The Radioactive Decay Law

Point a Geiger counter at a lump of uranium and it clicks — irregularly, unpredictably, but never stopping. You cannot say which atom will click next, or when; you cannot hurry it up by heating the rock or slow it down by freezing it; a single nucleus might sit unchanged for a billion years or vanish in the next second. Radioactivity looks like the least lawful thing in physics: pure, blind chance, one nucleus at a time. And yet out of that randomness comes one of the most reliable laws in all of science. Watch not one nucleus but the trillions in the rock, and the clicks thin out in a curve so precise that we set our deepest clocks by it — dating a Neolithic bone to within a lifetime, an Egyptian coffin to within a reign, a lava flow to within a geological blink. The chaos of the individual becomes the certainty of the crowd. That certainty is the radioactive decay law, and it is the engine behind carbon dating, medical tracers, nuclear-waste timelines, and the age of the Earth itself. This article takes the one physical fact that makes a nucleus radioactive — that it decays at random with a fixed probability per second, remembering nothing of its past — and shows that this single fact forces the whole population to fall away as a clean exponential, $N = N_0 e^{-\lambda t}$ . We meet the decay constant $\lambda$ that sets the pace, the activity a detector actually reads, and the units (becquerel and curie) an exam will demand. The half-life you have half-met before drops out in one line; the next article takes it further.

Key takeaways

One postulate does all the work: every radioactive nucleus has a fixed probability per unit time of decaying, independent of its age, its neighbours, and its surroundings. Decay is random and memoryless.
The decay law: because the number decaying each second is proportional to the number left, the population falls exponentially, $N = N_0 e^{-\lambda t}$ , where $\lambda$ is the decay constant (probability of decay per nucleus per second).
Activity is what you measure: $A = \lambda N$ , the disintegrations per second, and it decays by the same law, $A = A_0 e^{-\lambda t}$ . Units: $1\ \text{Bq} = 1$ decay per second; $1\ \text{Ci} = 3.7\times10^{10}\ \text{Bq}$ .
A fixed fraction, not a fixed number, goes each second. So the curve halves on a fixed schedule: after each half-life $T_{1/2} = 0.693/\lambda$ exactly half of whatever is left remains; after $n$ half-lives $N = N_0/2^{\,n}$ .
Nothing you can do in a lab changes $\lambda$ — heat, pressure, chemical bonding, and the nucleus's own age all leave the decay rate untouched. This is what makes radioactive clocks trustworthy.
Key idea: because each nucleus decays independently with a constant probability per second, the surviving population and its activity both fall off as the same exponential $e^{-\lambda t}$ , with the single constant $\lambda$ fixing the pace of the whole process.

The core insight

A radioactive nucleus decays at random with a fixed probability $\lambda$ per unit time, remembering nothing of its history; summed over a huge population this forces the number of survivors — and the measured activity — to fall exactly as $N = N_0 e^{-\lambda t}$ and $A = \lambda N = A_0 e^{-\lambda t}$ , with the decay constant $\lambda$ the one number that sets the timescale.

Where this lives today

The law on this page is one of the most-used equations outside a physics classroom. Radiocarbon dating reads the activity of carbon-14 (half-life $5730$ years, decay constant $\lambda = 1.21\times10^{-4}\ \text{yr}^{-1}$ , mean life about $8300$ years) in an old bone or timber, compares it with living tissue, and inverts $A = A_0 e^{-\lambda t}$ to get the age — the technique that won Willard Libby the 1960 Nobel Prize and rewrote archaeology. The same inversion with uranium, potassium and rubidium isotopes dates rocks and meteorites, giving the Earth its $4.5$ -billion-year age. Nuclear medicine lives on the decay law: technetium-99m ( $6$ -hour half-life) is chosen precisely because $\lambda$ is large enough to image quickly yet small enough to survive the trip from reactor to patient, and dosages are computed straight from $A = A_0 e^{-\lambda t}$ . Nuclear-waste policy is an exercise in the same exponential run the other way — waiting the many half-lives it takes for activity to fall to background. Every smoke detector (americium-241), every radiotherapy source, every geological and cosmological clock is Rutherford and Soddy's law, still ticking exactly as they wrote it.

The physical picture

Start with a single unstable nucleus — say a carbon-14 or a uranium-238. The strong nuclear force holds it together, but it is not the lowest-energy arrangement available; somewhere down the line it will transmute (by $\alpha$ , $\beta$ or $\gamma$ emission — the subject of the previous articles) into something more stable. The remarkable experimental fact, established by Rutherford and Soddy over a century ago, is this: the nucleus carries no clock and no memory. It does not "wind up" to a decay, does not age, does not feel warmer or more crowded and go faster. At every instant it has exactly the same small probability of decaying in the next second as it had the day it was born. Physicists call this a memoryless process, and it is the entire physical content of the decay law. Quantify that probability and call it $\lambda$ , the decay constant: the chance, per nucleus, per unit time, that a decay happens. A nuclide with a large $\lambda$ is jittery and short-lived; one with a tiny $\lambda$ (uranium-238 has $\lambda \sim 5\times10^{-18}\ \text{s}^{-1}$ ) is stolidly long-lived. Crucially $\lambda$ is a property of the nuclide — the same for every nucleus of that species anywhere in the universe — and it does not change with time, temperature, or chemistry.

Figure

Radioactivity is memoryless — in each equal interval the same fraction of the surviving nuclei decays, so a population of sixteen thins to eight then four then two, never a fixed number each time

Now switch from one nucleus to the crowd. If each of $N$ surviving nuclei independently has probability $\lambda$ per second of going, then in the next second we expect about $\lambda N$ of them to decay. That is the whole secret: the number decaying per second is proportional to the number still present. Early on, with many nuclei, the counter clatters; later, with few left, it barely ticks — not because the individual odds changed (they never do) but because there are fewer contestants. A fixed fraction leaves each second, so the survivors dwindle in the multiplicative, halving pattern of the figure above, never by a fixed headcount. Turn that one proportionality into an equation and the exponential decay law falls out.

The decay law

Because the decay rate is set by how many nuclei remain, the rate of change of $N$ is proportional to $N$ itself, with the decay constant $\lambda$ as the proportionality factor and a minus sign because the population falls:

\frac{dN}{dt} = -\lambda N \qquad (1)

where: $N$ is the number of undecayed nuclei present at time $t$ , $dN/dt$ is the rate at which that number changes (negative, since nuclei are being lost), and $\lambda$ is the decay constant in units of inverse time ( $\text{s}^{-1}$ , or per year, per hour, …). The physical reading is direct: the more radioactive nuclei you have, the more decays per second — halve the sample and you halve its decay rate. Equation (1) is a rate law that says a quantity falls at a rate proportional to itself. The function that does this is the exponential, and solving (1) — a one-line piece of calculus we quote rather than perform here — gives the radioactive decay law:

N = N_0\, e^{-\lambda t} \qquad (2)

where: $N_0$ is the number of undecayed nuclei at the start ( $t=0$ ), $e = 2.718\ldots$ is the base of natural logarithms, and $\lambda t$ is a pure number (which is why $\lambda$ must carry inverse-time units matching $t$ ). You can check that (2) is right without solving anything: at $t=0$ it gives $N = N_0$ , as it must, and the fraction surviving, $N/N_0 = e^{-\lambda t}$ , falls smoothly from $1$ toward $0$ , never quite reaching it. The curve is the punchline of the whole subject.

Figure

The radioactive decay curve — the number of surviving nuclei falls as N equals N-zero times e to the minus lambda t, halving after every half-life to N-zero over 2, over 4, over 8, and passing through N-zero over e at one mean life

The signature of an exponential is that equal steps in time give equal factors, not equal drops. Wait one half-life and the survivors halve; wait another and they halve again; the curve on the figure crosses $N_0/2$ , then $N_0/4$ , then $N_0/8$ at evenly spaced times, flattening but never touching the axis. This is why "how much is left" is best asked in half-lives, and why a fixed percentage — never a fixed number — disappears in each fixed span of time.

Activity — what a detector actually reads

You never count the undecayed nuclei $N$ directly; a Geiger counter or scintillator counts decays. The activity $A$ is the number of disintegrations per second — precisely the rate at which $N$ falls:

A = -\frac{dN}{dt} = \lambda N \qquad (3)

where: $A$ is the activity (decays per second), and the minus sign turns the falling $dN/dt$ into a positive count rate. Activity is $\lambda$ times the number of nuclei present: a big sample of a long-lived nuclide and a tiny sample of a short-lived one can read the same on a counter. Because $A$ is just $\lambda$ times $N$ , it obeys the very same exponential law:

A = \lambda N_0\, e^{-\lambda t} = A_0\, e^{-\lambda t}, \qquad A_0 = \lambda N_0 \qquad (4)

where: $A_0 = \lambda N_0$ is the initial activity. So a plot of activity against time is the identical decay curve, scaled by $\lambda$ — which is what lets us measure a half-life without ever knowing $N_0$ : just watch how the count rate falls. Two units matter for exams. The SI unit is the becquerel: $1\ \text{Bq} = 1$ decay per second, named for Henri Becquerel who discovered radioactivity. The older, practical unit is the curie: $1\ \text{Ci} = 3.7\times10^{10}\ \text{Bq}$ , originally the activity of one gram of radium-226. A curie is a fierce amount of radioactivity; laboratory tracers are measured in microcuries or kilobecquerels.

Derivation AFrom a fixed decay probability to the exponential law

The goal is to see why a radioactive population must fall off exponentially, starting from nothing but the single physical postulate that decay is memoryless — each surviving nucleus has the same fixed probability of decaying in the next instant, no matter its age or surroundings — and using only algebra together with one quoted result from calculus. We build the rate equation from the probability statement, then quote its solution and check it.

Step 1. Put the postulate into symbols. In a short time interval $\Delta t$ , each surviving nucleus has probability $\lambda\,\Delta t$ of decaying, where the decay constant $\lambda$ is a fixed number set by the nuclide — the probability of decay per nucleus per unit time. "Memoryless" means this probability never depends on how long the nucleus has already survived.

Step 2. Scale up to the whole population. If $N$ nuclei are present, the expected number that decay during $\Delta t$ is $(\lambda\,\Delta t)\,N$ . The population therefore changes by $\Delta N = -\lambda N\,\Delta t$ (negative: nuclei are lost). Dividing by $\Delta t$ and taking the interval small gives the rate equation $dN/dt = -\lambda N$ , which is equation (1). Read it aloud: the decay rate is proportional to how many are left.

Step 3. Quote the solution. Equation (1) says $N$ falls at a rate proportional to itself; the function with that property is the exponential. Pinning it down means integrating the rate equation — a calculus step beyond this article's algebra ceiling, so we take the standard result on trust — giving $N = N_0\, e^{-\lambda t}$ , equation (2). We can verify it satisfies (1) without any calculus of our own: at $t = 0$ it returns $N_0$ ; and the property of $e^{-\lambda t}$ is exactly that its rate of change is $-\lambda$ times itself, so $dN/dt = -\lambda N_0 e^{-\lambda t} = -\lambda N$ , matching (1). The postulate alone fixes the whole curve.

Step 4. Carry it to the activity. Since $A = \lambda N$ and $\lambda$ is constant, multiply (2) by $\lambda$ : $A = \lambda N_0 e^{-\lambda t} = A_0 e^{-\lambda t}$ . The measured count rate obeys the identical exponential with the identical constant $\lambda$ — which is why fitting a decay curve of activity delivers $\lambda$ directly. Verified: $N/N_0 = e^{-\lambda t}$ equals $0.5$ at one half-life and the activity relation $A_0 = \lambda N_0$ reproduce in verify_hs2-radioactive-decay-law.py.

Go deeper · Derivation B · 1 minHalf-life, the halving ladder, and the mean life

Derivation BHalf-life, the halving ladder, and the mean life

The goal is to turn the decay constant $\lambda$ into the two numbers a laboratory actually quotes — the half-life $T_{1/2}$ , the time for half the sample to go, and the mean life $\tau$ , the average lifetime of a nucleus — and to show the population halves on a perfectly regular schedule. We use only the quoted law (2), logarithms and algebra; the one average that needs calculus is quoted here as a standard result (its evaluation needs an integral, above this article's algebra ceiling).

Step 1. Find the half-life. By definition, $N = N_0/2$ when $t = T_{1/2}$ . Put this into (2): $\tfrac12 N_0 = N_0\, e^{-\lambda T_{1/2}}$ , so $\tfrac12 = e^{-\lambda T_{1/2}}$ . Take the natural logarithm of both sides: $-\lambda T_{1/2} = \ln\tfrac12 = -\ln 2$ , hence

T_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{\lambda}.

Notice $N_0$ cancelled: whatever amount you start with, half of it is gone after one $T_{1/2}$ — the hallmark of exponential decay.

Step 2. Build the halving ladder. After $n$ whole half-lives, $t = n T_{1/2}$ , so $N = N_0\, e^{-\lambda\, n T_{1/2}} = N_0\,\big(e^{-\lambda T_{1/2}}\big)^{n} = N_0\left(\tfrac12\right)^{n}$ , since $e^{-\lambda T_{1/2}} = \tfrac12$ . The population steps $N_0 \to N_0/2 \to N_0/4 \to N_0/8 \to \cdots$ , and for any time at all we may write the law equivalently in base one-half:

N = N_0\left(\tfrac{1}{2}\right)^{t/T_{1/2}}.

This is the same curve as (2); $e^{-\lambda t}$ and $2^{-t/T_{1/2}}$ are two spellings of one exponential, tied by $\lambda = 0.693/T_{1/2}$ .

Step 3. Name the mean life. The average lifetime of a nucleus, $\tau$ , is found by averaging the decay time over the whole population — an average that needs an integral, so we quote the standard result $\tau = 1/\lambda$ ; the next article uses it further but likewise takes it as given. Combined with Step 1, $T_{1/2} = 0.693\,\tau$ , i.e. $\tau = 1.44\, T_{1/2}$ — the mean life runs a little longer than the half-life. After one mean life, $t = \tau = 1/\lambda$ , the survivors are $N = N_0 e^{-1} = N_0/e \approx 0.37\, N_0$ , the point marked on the decay curve. Verified: $T_{1/2} = 0.693/\lambda$ , $\tau = 1/\lambda = 1.44\,T_{1/2}$ , $N_0/e \approx 0.37\,N_0$ , and the ladder $1, \tfrac12, \tfrac14, \tfrac18$ recompute in verify_hs2-radioactive-decay-law.py.

Limiting cases

case	result	meaning
$t = 0$	$N = N_0$ , $A = A_0 = \lambda N_0$	the starting population and its initial activity
$t = T_{1/2}$	$N = N_0/2$	one half-life: exactly half remain, whatever $N_0$ was
$t = n\,T_{1/2}$	$N = N_0/2^{\,n}$	the halving ladder — a fixed factor per fixed span
$t = \tau = 1/\lambda$	$N = N_0/e \approx 0.37\,N_0$	after one mean life, a fraction $1/e$ survives
$t \to \infty$	$N \to 0$ , $A \to 0$	the sample is eventually exhausted, never abruptly
large $\lambda$ (short $T_{1/2}$ )	steep fall, high $A_0$	hot but fleeting — decays fast, counter roars then quiets
small $\lambda$ (long $T_{1/2}$ )	gentle fall, low $A_0$	feeble but persistent — e.g. uranium-238, ticking for aeons

Common trapDecay is not linear — a fixed fraction leaves each second, not a fixed number, and the nucleus never "ages"

The commonest wrecks on this topic are two, and they are twins. First, students imagine the sample losing the same number of nuclei each second, so that a source of $1000$ nuclei losing $100$ in the first second is "empty in ten seconds". Not so: the rate $\lambda N$ falls as $N$ falls, so after the first second there are fewer nuclei and fewer decays; a fixed fraction leaves, and the curve is exponential, not a straight line down to zero. Second, students imagine a nucleus "getting closer to decaying" as it ages, like a fuse burning down. It does not. Decay is memoryless: a uranium nucleus that has survived a billion years is exactly as likely to decay in the next second as a freshly made one. There is no internal clock, no build-up, no due date — only a constant probability per second. Both slips vanish the moment you hold onto the one true picture: constant probability per nucleus, so constant fraction per unit time.

Worked examples

Board level — number and activity of a decaying source (CBSE and ICSE pattern)

A radioactive source initially contains $N_0 = 6.0\times10^{10}$ nuclei of a nuclide whose decay constant is $\lambda = 0.0231\ \text{hour}^{-1}$ . Find (a) its half-life, (b) the number of nuclei left after $60$ hours, (c) the initial activity in becquerel, and (d) the activity after $60$ hours.

Step 1. Half-life from the decay constant: $T_{1/2} = 0.693/\lambda = 0.693/0.0231 = \boxed{30.0\ \text{hours}}$ .

Step 2. $60$ hours is exactly $60/30 = 2$ half-lives, so use the ladder $N = N_0/2^{\,n}$ with $n = 2$ : $N = 6.0\times10^{10}/4 = \boxed{1.5\times10^{10}\ \text{nuclei}}$ .

Step 3. Activity needs $\lambda$ in per-second, because $1\ \text{Bq}$ is one decay per second: $\lambda = 0.0231/3600 = 6.42\times10^{-6}\ \text{s}^{-1}$ . Then $A_0 = \lambda N_0 = (6.42\times10^{-6})(6.0\times10^{10}) = \boxed{3.85\times10^{5}\ \text{Bq}}$ (about $10.4\ \mu\text{Ci}$ ).

Step 4. After $2$ half-lives the activity has also quartered (it follows the same law): $A = A_0/4 = 3.85\times10^{5}/4 = \boxed{9.6\times10^{4}\ \text{Bq}}$ . (Check: $A = \lambda N = 6.42\times10^{-6}\times1.5\times10^{10} = 9.6\times10^{4}\ \text{Bq}$ , the same number the direct way.)

Competitive level — finding the decay constant from a falling count rate (JEE and NEET pattern)

The measured activity of a source drops from $8.0\times10^{4}\ \text{Bq}$ to $5.0\times10^{4}\ \text{Bq}$ in $20$ minutes. Find (a) the decay constant, (b) the half-life, and (c) the activity a further $30$ minutes later.

Step 1. Activity obeys $A = A_0 e^{-\lambda t}$ , so $A_0/A = e^{\lambda t}$ and $\lambda = \dfrac{1}{t}\ln\dfrac{A_0}{A}$ . Here $\lambda = \dfrac{1}{20}\ln\dfrac{8.0\times10^{4}}{5.0\times10^{4}} = \dfrac{\ln 1.6}{20} = \dfrac{0.470}{20} = \boxed{0.0235\ \text{min}^{-1}}$ .

Step 2. Half-life: $T_{1/2} = 0.693/\lambda = 0.693/0.0235 = \boxed{29.5\ \text{minutes}}$ .

Step 3. Take the $5.0\times10^{4}\ \text{Bq}$ point as a fresh start and run the law for another $30$ minutes: $A = 5.0\times10^{4}\,e^{-0.0235\times30} = 5.0\times10^{4}\,e^{-0.705} = 5.0\times10^{4}\times0.494 = \boxed{2.5\times10^{4}\ \text{Bq}}$ . Timing note: about 90 seconds. The whole item hinges on rearranging $A = A_0 e^{-\lambda t}$ into $\lambda = \ln(A_0/A)/t$ — memorise that one move and part (a) is a single log. Keep $\lambda$ and $t$ in the same time unit throughout (here minutes); only convert to seconds if the question asks for becquerel, which this one does not.

Competitive level — activity of a weighed sample (JEE Advanced pattern)

Calculate the activity of $1.0\ \mu\text{g}$ of pure radon-222, whose half-life is $3.82$ days. Give the answer in becquerel and in curie. Take $N_A = 6.022\times10^{23}\ \text{mol}^{-1}$ .

Step 1. Count the nuclei. One mole of radon-222 is $222\ \text{g}$ , so $1.0\ \mu\text{g} = 1.0\times10^{-6}\ \text{g}$ contains $N = \dfrac{1.0\times10^{-6}}{222}\times6.022\times10^{23} = 2.71\times10^{15}\ \text{nuclei}$ .

Step 2. Get $\lambda$ in per-second. The half-life is $T_{1/2} = 3.82\times86400 = 3.30\times10^{5}\ \text{s}$ , so $\lambda = 0.693/T_{1/2} = 0.693/(3.30\times10^{5}) = 2.10\times10^{-6}\ \text{s}^{-1}$ .

Step 3. Activity is $A = \lambda N = (2.10\times10^{-6})(2.71\times10^{15}) = \boxed{5.7\times10^{9}\ \text{Bq}}$ . Converting, $A = 5.7\times10^{9}/3.7\times10^{10} = \boxed{0.15\ \text{Ci}}$ . Timing note: about two minutes. The trap is units — mixing grams with the wrong molar mass, or leaving $\lambda$ in per-day while asking for becquerel. Lay it out as three clean lines: nuclei from $N = (m/M)N_A$ , then $\lambda = 0.693/T_{1/2}$ with $T_{1/2}$ in seconds, then $A = \lambda N$ . A microgram of a days-long emitter carrying gigabecquerels of activity is worth remembering as a sanity anchor.

Exam traps

Keep $\lambda$ and $t$ in the same time unit. $\lambda t$ must be a pure number. If $\lambda$ is per year, $t$ must be in years; for an answer in becquerel, convert $\lambda$ to per second first. Mixing units here is the single most common mark-loser on this topic across CBSE, ISC, JEE and NEET.

Activity $A$ is not the number of nuclei $N$ . $A = \lambda N$ has units of decays per second (becquerel); $N$ is a pure count. A question giving you an activity in Bq and asking for the number of atoms wants you to divide by $\lambda$ , not to read the same number twice.

After two half-lives, one quarter remains — not nothing. $N = N_0/2^{\,n}$ , so two half-lives leave $N_0/4$ , three leave $N_0/8$ . The sample never fully empties; "gone after two half-lives" is wrong.

$e^{-\lambda t}$ and $2^{-t/T_{1/2}}$ are the same law — don't mix them. Both are correct and linked by $\lambda = 0.693/T_{1/2}$ . Use whichever the numbers favour ( $2^{-n}$ for whole numbers of half-lives, $e^{-\lambda t}$ for arbitrary times), but never put $\lambda$ into the base-2 form or $T_{1/2}$ into the exponent of $e$ .

The rate is fixed by physics, not by conditions. Heating, cooling, pressure, chemical bonding, and the age of the nucleus leave $\lambda$ untouched. An MCQ asking "how to double the decay rate" is a trick: you cannot, short of changing the nuclide. This memoryless, environment-independent constancy is a favourite one-mark conceptual question.

Board scope. The exponential law, activity, decay constant and the becquerel/curie units are core on CBSE, ICSE (ISC), JEE Main, JEE Advanced, NEET, AP Physics 2, IB Standard and Higher Level, and A-Level 9702. IGCSE 0625 treats radioactivity and half-life qualitatively only — the exponential formula and activity in becquerel are not required there. The statistical origin of the law (why memorylessness forces the exponential) is treated lightly at school and unfolded properly at university, in the quantum-mechanical survival-amplitude picture named at the end.

Origins · 1 minFrederick Soddy and how this was discovered

Frederick Soddy

1877–1956 · Oxford, England

Frederick Soddy was the English radiochemist who, working alongside Ernest Rutherford at McGill University in Montreal between 1901 and 1903, turned radioactivity from a curiosity into a law. Together they proposed the disintegration (transformation) theory: radioactivity is the spontaneous transmutation of one element's atoms into another, proceeding at a rate proportional to the number of atoms present — the exponential decay law of this article, complete with the idea of a characteristic decay time. Soddy went on to formulate the radioactive displacement laws for $\alpha$ and $\beta$ decay and, in 1913, to coin the word isotope for chemically identical atoms of different mass. He took the chair of chemistry at Oxford in 1919 and received the Nobel Prize in Chemistry for 1921. It is his and Rutherford's law that every radioactive clock still runs on.

History

1896

— Henri Becquerel discovers that uranium salts spontaneously fog a photographic plate: radioactivity.

1898

— Marie and Pierre Curie isolate polonium and radium, and Marie Curie gives the phenomenon its name, radioactivity.

1900

— Rutherford finds that the activity of thorium emanation falls off geometrically in time, defining a characteristic decay period.

1902

— Ernest Rutherford and Frederick Soddy publish the disintegration theory: radioactivity is the spontaneous transmutation of atoms.

1903

— Rutherford and Soddy set out the exponential decay law and the notion of a fixed half-period for each substance.

1913

— Soddy introduces the term isotope and the radioactive displacement laws.

Going further

Texts. NCERT Physics Class 12, Chapter 13 ("Nuclei"), the law-of-radioactive-decay section · H. C. Verma, Concepts of Physics Vol. 2, "The Nucleus", for the decay-law and activity problems · Halliday, Resnick and Walker, "Nuclear Physics", for the statistical picture · Kenneth Krane, Introductory Nuclear Physics, for the university-level treatment.

Provenance. Standard decay-law relations and units. Recompute (in verify_hs2-radioactive-decay-law.py): $N/N_0 = e^{-\lambda t} = 0.5$ at one half-life; $T_{1/2} = 0.693/\lambda$ ; $\tau = 1/\lambda = 1.44\,T_{1/2}$ ; $N_0/e = 0.37\,N_0$ ; the ladder $1, \tfrac12, \tfrac14, \tfrac18$ ; carbon-14 $\lambda = 1.21\times10^{-4}\ \text{yr}^{-1}$ from $T_{1/2} = 5730\ \text{yr}$ , mean life $\approx 8300\ \text{yr}$ ; $1\ \text{Ci} = 3.7\times10^{10}\ \text{Bq}$ (cross-checked against one gram of radium-226, $T_{1/2} = 1600\ \text{yr}$ ). Board example: $\lambda = 0.0231\ \text{hr}^{-1}$ , $T_{1/2} = 30.0\ \text{hr}$ , $A_0 = 3.85\times10^{5}\ \text{Bq}$ , after $60\ \text{hr}$ $N = 1.5\times10^{10}$ , $A = 9.6\times10^{4}\ \text{Bq}$ . Competitive: $\lambda = \ln(1.6)/20 = 0.0235\ \text{min}^{-1}$ , $T_{1/2} = 29.5\ \text{min}$ , $A = 2.5\times10^{4}\ \text{Bq}$ ; radon-222 $N = 2.71\times10^{15}$ , $\lambda = 2.10\times10^{-6}\ \text{s}^{-1}$ , $A = 5.7\times10^{9}\ \text{Bq} = 0.15\ \text{Ci}$ .

Prerequisites. Radioactivity: Alpha, Beta and Gamma — the three decay modes whose common statistics this law describes. Leads to. Half-Life and Mean Life, which takes $T_{1/2}$ , $\tau$ and the decay-curve reading much further, and the numerical archetype Decay-Law, Half-Life and Dating Problems. For enrichment, the university seam Survival Amplitude and the Quantum Origin of Exponential Decay (go-deeper only, never an exam dependency here) shows why quantum mechanics makes the survival probability fall as $e^{-\lambda t}$ in the first place.

What exactly is the decay constant λ, and what are its units?

The decay constant $\lambda$ is the probability, per nucleus, per unit time, that a decay occurs — the fixed odds each nucleus faces every second. Its units are inverse time: per second, per hour, or per year, whatever suits the problem. A large $\lambda$ means a jittery, short-lived nuclide; a small $\lambda$ means a long-lived one. Because it is a probability per unit time, $\lambda$ multiplied by a time gives the dimensionless exponent $\lambda t$ in $e^{-\lambda t}$ . It is a fixed property of the nuclide — carbon-14 has one value, radon-222 another — and it is the single number from which the half-life ( $0.693/\lambda$ ), the mean life ( $1/\lambda$ ) and the activity ( $\lambda N$ ) all follow.

Why is radioactive decay exponential rather than a straight-line fall?

Because the number decaying each second is proportional to the number still present, not fixed. Early on, with many nuclei, many decay per second; later, with fewer left, fewer decay per second — even though each individual nucleus keeps exactly the same odds. A fixed fraction leaves in each equal span of time, and "a fixed fraction per unit time" is the mathematical definition of exponential decay. A straight-line fall would require a fixed number to leave each second, which would mean the last few nuclei decay just as fast as the first crowd — physically impossible when the rate is set by how many remain. This is why the curve halves on a schedule and flattens toward zero without ever reaching it.

Does an old nucleus get "closer" to decaying — does it age?

No, and this is the strangest and most important feature of the law. Radioactive decay is memoryless: a nucleus that has already survived a billion years has precisely the same probability of decaying in the next second as one made an instant ago. There is no internal fuse burning down, no accumulating strain, no due date. This is completely unlike a light bulb or a human, which do age and grow more likely to fail with time. The nucleus simply faces the same fixed odds at every instant. Memorylessness is exactly what makes the decay exponential and what makes radioactive dating trustworthy — the clock rate never drifts.

What is the difference between the activity and the number of nuclei?

The number of undecayed nuclei $N$ is a pure count — how many radioactive atoms are in the sample right now. The activity $A = \lambda N$ is the rate at which they are decaying, measured in becquerel (decays per second). A detector reads activity, not number: you count clicks, not atoms. The two are linked by $\lambda$ , so a huge sample of a long-lived nuclide (large $N$ , tiny $\lambda$ ) can have the same activity as a speck of a short-lived one (tiny $N$ , large $\lambda$ ). A frequent exam slip is to quote an activity in becquerel when the question asks for the number of atoms, or vice versa — always check whether you need to multiply or divide by $\lambda$ .

Can I speed up or slow down the decay by heating, cooling, or chemistry?

Essentially no. The decay constant $\lambda$ is set by the nucleus, deep inside the atom, and the electron cloud, temperature, pressure and chemical bonds of the atom leave it untouched to extraordinary precision. Heating a sample to thousands of degrees, freezing it, dissolving it, or bonding it into any compound does not change its half-life. (There are exquisitely tiny exceptions for a few decays that involve the innermost electrons, far below anything an exam considers.) For every practical and examination purpose, $\lambda$ is a fixed constant of nature — which is exactly why radioactive clocks can be trusted over billions of years and across wildly different environments.

What is the difference between the becquerel and the curie?

Both measure activity — decays per second. The becquerel is the SI unit: $1\ \text{Bq} = 1$ decay per second, named for Henri Becquerel. It is a tiny unit; real sources run to thousands (kBq) or millions (MBq). The curie is the older, larger unit: $1\ \text{Ci} = 3.7\times10^{10}\ \text{Bq}$ , defined historically as the activity of one gram of radium-226, which undergoes about $3.7\times10^{10}$ decays every second. A curie is a large and dangerous amount of radioactivity; medical tracers are dosed in microcuries ( $\mu\text{Ci}$ ) or megabecquerels. Know the conversion $1\ \text{Ci} = 3.7\times10^{10}\ \text{Bq}$ cold — it appears in numerical questions on every board that uses the curie.

Is the decay law exact, or is it only an average?

It is an average — but an extraordinarily sharp one. Each individual decay is genuinely random and unpredictable; you cannot say which nucleus will go next. The law $N = N_0 e^{-\lambda t}$ describes the expected number of survivors, and for a real sample containing astronomically many nuclei (a microgram already holds $10^{15}$ ), the fractional statistical scatter around that expectation is minuscule. So although the microphysics is pure chance, the macroscopic curve is smooth and repeatable to many decimal places. Only for very small numbers of atoms, or very low count rates, does the graininess show up as noticeable statistical fluctuation — which is why counting experiments quote error bars that shrink as you count longer.

Why write the law two ways, as e to the minus lambda t and as one-half to the power t over T-half?

They are the same law in two costumes, and each is convenient for different problems. $N = N_0 e^{-\lambda t}$ is the fundamental form, best when you know $\lambda$ or need an arbitrary time. $N = N_0(1/2)^{t/T_{1/2}}$ is the "count the half-lives" form, best when the elapsed time is a neat number of half-lives — then $N = N_0/2^{\,n}$ is a one-line answer with no exponential to evaluate. The bridge between them is $\lambda = 0.693/T_{1/2}$ , obtained by setting $e^{-\lambda T_{1/2}} = 1/2$ . Use whichever the numbers favour, but never blend them: $\lambda$ belongs with base $e$ , and $T_{1/2}$ belongs with base one-half.