappan site of Dholavira in Kutch, Gujarat — a major Indus Valley city excavated by the Archaeological Survey of India since the 1990s. The sample is sent to the Physical Research Laboratory in Ahmedabad for radiocarbon dating. The specific activity of ^{14}C in the charcoal, measured as decays per minute per gram of carbon, is 8.1 dpm/g. Living plants today show 13.6 dpm/g (this is the contemporary atmospheric baseline). The half-life of carbon-14 is 5730 years. How old is the charcoal?

Carbon-14 activity drops from 13.6 to 8.1 dpm/g over the age of the sampleA horizontal axis labelled t in years, vertical axis labelled activity A in dpm per gram. A red exponential curve starts at 13.6 at t equals zero, falls through the point eight point one at roughly four thousand three hundred years. A dashed horizontal line at 8.1 intersects the curve at the sample age. t (yr) A (dpm/g) 0 6.8 13.6 2000 4000 6000 8000 8.1 t ≈ 4290 yr
The activity of $^{14}$C in the charcoal, $A = 13.6\, e^{-\lambda t}$ with $\lambda = (\ln 2)/5730$ yr$^{-1}$, falls from 13.6 (today) to 8.1 at the sample age. Reading off the intersection gives roughly 4290 years — consistent with the mature Harappan phase (c. 2600–1900 BCE).

Step 1. Identify the ratio of activities — the only experimental number that matters.

\frac{A(t)}{A_0} \;=\; \frac{8.1}{13.6} \;=\; 0.5956.

Why: the sample was a living plant at the moment the tree it came from was cut down (or burned). At that instant, it had the atmospheric carbon-14 activity A_0 = 13.6 dpm/g. Since then, its carbon-14 has been decaying without replenishment (dead plants stop taking in new CO_2). So the ratio of today's activity to A_0 is exactly the survival fraction e^{-\lambda t}.

Step 2. Compute the decay constant \lambda from the half-life.

\lambda \;=\; \frac{\ln 2}{t_{1/2}} \;=\; \frac{0.693}{5730\ \text{yr}} \;=\; 1.210 \times 10^{-4}\ \text{yr}^{-1}.

Why: you need \lambda to solve for time in the decay law. One conversion, done once.

Step 3. Invert the decay law to solve for t.

Starting from A(t)/A_0 = e^{-\lambda t}, take the natural log of both sides:

\ln\!\left(\frac{A}{A_0}\right) \;=\; -\lambda t \quad\Longrightarrow\quad t \;=\; -\frac{1}{\lambda}\ln\!\left(\frac{A}{A_0}\right).

Why: the minus sign flips because \ln(\text{number less than 1}) is negative, and age is positive. When you compute it out, the answer comes out positive without needing to track signs.

Step 4. Plug in.

t \;=\; -\frac{1}{1.210 \times 10^{-4}}\, \ln(0.5956) \;=\; -\frac{1}{1.210 \times 10^{-4}}\, (-0.5183) \;=\; 4284\ \text{yr}.

Why: \ln(0.5956) \approx -0.5183. Divide by \lambda. The answer lands in the Bronze Age, which is exactly where the Harappan civilisation sits in the radiocarbon record.

Result: The charcoal is approximately 4290 years old — consistent with the Mature Harappan period at Dholavira (ca. 2600–1900 BCE, which is 4600–3900 years before 2026 CE).

What this shows: Radiocarbon is the go-to archaeological clock for organic materials younger than about 60 000 years (roughly ten half-lives — beyond that, the surviving ^{14}C activity is too low to measure reliably against background). Dholavira's occupation, like that of Mohenjo-daro and Harappa, has been tightly pinned down by exactly this kind of analysis.

Example 2: A vial of technetium-99m from BARC for a PET scan

Technetium-99m (a metastable nuclear isomer of Tc-99, denoted ^{99\text{m}}Tc) is the most-used medical imaging isotope in the world. It gamma-decays with a half-life of 6.01 hours and is produced at the BARC research reactor in Trombay via the decay of molybdenum-99. A hospital in Mumbai orders a vial with specific activity 10 mCi at 08:00 on Tuesday. The patient's PET scan is scheduled at 14:00 the same day. What is the activity at scan time? And what is \lambda in SI units?

Activity of a Tc-99m vial over 24 hoursA horizontal axis labelled t in hours from zero to twenty-four, vertical axis labelled activity in millicurie. A red exponential curve starts at ten at t equals zero, falls through roughly five at six hours, two point five at twelve hours, and nearly touches zero at twenty-four hours. A vertical dashed line marks t equals six hours, where the activity is five milicurie — the patient's scan time. t (h) A (mCi) 0 5 10 6 12 18 24 5 mCi at 14:00
Activity of a 10 mCi Tc-99m vial over 24 hours. Half-life is 6 hours, so after 6 hours the activity is half (5 mCi); after 24 hours it is $(1/2)^4 = 1/16$, or 0.625 mCi — effectively spent.

Step 1. Compute the elapsed time and the number of half-lives.

\Delta t \;=\; 14{:}00 - 08{:}00 \;=\; 6\ \text{h} \;=\; 1.000\ \text{half-lives}.

Why: the scan is exactly one half-life after the delivery. This is not an accident — hospitals schedule Tc-99m scans within a few hours of delivery so that you get good signal without needing a huge initial dose.

Step 2. Apply the half-life rule.

A(\text{scan}) \;=\; \frac{A_0}{2^{\Delta t/t_{1/2}}} \;=\; \frac{10\ \text{mCi}}{2^1} \;=\; 5\ \text{mCi}.

Why: since the elapsed time is exactly one half-life, you can skip the exponential entirely — one half-life means divide by two. When the elapsed time is a whole number of half-lives, this shortcut is the fastest route.

Step 3. Convert \lambda to SI units (per second).

\lambda \;=\; \frac{\ln 2}{t_{1/2}} \;=\; \frac{0.693}{6.01\ \text{h}\ \times\ 3600\ \text{s/h}} \;=\; \frac{0.693}{21\,636\ \text{s}} \;=\; 3.20 \times 10^{-5}\ \text{s}^{-1}.

Why: SI activity is decays per second, so \lambda must be per second. Compare with the carbon-14 value \lambda \approx 1.21 \times 10^{-4} yr^{-1} from Example 1 — the Tc-99m decay constant is about 10^{13} times larger, which is exactly what you want for a short-lived medical isotope.

Step 4. Sanity check — the number of atoms in the vial at delivery.

Starting activity A_0 = 10\ \text{mCi} = 10^{-2} \times 3.7 \times 10^{10}\ \text{Bq} = 3.7 \times 10^8\ \text{Bq}.

N_0 \;=\; \frac{A_0}{\lambda} \;=\; \frac{3.7 \times 10^8}{3.20 \times 10^{-5}} \;\approx\; 1.16 \times 10^{13}\ \text{atoms}.

Why: N_0 = A_0/\lambda rearranges equation (8). The result is about twelve trillion atoms — a vanishingly small mass (around 2 picograms of Tc-99m), yet enough to image a patient. This is a recurring lesson of nuclear medicine: the activity is what matters, not the mass.

Result: The vial delivers 5 mCi at 14:00, exactly half its morning value. The decay constant is \lambda = 3.20 \times 10^{-5}\ \text{s}^{-1}, corresponding to a mean life \tau = 1/\lambda \approx 31\ 250\ \text{s} \approx 8.68\ \text{h}.

What this shows: Medical isotopes are engineered to have half-lives comparable to the clinical timeline — long enough to survive the trip from BARC's reactor to a Mumbai hospital, short enough that the patient is no longer radioactive by the next day. A half-life of a few hours is the Goldilocks zone for diagnostic PET. Iodine-131 (used for thyroid therapy) has t_{1/2} = 8 days — a longer half-life because the therapeutic dose must stay in the thyroid long enough to do its work.

Example 3: How long until a uranium-238 lump has lost 1% of its original mass?

A one-gram lump of freshly-purified ^{238}U sits in a lab at IIT Bombay. How long until 1% of the atoms have decayed? Half-life of ^{238}U is 4.47 \times 10^9 years.

Step 1. Compute \lambda.

\lambda \;=\; \frac{\ln 2}{4.47 \times 10^9\ \text{yr}} \;=\; 1.55 \times 10^{-10}\ \text{yr}^{-1}.

Step 2. Set N/N_0 = 0.99 and solve for t.

0.99 \;=\; e^{-\lambda t} \quad\Longrightarrow\quad t \;=\; -\frac{\ln 0.99}{\lambda} \;=\; -\frac{-0.01005}{1.55 \times 10^{-10}} \;\approx\; 6.5 \times 10^{7}\ \text{yr}.

Why: \ln(0.99) \approx -0.01005 (using the approximation \ln(1-x) \approx -x for small x, which is excellent here). Dividing by \lambda converts the dimensionless log into a time.

Step 3. Interpret.

Sixty-five million years. That is roughly the time since the extinction of the non-avian dinosaurs. In all of that geological time, a lump of uranium loses one percent of its mass to alpha decay. This is why uranium is the workhorse of radiometric dating for rocks older than a few hundred million years — it decays slowly enough to still be around in significant quantities, but fast enough that a measurable fraction has decayed over the age of the Earth. Samarium-neodymium and potassium-argon cover even longer timescales; rubidium-strontium and uranium-lead together pin down the age of the solar system at 4.57 \times 10^9 years — roughly one ^{238}U half-life ago.

Result: A 1% depletion of ^{238}U takes about 6.5 \times 10^7 years — useful as a scale for how slowly this isotope decays.

Successive decays — a chain in brief

Many naturally-occurring radioactive nuclei do not decay straight to a stable daughter. They decay to an unstable daughter, which decays to another unstable nucleus, which decays again, and so on, through a decay chain of eight to twenty steps. Uranium-238 ends at stable ^{206}Pb after fourteen intermediate alpha and beta decays. Thorium-232 ends at ^{208}Pb after ten steps. Uranium-235 ends at ^{207}Pb after eleven steps. Each natural radionuclide found on Earth today is either a primordial long-lived survivor (half-life comparable to the age of the Earth) or a member of one of these chains, continually regenerated by its ancestor.

Let N_1(t) be the parent (decay constant \lambda_1), N_2(t) the daughter (decay constant \lambda_2). The parent obeys the ordinary decay law. The daughter population gains from parent decays and loses from its own decays:

\frac{dN_2}{dt} \;=\; \lambda_1 N_1 - \lambda_2 N_2.

Why: the first term is the source (each parent decay creates one daughter). The second term is the sink (each daughter decay removes one). Same bookkeeping as before, with one extra term.

This is a linear ODE; its solution (with N_2(0) = 0) is the Bateman equation:

N_2(t) \;=\; \frac{\lambda_1 N_{1,0}}{\lambda_2 - \lambda_1}\left(e^{-\lambda_1 t} - e^{-\lambda_2 t}\right).

The full derivation (multiplying by an integrating factor e^{\lambda_2 t}) is left to the going-deeper section below. For most purposes you need only two qualitative features.

Secular equilibrium. If the parent has a much longer half-life than the daughter (\lambda_1 \ll \lambda_2), after a few daughter half-lives the two activities become equal: \lambda_1 N_1 = \lambda_2 N_2. The daughter decays as fast as the parent creates it, so its population stabilises. This is how radium-226 sits in equilibrium with its ^{222}Rn daughter inside a closed uranium ore sample.

Transient equilibrium. If the parent is longer-lived than the daughter but not by much, the daughter activity approaches a value slightly larger than the parent's and then decays with the parent's half-life. This is the regime of the famous ^{99}Mo / ^{99\text{m}}Tc generator — BARC ships a column of Mo-99 (t_{1/2} = 66 h) to hospitals, which "milk" the column daily to extract fresh Tc-99m (t_{1/2} = 6 h). Each day, Mo-99 has regenerated enough Tc-99m for another batch of scans.

Common confusions

If you came here to compute activities and half-lives, equations (5), (6), (7), and (9) are all you need. What follows is for readers who want to see why "memorylessness" uniquely specifies the exponential, the formal derivation of the Bateman equation, and the statistical basis of the decay law as a Poisson process.

Why memorylessness forces the exponential

The decay law N(t) = N_0\, e^{-\lambda t} was obtained from the one-line assumption that each nucleus has a fixed probability per unit time \lambda of decaying. Here is a more fundamental statement: the exponential is the only distribution that is memoryless.

Let T be the random variable "lifetime of a nucleus." Let S(t) = P(T > t) = N(t)/N_0 be the survival probability — the probability a nucleus makes it past time t. Memorylessness says: given that the nucleus has survived to time s, the probability it survives a further time t is the same as the unconditional probability of surviving time t from the start. In formulae:

P(T > s + t \mid T > s) \;=\; P(T > t).

By the definition of conditional probability, the left side is S(s + t)/S(s). Therefore:

S(s + t) \;=\; S(s)\, S(t).

This is Cauchy's multiplicative functional equation. The only continuous solutions with S(0) = 1 and S(t) \to 0 as t \to \infty are S(t) = e^{-\lambda t} for some \lambda > 0.

Why: take the log of both sides — \ln S(s+t) = \ln S(s) + \ln S(t) — and the only continuous solution is \ln S(t) = -\lambda t (a linear function of t, pinned to zero at t=0, decreasing). Exponentiate and you get e^{-\lambda t}.

So the decay law is not a physical postulate about radioactivity — it is a consequence of memorylessness, which in turn is a consequence of the quantum-mechanical nature of the decay process. If a decay depended on some internal "clock," it would not be memoryless, and the decay law would not be exponential. That we observe exponential decay over ten half-lives is direct evidence that no such internal clock exists.

The Bateman equation derived

Solve \dot N_2 + \lambda_2 N_2 = \lambda_1 N_1 = \lambda_1 N_{1,0}\, e^{-\lambda_1 t}. Multiply by the integrating factor e^{\lambda_2 t}:

\frac{d}{dt}\!\left(N_2\, e^{\lambda_2 t}\right) \;=\; \lambda_1 N_{1,0}\, e^{(\lambda_2 - \lambda_1) t}.

Why: the integrating factor is chosen so that the left side becomes a single derivative by the product rule. Now the equation is a pure integral.

Integrate from 0 to t, using N_2(0) = 0:

N_2\, e^{\lambda_2 t} \;=\; \frac{\lambda_1 N_{1,0}}{\lambda_2 - \lambda_1}\left(e^{(\lambda_2 - \lambda_1) t} - 1\right).

Divide by e^{\lambda_2 t}:

N_2(t) \;=\; \frac{\lambda_1 N_{1,0}}{\lambda_2 - \lambda_1}\left(e^{-\lambda_1 t} - e^{-\lambda_2 t}\right). \tag{10}

Why: this is the Bateman equation for a two-step chain. It has the symmetry \lambda_1 \leftrightarrow \lambda_2 (the two exponentials swap and the prefactor flips sign), which is a useful cross-check.

Secular equilibrium. When \lambda_1 \ll \lambda_2, after times t \gg 1/\lambda_2 the second exponential dies and equation (10) becomes

N_2(t) \;\approx\; \frac{\lambda_1 N_{1,0}}{\lambda_2}\, e^{-\lambda_1 t} \;=\; \frac{\lambda_1}{\lambda_2} N_1(t),

so \lambda_2 N_2 = \lambda_1 N_1 — the daughter's activity equals the parent's. The daughter is produced as fast as it decays.

Transient equilibrium. When \lambda_1 < \lambda_2 but the ratio is only modest (say Mo-99 and Tc-99m, with t_{1/2} ratios 66/6 = 11), equation (10) gives a ratio of activities

\frac{\lambda_2 N_2}{\lambda_1 N_1} \;=\; \frac{\lambda_2}{\lambda_2 - \lambda_1}\left(1 - e^{-(\lambda_2 - \lambda_1)t}\right),

which approaches the constant \lambda_2/(\lambda_2 - \lambda_1) > 1 for large t. The daughter activity slightly exceeds the parent's and then decays at the parent's rate.

The decay law as a Poisson process

The ticks of a Geiger counter near a radioactive source are a Poisson process. In any interval of length \Delta t, the probability of exactly k clicks is

P(k; \mu) \;=\; \frac{\mu^k\, e^{-\mu}}{k!}, \qquad \mu \;=\; A\cdot \Delta t,

where A = \lambda N is the activity at the midpoint of the interval (treating it as constant over \Delta t, valid when \Delta t \ll 1/\lambda). Two consequences are useful to know.

Counting statistics. The standard deviation of the count k in a Poisson distribution equals \sqrt{\mu}. If you count for long enough to get 10 000 clicks, the uncertainty is \sqrt{10\,000} = 100 — a 1% measurement. To halve the relative uncertainty you have to count four times as long. This is why low-activity dating (old carbon samples) demands either massive detectors or very long counting times, and why accelerator mass spectrometry — which counts ^{14}C atoms rather than decays — dominates modern radiocarbon work.

Waiting times. The time T between successive clicks is exponentially distributed with rate parameter A. The mean waiting time is 1/A. This is consistent with memorylessness: if you just heard a click, the time to wait for the next one is still exponentially distributed, with the same 1/A — the counter has no memory of when the previous click was.

Multi-step chains and geological dating

A decay chain of n steps, each with decay constant \lambda_i, generalises the Bateman equation to a sum of n exponentials with specific coefficients. After sufficient time, the longest-lived member of the chain dominates the population, and every downstream member reaches secular equilibrium with it — a beautiful statistical balance that makes closed uranium ore a natural reactor of neutrino emission.

Geological dating by the uranium-lead method uses the fact that each uranium atom in a zircon crystal eventually becomes a lead atom, and both are locked in the crystal lattice. The ratio ^{206}Pb/^{238}U in the mineral is

\frac{N_{\text{Pb}}}{N_\text{U}(t)} \;=\; e^{\lambda t} - 1,

(derived from N_\text{U}(t) = N_{\text{U},0}\, e^{-\lambda t} and N_{\text{Pb}}(t) = N_{\text{U},0}(1 - e^{-\lambda t}), assuming the daughter chain reaches equilibrium so each U decay produces one Pb atom). Measure the ratio, solve for t, and you have the crystal's age. The oldest dated zircons (from the Jack Hills in Western Australia) give t \approx 4.4 \times 10^9 years — very nearly the age of the Earth itself.

Multiple decay modes — branching ratios

Some nuclei have more than one way to decay. Potassium-40 beta-minus decays to ^{40}Ca with probability 89% per decay and electron-captures to ^{40}Ar with probability 11%. The overall decay constant is the sum: \lambda = \lambda_{\beta^-} + \lambda_{\text{EC}}. The half-life that appears in formulas is always the total; the individual branching ratios give the fraction of decays going through each channel. The ^{40}K \to ^{40}Ar channel is the basis of potassium-argon dating — useful for rocks from a few thousand to a few billion years old, filling the gap between carbon dating and uranium-lead.

Where this leads next