ARCHETYPE: Decay-Law, Half-Life & Dating Problems

Hand an examiner one radioactive sample and a single half-life, and they can mint an entire exam paper from it. Find the decay constant. Find the mean life. Find how many nuclei survive after ten days, and how many becquerels the sample still ticks over at. Find how long ago a wooden coffin's carbon stopped breathing. Find what fraction of a medical tracer is left when the patient reaches the ward. Every one of these is the same physics — one nucleus, decaying with a fixed probability per second, so a whole population thinning along the curve $N = N_0 e^{-\lambda t}$ — wearing a different exam costume. Learn the one curve and the three clocks that read it, and the entire radioactivity-numerical family becomes fill-in-the-blanks. This is a numerical gym, so the bookkeeping is the article. At its centre sit two locked facts. First, activity and population fall along the same exponential: $A = \lambda N$ , and both $N$ and $A$ decay as $e^{-\lambda t}$ . Second, the one decay constant $\lambda$ can be dressed as three different clocks that all say the same thing — the half-life $T_{1/2} = 0.693/\lambda$ , the mean life $\tau = 1/\lambda$ , and their fixed ratio $T_{1/2} = 0.693\,\tau$ . Convert freely between those three, count in powers of two when the time is a whole number of half-lives ( $N = N_0/2^{\,n}$ ), and reach for the logarithm when it is not ( $t = \tau\ln(A_0/A)$ — the carbon-dating master formula). The trap that costs half the class its marks is a units-and-clocks muddle: quoting a half-life where the formula wants a mean life, or leaving $\lambda$ in "per year" while $t$ is in seconds. Keep $\lambda$ , $T_{1/2}$ and $\tau$ straight, keep $0.693$ and $N_0/2^n$ on the tip of your pen, and every decay-law, activity and dating numerical on CBSE, ICSE, JEE and NEET — and on IB HL and A-Level 9702 — falls out of it.

Key takeaways

The engine of every decay numerical is one curve: $N = N_0 e^{-\lambda t}$ , and its twin activity $A = \lambda N = A_0 e^{-\lambda t}$ (nuclei and their ticking fall along the same exponential).
One constant, three clocks: $T_{1/2} = 0.693/\lambda$ , mean life $\tau = 1/\lambda$ , and the fixed ratio $T_{1/2} = 0.693\,\tau$ (so $\tau = 1.44\,T_{1/2}$ ) — convert between them before you touch anything else.
When the time is a whole number of half-lives, forget the exponential and count in halves: $N = N_0/2^{\,n}$ with $n = t/T_{1/2}$ (and $A = A_0/2^{\,n}$ ).
Activity is measured in becquerels (1 Bq = 1 decay per second) or curies (1 Ci = $3.7\times10^{10}$ Bq); get $A_0$ via $A_0 = \lambda N_0$ and $N_0 = (m/M)\,N_A$ .
Carbon-dating is one rearrangement: $t = \tau\ln(A_0/A)$ ; a sample at one-quarter of living activity is exactly two half-lives, $11\,460$ years, old.
The fraction remaining is $e^{-\lambda t} = (1/2)^{\,t/T_{1/2}}$ — a pure number between 0 and 1 that never depends on how big the sample was.
Key idea: one curve $N = N_0 e^{-\lambda t}$ , one constant $\lambda$ read as three clocks ( $T_{1/2}=0.693/\lambda$ , $\tau=1/\lambda$ ), plus power-of-two counting and the log for dating, solve the whole radioactivity-numerical family — provided you keep the clocks and the units straight.

The core insight

Every radioactivity numerical reduces to the decay curve $N = N_0 e^{-\lambda t}$ and its activity twin $A = \lambda N$ : the single constant $\lambda$ converts to a half-life $T_{1/2}=0.693/\lambda$ and a mean life $\tau=1/\lambda$ (with $T_{1/2}=0.693\,\tau$ ), a whole number of half-lives is counted as $N = N_0/2^{\,n}$ , activity is quoted in becquerels or curies, and an age comes from the one rearrangement $t = \tau\ln(A_0/A)$ .

Where this lives today

The one exponential you just drilled is quietly at work across a dozen sciences. Nuclear medicine runs on it: a technetium-99m tracer with a six-hour half-life is timed so it images a patient's organs and then decays to almost nothing before it can do harm, planned with exactly the $A = A_0 e^{-\lambda t}$ arithmetic of example 2. Archaeology and climate science date the Turin Shroud, Ötzi the Iceman and ancient ice cores by Libby's $t=\tau\ln(A_0/A)$ , while geology stretches the same formula to billion-year clocks — uranium-lead, potassium-argon, rubidium-strontium — that fix the age of rocks and of the Earth itself at 4.5 billion years. Reactor engineers track spent fuel down the same curve; smoke detectors exploit the steady activity of a speck of americium-241; and standards laboratories realise the becquerel and the curie by counting decays per second. A rule a student meets as " $N_0$ over two to the $n$ " turns out to timestamp everything from a heartbeat scan to the age of the solar system.

The setup

Bookkeeping first, because in a numerical gym the accounting is the physics. A radioactive sample is a huge population of identical unstable nuclei — say $N$ right now — each of which will, sooner or later, transmute by spitting out an $\alpha$ , a $\beta$ , or a $\gamma$ . The one physical law behind everything is disarmingly simple: each nucleus has the same fixed probability of decaying per unit time, regardless of how old it is or what its neighbours are doing. Call that probability-per-second the decay constant $\lambda$ . Then the number that decay in a short time is proportional to how many are present, and the population thins along a falling exponential. We quote the solution of that statement — obtaining it is a one-line piece of calculus above this article's algebra-and-trig level — and build the whole gym on the result:

N = N_0\,e^{-\lambda t}, \qquad A = \lambda N = A_0\,e^{-\lambda t}. \qquad (1)

where: $N_0$ — the number of undecayed nuclei at the start ( $t=0$ ); $N$ — the number left after time $t$ ; $\lambda$ — the decay constant, the probability per second (or per year) that any one nucleus decays; $A$ — the activity, the number of decays per second the whole sample produces; and $A_0 = \lambda N_0$ its starting value. Read the two equations as one picture: the number of nuclei and the rate at which they tick are strictly proportional ( $A = \lambda N$ ), so they fall along the identical exponential curve. Halve the nuclei and you halve the clicks per second. Activity is what a Geiger counter actually measures, which is why its form of the law is the one you meet most in real problems. The single most important habit is to notice that $\lambda$ never appears alone in a real question — it always arrives wearing one of three interchangeable disguises, and your first move is always to sort out which one you have been handed.

Figure

Radioactive decay curve showing the number of nuclei as a fraction of the starting number falling from one at time zero to one half after one half-life one quarter after two half-lives and one eighth after three half-lives with the smooth exponential drawn through the points and the mean life marked at 1.44 half-lives where the fraction is one over e about 0.37

Those three disguises of $\lambda$ are the beating heart of the topic, so pin them down before anything else. The half-life $T_{1/2}$ is the time for the sample to halve; the mean life (or average life) $\tau$ is the average lifespan of a single nucleus; and $\lambda$ itself is the decay probability per unit time. They are all the same information, related by fixed constants:

T_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{\lambda}, \qquad \tau = \frac{1}{\lambda}, \qquad T_{1/2} = 0.693\,\tau. \qquad (2)

where: $\ln 2 = 0.693$ appears because "halving" is a factor of two on an exponential; $\tau = 1/\lambda$ is the mean life, which comes from averaging each nucleus's lifetime (a one-line integral we quote, beyond this article's algebra); and dividing the first two gives $\tau = 1.44\,T_{1/2}$ . A nucleus lives, on average, about 44% longer than the time it takes half the sample to disappear, because the long-lived stragglers pull the average up.

The four master relations

Four tools run the entire family, and every question is one of them solved for whichever letter is hidden. The first is the decay law itself, equation (1): the falling exponential for $N$ and for $A = \lambda N$ , used whenever you are given a starting amount and a time and asked for what is left — or the reverse — and the time is not a clean number of half-lives. The second is the clock conversion, equation (2). Nine questions in ten hand you one of $\{\lambda, T_{1/2}, \tau\}$ and demand another; do this first, in consistent units. Keep $\lambda$ and $t$ in the same time unit — both seconds, or both years — or every later number is wrong by a factor of $3.16\times10^{7}$ (the seconds in a year). The third is power-of-two counting. When the elapsed time is exactly $n$ half-lives, you never need the exponential — each half-life simply halves what is there:

N = \frac{N_0}{2^{\,n}}, \qquad A = \frac{A_0}{2^{\,n}}, \qquad n = \frac{t}{T_{1/2}}. \qquad (3)

where: $n$ — the number of half-lives elapsed. After 1 half-life, half remains; after 2, a quarter; after 3, an eighth; after 10, about a thousandth ( $1/1024$ ). This is the fastest road whenever $n$ is a small integer, and it is exactly the same as the exponential because $2^{-n} = e^{-\lambda t}$ when $t = n\,T_{1/2}$ . The fourth is the dating rearrangement. When you know the ratio of present to original amount and want the elapsed time, invert the exponential by taking a logarithm:

t = \frac{1}{\lambda}\ln\!\frac{N_0}{N} = \tau\,\ln\!\frac{A_0}{A} = \frac{T_{1/2}}{0.693}\,\ln\!\frac{A_0}{A}. \qquad (4)

where: $A_0/A$ (equal to $N_0/N$ ) — the factor by which the activity has fallen. Because you supply a ratio, the size of the sample cancels out entirely: a milligram and a kilogram of the same ancient wood give the same age. This one line is the whole of radiometric dating, and the carbon-14 version — the clock that reads the age of anything once alive — is the star application of the entire chapter.

Figure

Toolkit map of the four master relations linking the decay constant lambda the half-life and the mean life with the conversion formulas and showing the decay-law activity relation the power-of-two counting rule and the carbon-dating logarithm formula

The variants examiners actually pose

Convert between the three clocks. Given one of $\lambda$ , $T_{1/2}$ , $\tau$ , get another from equation (2): $\lambda = 0.693/T_{1/2}$ , $\tau = 1/\lambda = 1.44\,T_{1/2}$ . Watch the time unit — carbon-14's half-life is $5730$ years, but the same $\lambda$ reads $1.21\times10^{-4}$ per year or $3.83\times10^{-12}$ per second (worked in full in example 1).

Find how much survives after a time $t$ . If $t$ is a clean number of half-lives, count with equation (3): $N = N_0/2^n$ . Otherwise use the exponential $N = N_0 e^{-\lambda t}$ , keeping $\lambda$ and $t$ in the same unit. The fraction remaining, $e^{-\lambda t}$ , is the useful thing to report — it is independent of sample size.

Find the activity in becquerels or curies. Activity is $A = \lambda N$ . For the starting activity of a weighed sample, first count the nuclei with $N_0 = (m/M)\,N_A$ — mass over molar mass, times Avogadro's number — then $A_0 = \lambda N_0$ , in becquerels (1 Bq = 1 decay per second) or curies (1 Ci = $3.7\times10^{10}$ Bq).

Date something from its residual activity. Knowing the present activity $A$ and the original $A_0$ , find the age with equation (4): $t = \tau\ln(A_0/A)$ . If the ratio is a clean power of two, skip the logarithm and count half-lives: one-quarter of the living carbon-14 activity is exactly two half-lives, $2 \times 5730 = 11\,460$ years. And sometimes you simply read $T_{1/2}$ off a plotted decay curve — the point where the activity has halved.

Derivation AThe three clocks from one decay curve: where 0.693, the mean-life relation and the power-of-two rule come from

The goal is to build every conversion in equations (2) and (3) — the half-life $T_{1/2} = 0.693/\lambda$ , the mean-life relation $T_{1/2} = 0.693\,\tau$ , and the counting rule $N = N_0/2^n$ — starting from the quoted decay curve $N = N_0 e^{-\lambda t}$ , using nothing heavier than school algebra and the definition of a logarithm. The strategy is to feed each definition ("the time to halve", " $n$ half-lives elapsed") straight into the curve and solve. We do not re-derive the exponential itself: obtaining $N = N_0 e^{-\lambda t}$ from the decay statement is a one-line integration that lies above this article's algebra-and-trig ceiling, so we take it as given and work downstream of it.

Step 1. Pin down the half-life. By definition, after one half-life the sample has fallen to $N = N_0/2$ . Put that into the curve: $\frac{N_0}{2} = N_0\,e^{-\lambda T_{1/2}}$ . Cancel $N_0$ and you have $\tfrac12 = e^{-\lambda T_{1/2}}$ , i.e. $2 = e^{\lambda T_{1/2}}$ . Take the natural logarithm of both sides: $\ln 2 = \lambda T_{1/2}$ , so $T_{1/2} = \ln 2/\lambda = 0.693/\lambda$ . The mysterious $0.693$ is nothing but $\ln 2$ , and it enters purely because a half-life asks the exponential to drop by a factor of two.

Step 2. Bring in the mean life. The average lifetime of a single nucleus (a short integral we quote, not derive, since it needs calculus) collapses to the clean result $\tau = 1/\lambda$ . Eliminate $\lambda$ using Step 1: from $T_{1/2} = 0.693/\lambda$ and $\lambda = 1/\tau$ we get $T_{1/2} = 0.693\,\tau$ , and inverting, $\tau = 1.44\,T_{1/2}$ — always the longer of the two, because a few long-lived nuclei drag the average up past the halving time.

Step 3. Turn the exponential into power-of-two counting. Substitute Step 1's result $\lambda = 0.693/T_{1/2}= \ln2/T_{1/2}$ back into the curve: $N = N_0\,e^{-(\ln 2)\,t/T_{1/2}}$ . But $e^{\ln 2} = 2$ , so $e^{(\ln 2)\,x} = 2^{\,x}$ , which turns the exponent into a clean power of two: $N = N_0\,2^{-t/T_{1/2}} = N_0/2^{\,n}$ with $n = t/T_{1/2}$ . This is why, after a whole number of half-lives, you can throw the exponential away and just keep halving — the two roads are algebraically identical.

Step 4. Check that activity rides the same curve. Since $\lambda$ is constant, multiply the whole of Step 3 by $\lambda$ : $A = \lambda N_0\,2^{-n} = A_0/2^{\,n}$ , and equally $A = A_0 e^{-\lambda t}$ . Number and activity fall in lockstep — which is what lets a Geiger counter (which reads $A$ ) date a sample (which is about $N$ ). Verified: $T_{1/2}=0.693/\lambda$ , $\tau=1/\lambda$ , $T_{1/2}=0.693\,\tau$ and the equivalence $N_0/2^n = N_0 e^{-\lambda t}$ at $t=n\,T_{1/2}$ are recomputed for carbon-14 and checked against each other in verify_hs2-radioactivity-decay-numericals.py.

Go deeper · Derivation B · 1 minFrom residual activity to an age: the carbon-dating master formula

Derivation BFrom residual activity to an age: the carbon-dating master formula

The goal is to turn a ratio of present-to-original activity into an elapsed time — equation (4), $t = \tau\ln(A_0/A)$ — and to see why the size of the sample never enters, which is what makes radiocarbon dating possible from a fingernail-sized fleck of charcoal. The strategy is a single inversion of the activity law by a logarithm: pure algebra, no calculus, once the exponential is granted.

Step 1. Start from the activity law, equation (1): $A = A_0\,e^{-\lambda t}$ . While an organism lives it keeps exchanging carbon with the atmosphere, so its carbon-14 activity stays pinned at the living value $A_0$ . At death the exchange stops and the clock starts: the trapped carbon-14 decays with no resupply, and today, a time $t$ later, the sample reads a smaller activity $A$ .

Step 2. Isolate the exponential. Divide both sides by $A_0$ : $\dfrac{A}{A_0} = e^{-\lambda t}$ , or flipping the fraction, $\dfrac{A_0}{A} = e^{+\lambda t}$ . Notice that $A_0$ and $A$ are both proportional to the amount of carbon in the sample, so their ratio is a pure number — the sample's mass has cancelled. This is why you may date a splinter as well as a beam.

Step 3. Take the logarithm. Applying $\ln$ to $A_0/A = e^{\lambda t}$ gives $\ln(A_0/A) = \lambda t$ , so $t = \dfrac{1}{\lambda}\ln\dfrac{A_0}{A}$ . Dress $\lambda$ in whichever clock the data offer: since $1/\lambda = \tau = T_{1/2}/0.693$ , the same age reads $t = \tau\ln(A_0/A) = \dfrac{T_{1/2}}{0.693}\ln\dfrac{A_0}{A}$ . For carbon-14, $\tau = 8267$ years.

Step 4. Sanity-check against the counting rule. If the sample sits at exactly a quarter of the living activity, $A_0/A = 4$ , the formula gives $t = 8267\ln 4 = 8267 \times 1.386 = 11\,460$ years — and $\ln 4 = 2\ln 2$ , so this is precisely $2\,T_{1/2} = 2\times5730$ years, the answer power-of-two counting gives for "a quarter left." The logarithm road and the halving road agree exactly, as they must, which is your in-exam cross-check. Verified: $t = \tau\ln(A_0/A)$ is recomputed for a quarter-activity sample (giving $11\,460$ yr, cross-checked against $2\,T_{1/2}$ ) and for a 60%-activity sample in verify_hs2-radioactivity-decay-numericals.py.

Trap table and limiting cases

pitfall or limiting case	what actually happens	the fix
given $T_{1/2}$ , formula wants $\tau$	they differ by 44%: $\tau = 1.44\,T_{1/2}$	convert with equation (2) before substituting
$\lambda$ in per-year, $t$ in seconds	answer wrong by $3.16\times10^{7}$	keep $\lambda$ and $t$ in the same time unit
after 1 half-life, is any nucleus "used up"?	no single nucleus ages — the population halves	decay is memoryless; $\lambda$ is fixed, not growing
$t \to 0$ (just made)	$N \to N_0$ , $A \to A_0$	$e^{0}=1$ : the exponential starts at the full amount
$t \to \infty$ (very old)	$N \to 0$ , $A \to 0$	the curve approaches zero but never reaches it
$t = n\,T_{1/2}$ with $n$ a whole number	use $N_0/2^n$ , skip the exponential	$2^{-n}=e^{-\lambda t}$ exactly here
fraction remaining depends on sample size?	no — $e^{-\lambda t}$ is a pure number	mass cancels; a mg and a kg of one nuclide date alike
half a half-life ( $n=\tfrac12$ )	$N/N_0 = 2^{-1/2} = 0.707$ , not $0.75$	halving is multiplicative, never linear in time
activity "runs out" faster than nuclei?	no — $A=\lambda N$ , they share one curve	halve the nuclei and you halve the clicks per second

Common trapHalf-life and mean life are not the same number — and neither is a nucleus's "expiry date"

Two errors live here, and examiners bait both. The first is treating $T_{1/2}$ and $\tau$ as interchangeable. They are not: the mean life is 44% longer, $\tau = 1.44\,T_{1/2}$ , because the average lifetime of a nucleus is pulled up by the long-lived stragglers. If a formula asks for the mean life — as the dating formula $t=\tau\ln(A_0/A)$ does — and you feed it the half-life, every answer is 44% too small. Always convert first with equation (2). The second, deeper error is imagining that a half-life is some kind of expiry date after which a nucleus is "due" to decay. Radioactivity is memoryless: a nucleus that has sat unchanged for a billion years has exactly the same probability of decaying in the next second as a freshly made one. The half-life is a statement about the population, not any individual nucleus — the time for half of a large sample to go, working only because astronomically many nuclei average out. This is why $\lambda$ is a genuine constant: it does not creep up as the sample ages, and it is unmoved by temperature, pressure, or chemical state. Heat a sample, freeze it, dissolve it in acid — its half-life does not budge, because decay is a property of the nucleus, sealed off from the chemical world outside it.

Worked examples

1 · Board level — the three clocks of carbon-14 (CBSE and NEET pattern)

Carbon-14 has a half-life of $5730$ years. Find its decay constant $\lambda$ (in per year and in per second) and its mean life $\tau$ .

Step 1. Decay constant in per year, from equation (2): $\lambda = 0.693/T_{1/2} = 0.693/5730 = \boxed{1.21\times10^{-4}\ \text{per year}}$ . This is the probability that any one carbon-14 nucleus decays in a given year — about one chance in eight thousand.

Step 2. Decay constant in per second. First rewrite the half-life in seconds: $T_{1/2} = 5730 \times 3.156\times10^{7} = 1.81\times10^{11}$ s. Then $\lambda = 0.693/1.81\times10^{11} = \boxed{3.83\times10^{-12}\ \text{per second}}$ . It is the same physical constant — only the time unit changed.

Step 3. Mean life, from equation (2): $\tau = 1/\lambda = 1/(1.21\times10^{-4}) = \boxed{8267\ \text{years}}$ , in line with the direct relation $\tau = 1.44\,T_{1/2}$ (the mean life runs about 44% longer than the half-life).

Step 4. Sense-check: the mean life (8267 yr) is longer than the half-life (5730 yr), exactly as it must be — the average nucleus outlives the halving time because the survivors linger.

2 · Board level — counting in half-lives and the halving of activity (ICSE and CBSE pattern)

A fresh sample of iodine-131 (half-life $8$ days) contains $6.0\times10^{10}$ nuclei and has an activity of $3.2\times10^{5}$ Bq. After $24$ days, how many nuclei remain, what fraction has decayed, and what is the activity?

Step 1. Count the half-lives: $n = t/T_{1/2} = 24/8 = 3$ — a clean whole number, so use power-of-two counting, equation (3), and skip the exponential entirely.

Step 2. Nuclei remaining: $N = N_0/2^{\,n} = 6.0\times10^{10}/2^{3} = 6.0\times10^{10}/8 = \boxed{7.5\times10^{9}}$ nuclei.

Step 3. Fraction: fraction remaining $= 1/2^{3} = 1/8 = 0.125$ , so the fraction decayed is $1 - 0.125 = 0.875$ , i.e. $\boxed{87.5\%}$ of the sample has gone.

Step 4. Activity: because $A = \lambda N$ rides the same curve, $A = A_0/2^{\,n} = 3.2\times10^{5}/8 = \boxed{4.0\times10^{4}\ \text{Bq}}$ . No new calculation was needed — the activity halves in step with the nuclei.

3 · Competitive level — the activity of one gram of radium, and the birth of the curie (JEE-Advanced and NEET pattern)

Radium-226 has a half-life of $1600$ years and a molar mass of $226$ g/mol. Find the activity of a $1.0$ g sample in becquerels, and express it in curies. (This is the calculation by which the Curies' unit was originally defined.)

Step 1. Count the nuclei, $N_0 = (m/M)\,N_A = (1.0/226)\times6.022\times10^{23} = \boxed{2.66\times10^{21}}$ nuclei. One gram of a heavy element still holds a few thousand billion billion atoms.

Step 2. Get $\lambda$ in per second. The half-life in seconds is $1600 \times 3.156\times10^{7} = 5.05\times10^{10}$ s, so $\lambda = 0.693/5.05\times10^{10} = 1.37\times10^{-11}$ per second.

Step 3. Activity, from $A = \lambda N_0 = 1.37\times10^{-11}\times2.66\times10^{21} = \boxed{3.66\times10^{10}\ \text{Bq}}$ — that is 37 billion decays every second from a sample you could balance on a fingertip.

Step 4. In curies: $A = 3.66\times10^{10}/3.7\times10^{10} = \boxed{0.99\ \text{Ci} \approx 1\ \text{Ci}}$ . No accident: the curie was defined as the activity of one gram of radium-226. Timing note: about two minutes. The whole problem is $A_0 = \lambda N_0$ with $N_0 = (m/M)N_A$ — the two sub-steps are counting the atoms and converting the half-life to seconds. Do the seconds conversion first and keep it on the page; the single commonest slip here is leaving $\lambda$ in per-year while quoting activity in becquerels (per second).

4 · Competitive level — dating a wooden relic two ways (JEE-Main and A-Level pattern)

A wooden artefact from an archaeological dig has a carbon-14 activity that is (a) exactly one-quarter, and (b) $0.60$ times, the activity of a modern living sample of the same mass. Find the age in each case. Take $T_{1/2} = 5730$ yr, $\tau = 8267$ yr.

Step 1. Case (a), by counting. A quarter left means $A/A_0 = 1/4 = 1/2^{2}$ , i.e. exactly $n=2$ half-lives, so $t = 2\,T_{1/2} = 2 \times 5730 = \boxed{11\,460\ \text{years}}$ . When the ratio is a clean power of two, never reach for the logarithm.

Step 2. Case (a), cross-checked by the master formula, equation (4): $t = \tau\ln(A_0/A) = 8267 \times \ln 4 = 8267 \times 1.386 = 11\,460$ years — the same answer by the general road, because $\ln 4 = 2\ln 2$ .

Step 3. Case (b), where the ratio is not a power of two, so the logarithm is unavoidable: $A_0/A = 1/0.60 = 1.667$ , hence $t = \tau\ln(1.667) = 8267 \times 0.511 = \boxed{4220\ \text{years}}$ (to three figures, $4223$ yr).

Step 4. Sense-check: $0.60$ of the activity is a little more than half, so the age should be a little less than one half-life ( $5730$ yr) — and $4220$ yr duly is. If you had wrongly used $T_{1/2}$ in place of $\tau$ inside the logarithm, the age would have come out about 44% too small — the mean life is the right clock for the dating formula. Timing note: about 90 seconds. Read the ratio first: if it is a half, quarter or eighth, count half-lives (a few seconds); only if it is an "ugly" ratio do you spend a line on $t=\tau\ln(A_0/A)$ . Always use $\tau$ , never $T_{1/2}$ , inside the logarithm — that single substitution is where most of the lost marks go.

Exam traps

Muddling the three clocks. $\lambda$ , $T_{1/2}$ and $\tau$ are one piece of information in three costumes: $\lambda = 0.693/T_{1/2}$ , $\tau = 1/\lambda = 1.44\,T_{1/2}$ . The dating formula $t=\tau\ln(A_0/A)$ wants the mean life; feed it the half-life and every age is 44% too small. Convert first, substitute second.

Mixing time units. Keep $\lambda$ and $t$ in the same unit throughout — both per second and seconds, or both per year and years. A half-life quoted in years but an activity wanted in becquerels (per second) is the classic setup for a factor-of- $3.16\times10^{7}$ blunder; convert the half-life to seconds on the page before you compute $\lambda$ .

Treating decay as linear. After half a half-life, three-quarters is not left — $2^{-1/2} = 0.707$ is. Halving is multiplicative: two half-lives leave a quarter (not zero), three leave an eighth, ten leave about a thousandth. Never subtract "half per half-life."

Thinking a nucleus has a memory. A half-life is a property of the whole population, not a countdown clock on each nucleus. A nucleus that has already survived a billion years is no more "due" than a fresh one — $\lambda$ is fixed. And it is unmoved by temperature, pressure or chemistry; you cannot speed up or slow down a half-life.

Forgetting $A = \lambda N$ . Activity and number are proportional and fall along the same curve, so $A = A_0/2^n = A_0 e^{-\lambda t}$ just like $N$ . You never recompute the activity's decay separately — whatever factor cuts the nuclei cuts the clicks per second identically.

Board scope. The decay law, activity, half-life/mean-life conversions and carbon-dating numericals are firmly examined on CBSE, ICSE, JEE-Main, JEE-Advanced and NEET, and on IB Physics HL and A-Level 9702. AP Physics 2 takes decay largely qualitatively — half-life and random decay — so the quantitative decay constant, becquerel activity and dating logarithm here sit above its core. IGCSE 0625 carries half-life only qualitatively — reading it off a graph, "how many half-lives", the idea of random decay — but not the exponential formula $N=N_0e^{-\lambda t}$ , the decay constant, activity in becquerels, or the dating logarithm, so this quantitative gym is beyond an IGCSE paper. Mean life $\tau$ and the $\tau\ln(A_0/A)$ dating form are emphasised most on the Indian boards and A-Level.

Origins · 1 minWillard Libby and how this was discovered

Willard Libby

1908–1980 · Chicago, USA

Willard Frank Libby was the American physical chemist who turned the decay law into a clock for archaeology. Reasoning that cosmic rays continually make carbon-14 in the upper atmosphere, that living things absorb it at a steady level while they breathe and eat, and that this intake stops at death, Libby saw that the residual carbon-14 activity of any once-living material — wood, bone, cloth, charcoal — measures the time since it died, through exactly the formula $t=\tau\ln(A_0/A)$ of this article. In 1949 he and his co-workers demonstrated the method on samples of known age, pegging the living activity at roughly 15 disintegrations per minute per gram of carbon and the half-life at about 5568 years (the "Libby half-life"), later refined to the modern 5730 years. Radiocarbon dating revolutionised archaeology, geology and climate science, giving absolute ages back some 50,000 years where before there had been only guesswork, and it won Libby the 1960 Nobel Prize in Chemistry. Later refinements — tree-ring calibration and accelerator mass spectrometry that counts the carbon-14 atoms directly — rest on the very same exponential he first exploited.

History

1896

— Henri Becquerel discovers that uranium salts fog a photographic plate, revealing radioactivity.

1898

— Marie and Pierre Curie isolate polonium and radium, and Marie Curie coins the word radioactivity.

1900

— Paul Villard identifies gamma rays; Rutherford had named the alpha and beta rays the year before.

1902

— Ernest Rutherford and Frederick Soddy formulate the exponential law of radioactive decay and the idea of transmutation.

1910

— the curie is fixed near the activity of one gram of radium, about 3.7 times ten to the tenth decays per second.

1949

— Willard Libby demonstrates radiocarbon dating, reading the age of once-living matter from its residual carbon-14 activity.

Going further

Texts. NCERT Physics Class 12, Ch. 13 ("Nuclei"), the law of radioactive decay, activity and half-life, with the worked exercises · H. C. Verma, Concepts of Physics Vol. 2, "The Nucleus," decay-law and activity problems · D. C. Pandey / Irodov modern-physics problem sets (for the activity-from-mass and multi-half-life variants) · Cambridge A-Level 9702, Topic 23 (radioactivity) decay-constant and half-life questions.

Provenance. Standard exam-archive problems on the radioactive decay law. Recompute: carbon-14 clocks $T_{1/2}=5730$ yr $\Rightarrow \lambda = 1.21\times10^{-4}$ /yr $= 3.83\times10^{-12}$ /s, $\tau = 8267$ yr (with $1\ \text{yr}=3.156\times10^{7}$ s); iodine-131 example $n=3$ half-lives $\Rightarrow N = N_0/8 = 7.5\times10^{9}$ , 87.5% decayed, $A = 4.0\times10^{4}$ Bq from $A_0 = 3.2\times10^{5}$ Bq; radium-226 ( $T_{1/2}=1600$ yr, $M=226$ ) $\Rightarrow N_0 = 2.66\times10^{21}$ , $\lambda = 1.37\times10^{-11}$ /s, $A = 3.66\times10^{10}$ Bq $= 0.99$ Ci $\approx 1$ Ci; dating at one-quarter activity $\Rightarrow 2\,T_{1/2} = 11\,460$ yr (cross-check $8267\ln 4$ ); dating at 0.60 activity $\Rightarrow 8267\ln(1.667) = 4220$ yr; half a half-life $\Rightarrow 2^{-1/2}=0.707$ ; ten half-lives $\Rightarrow 1/1024$ ; 1 Ci $= 3.7\times10^{10}$ Bq; living-carbon reference $\approx 0.25$ Bq per gram $\approx 15$ counts per minute per gram.

Prerequisites. The Radioactive Decay Law — the concept article that builds $N=N_0e^{-\lambda t}$ and $A=\lambda N$ from the memoryless per-nucleus probability — and Half-Life and Mean Life, which develops $T_{1/2}=0.693/\lambda$ , $\tau=1/\lambda$ and the reading of a decay curve. This gym assumes both and drills using them at speed. Leads to. the university seam Survival Amplitude and the Decay of Quantum States (`aq-qm-survival-amplitude-decay`), which re-derives the exponential law from quantum mechanics and asks when it fails.

Go deeper (UG). The exponential decay law is quoted here; obtaining $N=N_0e^{-\lambda t}$ from the statement "constant probability per unit time" is a one-line integration of $dN/dt = -\lambda N$ , and the mean life $\tau=1/\lambda$ is the average of $t$ over that distribution — both are elementary calculus that sits just above this article's algebra-and-trig ceiling and is developed in The Radioactive Decay Law and the university seam above. The deeper question of why an unstable quantum state should decay exponentially at all — and the subtle ways real systems deviate from it at very short and very long times — is the subject of the survival-amplitude seam. Enrichment only; every numerical in this gym stands on the quoted law alone.

What exactly is the decay constant, and how is it different from the half-life?

The decay constant $\lambda$ is the probability, per unit time, that any single nucleus decays — a fundamental property of that nuclide, like a fixed "risk per second." The half-life $T_{1/2}$ is a more human-friendly repackaging of the same fact: the time for half a large sample to decay. They carry identical information and convert by $T_{1/2}=0.693/\lambda$ . A big $\lambda$ (high risk per second) means a short half-life, and vice versa. Use $\lambda$ when you need the raw exponential or an activity ( $A=\lambda N$ ); use $T_{1/2}$ when the data or the answer are naturally in "number of half-lives." The $0.693$ that links them is just $\ln 2$ , appearing because a half-life asks the exponential to fall by a factor of two.

Why is the mean life longer than the half-life — shouldn't half the nuclei set the average?

It feels as if the half-life should be the "typical" lifetime, but the mean life is longer: $\tau = 1.44\,T_{1/2}$ . The reason is that a decay curve has a long tail. By one half-life, half the nuclei have gone — but the survivors do not all vanish at once; some last two, three, five, ten half-lives before decaying, and those long-lived stragglers pull the average lifetime up beyond the halving time. Mathematically the average of $t$ over an exponential population works out to exactly $1/\lambda$ , which is $1.44\,T_{1/2}$ . A useful mental picture: the half-life is the median lifetime (half live longer, half shorter), while $\tau$ is the mean, and for a right-skewed distribution the mean always sits above the median.

When can I just halve repeatedly, and when do I actually need the exponential?

Halve repeatedly — $N = N_0/2^n$ — whenever the elapsed time is a whole number of half-lives, because then $n = t/T_{1/2}$ is a tidy integer and $2^{-n}$ is exact and quick. One half-life leaves a half, two leave a quarter, three an eighth, and so on. The moment the time is not a clean multiple of the half-life — say 7 days when the half-life is 8, or "0.60 of the activity remains" — the counting shortcut fails and you must use the full exponential $N = N_0 e^{-\lambda t}$ or, for an age, the logarithm $t=\tau\ln(A_0/A)$ . The two roads never disagree; power-of-two counting is simply the exponential evaluated at the convenient points where the arithmetic is easy.

What are becquerels and curies, and how do I get activity from a mass of material?

Activity is the number of decays a sample produces per second. Its SI unit is the becquerel: 1 Bq = 1 decay per second. The older unit is the curie: 1 Ci = $3.7\times10^{10}$ Bq, originally defined as the activity of one gram of radium-226 (see example 3). To find the activity of a weighed sample, first count its nuclei with $N_0 = (m/M)\,N_A$ — mass divided by molar mass, times Avogadro's number $6.022\times10^{23}$ — then multiply by the decay constant: $A_0 = \lambda N_0$ . The one thing to watch is the time unit: because a becquerel is per second, $\lambda$ must be in per-second too, so convert the half-life to seconds ( $1\ \text{year} \approx 3.156\times10^{7}$ s) before you compute $\lambda$ .

How does carbon-14 dating actually work, and what are its limits?

While a plant or animal is alive it constantly exchanges carbon with the atmosphere, so its carbon-14 sits at the same steady level as the air — a fixed activity $A_0$ per gram of carbon. At death the exchange stops and the trapped carbon-14 decays with a 5730-year half-life. Measuring the residual activity $A$ and comparing it to the living value gives the time since death from $t=\tau\ln(A_0/A)$ . The method works from a few hundred up to roughly 50,000 years — about nine or ten half-lives, beyond which too little carbon-14 is left to measure. It assumes the atmospheric level was constant, which is not quite true, so real dates are calibrated against tree-ring records. It dates only things that were once alive; it cannot date rock or metal.

Does temperature, pressure or chemical state change a half-life?

Essentially never, to any degree that matters for these problems. Radioactive decay is a process of the nucleus, which is a hundred-thousand times smaller than the atom and sealed off from the chemistry and physics happening among the outer electrons. Heating a sample to thousands of degrees, freezing it, squeezing it, or locking it into any chemical compound leaves the decay constant $\lambda$ — and hence the half-life — unchanged. This is exactly why radiometric dating is trustworthy: a buried, weathered, chemically altered artefact still keeps perfect time. (There are a few exotic exceptions, such as tiny shifts in decays that proceed by capturing an inner electron, but these are far outside school-exam scope.)