Half-Life & Mean Life

The last article handed you the radioactive decay law, $N = N_0 e^{-\lambda t}$ : a smooth curve sliding toward zero, governed by one number per nuclide, the decay constant $\lambda$ . But a decay constant of " $1.21\times10^{-4}$ per year" is a hard thing to feel. What a chemist, an archaeologist, a doctor injecting a tracer, or an examiner setting a numerical all actually reach for is a time — a stopwatch reading that says how fast this stuff disappears. That is what this article builds: two clocks, both hidden inside the same exponential. The first clock is the half-life $T_{1/2}$ , the time for any sample to fall to half of what it was. It is the single most useful number in all of radioactivity, precisely because it is the same no matter when you start counting: a fresh sample halves in one half-life, and so does a sample that is already a thousand years old. The second clock is the mean life $\tau$ , the average time an individual nucleus survives before it decays. Students meet both in the same breath and promptly confuse them — including the surprisingly common belief that the average lifetime must be shorter than the half-life. It is longer, by exactly the factor $1.44$ , and understanding why is understanding the whole shape of the curve. By the end you will be able to turn $\lambda$ into $T_{1/2}$ into $\tau$ and back with two constants and no panic; read the number remaining straight off a decay curve as $N_0/2^n$ ; and avoid the mark-losing trap that after two half-lives "it's all gone." This is the article every decay-law numerical secretly depends on.

Key takeaways

Half-life $T_{1/2}$ is the time for a sample to fall to half: set $N_0 e^{-\lambda T_{1/2}} = N_0/2$ and out drops $T_{1/2} = \ln 2/\lambda = 0.693/\lambda$ .
After $n$ half-lives the amount is $N = N_0/2^{\,n}$ : one half-life leaves $50\%$ , two leave $25\%$ , three leave $12.5\%$ — each half-life halves what remains, so the sample never quite reaches zero.
Mean life $\tau$ is the average time a nucleus survives, and equals the reciprocal of the decay constant, $\tau = 1/\lambda$ .
The link: $T_{1/2} = 0.693\,\tau$ , so $\tau = 1.44\,T_{1/2}$ — the mean life is always longer than the half-life, and after one mean life a fraction $1/e = 0.368$ (about $37\%$ ) is still left, not half.
Half-life is a property of the nuclide alone — independent of temperature, pressure, chemical bonding, and of how old the sample already is. A carbon-14 nucleus has no memory.
Activity halves in step with number: $A = \lambda N$ , so $A = A_0/2^{\,n}$ too — you can read the decay off a Geiger counter without ever counting atoms.
Key idea: half-life and mean life are just two convenient ways to name the single decay constant $\lambda$ — $T_{1/2} = 0.693/\lambda$ and $\tau = 1/\lambda$ — and both describe the same exponential fall, in which each half-life removes half of whatever is left.

The core insight

Every radioactive nuclide is characterised by one decay constant $\lambda$ , and its two everyday time-scales are fixed by it: the half-life $T_{1/2} = \ln 2/\lambda = 0.693/\lambda$ (the time to fall by half, after which $N = N_0/2^{\,n}$ over $n$ half-lives) and the mean life $\tau = 1/\lambda = 1.44\,T_{1/2}$ (the average lifetime of a single nucleus), both independent of the amount present or the sample's age.

Where this lives today

The half-life is one of the most widely used numbers in applied science, precisely because it is a fixed, portable property of each nuclide. In medicine it governs everything: technetium-99m, the workhorse of diagnostic imaging, has a half-life of about six hours — long enough to scan a patient, short enough to clear the body quickly — while a therapeutic or tracer dose is timed against its own decay clock so the radiation delivered is known to the becquerel. In archaeology and geology, half-lives are calendars: carbon-14 ( $T_{1/2} = 5730$ years) dates organic remains over tens of millennia, while uranium and potassium isotopes with half-lives of billions of years date rocks and the age of the Earth itself — the quantitative dating machinery is the very next thing you will build in the archetype problem set. Nuclear power and waste management live and die by half-lives: the fierce, short-lived fission products dominate a reactor's early heat, while a handful of long-lived isotopes set the timescale over which spent fuel must be isolated. Even smoke detectors use the steady, slow decay of americium-241 (half-life about $430$ years), effectively constant over the detector's whole service life. In every case the logic of this article — one decay constant, two clocks, an exponential that halves and halves again — is doing the work.

The physical picture

Radioactive decay is the purest gambling in nature. Take a lump of a radioactive nuclide — say carbon-14. Every one of its nuclei carries an identical, fixed probability per second of decaying, the decay constant $\lambda$ , and each one decides independently, like a coin that never remembers its past flips. There is no ageing, no wearing out, no "due to go soon." A carbon-14 nucleus that has already sat quietly for five thousand years is exactly as likely to decay in the next second as one made this morning. This memorylessness is the deep reason the decay is exponential, and it is what makes the half-life such a clean, portable number. Because each nucleus is a fair, independent coin, a huge collection of them thins out in a completely predictable way even though no single nucleus is predictable. In equal successive time intervals the population falls by equal fractions, not equal amounts. Pick the interval that happens to remove exactly half — that interval is the half-life — and the same interval later will remove half of the survivors, and half again after that. The curve therefore has the signature staircase: $N_0 \to N_0/2 \to N_0/4 \to N_0/8$ , one step per half-life, forever approaching but never touching zero.

Figure

Radioactive decay curve showing the number remaining falling to one half after each half-life so that after one two and three half-lives the fractions are one half one quarter and one eighth while the mean life sits at about 1.44 half-lives where a fraction one over e of about 0.37 survives

The two clocks are marked on that curve. The half-life $T_{1/2}$ is where the red curve crosses $N_0/2$ — one unit along the axis. The mean life $\tau$ is the gold line a little further out, at $1.44$ half-lives, where the curve has fallen not to a half but to $1/e \approx 0.368$ of the start. Read that off the picture and the single most common misconception dies on the spot: the average nucleus lives longer than one half-life, because the long, thin tail of late survivors drags the average out past the point where half the population has already gone.

Half-life: the time to fall by half

Start from the two results the previous article established and this one takes as given. The number of surviving nuclei follows the decay law, and the activity $A$ (decays per second, what a detector actually registers) is proportional to that number:

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

where: $N_0$ is the number of undecayed nuclei at $t=0$ , $N$ the number left at time $t$ , $\lambda$ the decay constant (probability per unit time that any one nucleus decays, units $\text{s}^{-1}$ or $\text{yr}^{-1}$ ), $A_0 = \lambda N_0$ the initial activity and $A$ the activity at time $t$ , measured in becquerel ( $1\ \text{Bq} = 1$ decay per second). The physical reading is that $N$ and $A$ are the same falling exponential up to the constant factor $\lambda$ — so anything true of the number of nuclei is equally true of the count rate on a Geiger tube, which is why we can study decay without ever isolating a single atom. The half-life is defined as the time $T_{1/2}$ at which $N$ has dropped to $N_0/2$ . Impose that condition on the decay law and the definition turns into a formula:

N_0\,e^{-\lambda T_{1/2}} = \frac{N_0}{2} \;\Longrightarrow\; e^{-\lambda T_{1/2}} = \tfrac12 \;\Longrightarrow\; T_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{\lambda}. \qquad (2)

where: $\ln 2 = 0.6931$ is the natural logarithm of two, the only new constant the half-life needs. The physical reading is a clean inverse relationship: a large $\lambda$ (a nucleus eager to decay) means a short half-life, and a tiny $\lambda$ means a long one. Nothing in equation (2) mentions $N_0$ — the half-life is completely independent of how much material you start with, which is exactly what lets one tabulated number ( $T_{1/2}$ ) characterise a nuclide for every experiment ever done on it. Because each half-life removes half of the survivors, after a whole number $n$ of half-lives the bookkeeping is pure powers of two, and for a general (possibly fractional) time it is a base-two exponential:

N = \frac{N_0}{2^{\,n}} \quad (n = t/T_{1/2}\ \text{half-lives}), \qquad N = N_0\left(\tfrac12\right)^{t/T_{1/2}} = N_0\,2^{-t/T_{1/2}}. \qquad (3)

where: $n = t/T_{1/2}$ is the number of half-lives elapsed. The two forms in (3) are one formula: for $n = 1, 2, 3, \dots$ the fraction remaining is $\tfrac12, \tfrac14, \tfrac18, \dots$ ; for a non-round time you use the continuous version $2^{-t/T_{1/2}}$ . And this base-two form is genuinely identical to the base- $e$ decay law, because $2^{-t/T_{1/2}} = e^{-(t/T_{1/2})\ln 2} = e^{-\lambda t}$ once you insert $\lambda = \ln 2/T_{1/2}$ — the two are the same curve wearing different clothes.

Figure

Bar chart of the fraction of a sample remaining after successive half-lives falling from 100 percent to 50 then 25 then 12.5 then 6.25 then 3.125 percent showing that each half-life halves what is left and the total never reaches zero

Mean life: the average a nucleus lives

The half-life answers "how long until half are gone." A different, equally natural question is "how long does a typical nucleus last?" — the average of the survival times of all the nuclei in the sample. That average is the mean life (or lifetime) $\tau$ . Some nuclei decay almost at once; a few linger far out in the tail; the mean life is the balance point of that whole distribution. It comes out beautifully simple:

\tau = \frac{1}{\lambda}. \qquad (4)

where: $\lambda$ is again the decay constant. The physical reading is that if a nucleus has a probability $\lambda$ of decaying per unit time, its expected time-to-decay is just the reciprocal $1/\lambda$ — the same reason a fair die with a $1/6$ chance per roll takes six rolls on average. (Pinning that average down rigorously means integrating the survival time $t$ against the decay distribution $\lambda e^{-\lambda t}$ , a calculus step that sits above this article's algebra-and-trig ceiling; we simply quote the clean result $\tau = 1/\lambda$ , which the higher boards that work with the decay constant — CBSE, ICSE, JEE, NEET, IB, and A-Level — treat as a standard formula.) Now put the two clocks side by side. Both are built from the one constant $\lambda$ , so they are rigidly linked:

T_{1/2} = \frac{\ln 2}{\lambda} = \ln 2 \cdot \tau = 0.693\,\tau, \qquad \tau = \frac{T_{1/2}}{0.693} = 1.44\,T_{1/2}. \qquad (5)

where: the factor is $\ln 2 = 0.693$ one way and $1/\ln 2 = 1.44$ the other. The physical reading is the fact that trips up half of all students: the mean life is longer than the half-life, by the fixed factor $1.44$ . It has to be — by the time half the sample has already decayed (one half-life), the survivors still waiting their turn pull the average survival time out past that mark. Concretely, evaluate the decay law at $t = \tau$ : $N = N_0 e^{-\lambda \tau} = N_0 e^{-1} = 0.368\,N_0$ . So after one mean life about $37\%$ of the sample survives — not $50\%$ . One mean life is the moment the sample falls to $1/e$ of its start, just as one half-life is the moment it falls to $1/2$ .

Derivation ADeriving the half-life T½ = 0.693/λ from the decay law

The goal is to obtain the half-life formula (2) directly from the radioactive decay law $N = N_0 e^{-\lambda t}$ quoted from the previous article, using nothing beyond the definition of the half-life and the ordinary algebra of logarithms — no calculus, since the decay law itself is taken as given. The strategy is to write the definition of the half-life as an equation and solve it for the time.

Step 1. State the definition as an equation. The half-life $T_{1/2}$ is the instant at which the surviving number has fallen to one half of its starting value, so substitute $t = T_{1/2}$ and $N = N_0/2$ into the decay law: $\;N_0/2 = N_0\,e^{-\lambda T_{1/2}}$ .

Step 2. The starting number $N_0$ appears on both sides and cancels — the first sign that the answer will not depend on how much material we began with. Dividing through gives the compact condition $e^{-\lambda T_{1/2}} = \tfrac12$ .

Step 3. Take the natural logarithm of both sides to bring the exponent down: $-\lambda T_{1/2} = \ln(\tfrac12) = -\ln 2$ . The minus signs match, leaving $\lambda T_{1/2} = \ln 2$ .

Step 4. Solve for the half-life: $T_{1/2} = \ln 2/\lambda = 0.693/\lambda$ , which is equation (2). As a check on a real nuclide, carbon-14 has $\lambda = 1.21\times10^{-4}\ \text{yr}^{-1}$ , so $T_{1/2} = 0.693/(1.21\times10^{-4}) = 5730\ \text{yr}$ — the tabulated value, recovered exactly. Because only $\lambda$ enters, the half-life is an intrinsic property of the nuclide, the same in a gram or a microgram, today or a century hence. Verified: $T_{1/2} = \ln 2/\lambda$ with $\lambda = 1.21\times10^{-4}\ \text{yr}^{-1}$ gives $5730\ \text{yr}$ ; $\ln 2 = 0.693$ — recomputed in verify_hs2-half-life-mean-life.py.

Go deeper · Derivation B · 1 minAfter n half-lives the fraction is 1/2 to the n, and how half-life links to mean life

Derivation BAfter n half-lives the fraction is 1/2 to the n, and how half-life links to mean life

The goal is to establish the two working relations most numericals lean on: the powers-of-two rule $N = N_0/2^{\,n}$ after $n$ half-lives (equation 3), and the bridge $T_{1/2} = 0.693\,\tau$ between the two clocks (equation 5), starting from the decay law and the definitions of $T_{1/2}$ and $\tau$ , with pure algebra. The strategy is to iterate the halving, then eliminate $\lambda$ between the two time-scales.

Step 1. One half-life takes $N_0$ to $N_0/2$ , by definition. Because the process is memoryless, the next half-life acts on the survivors identically, taking $N_0/2$ to $N_0/4$ ; the third takes it to $N_0/8$ . After $n$ such steps the sample has been halved $n$ times over, so $N = N_0 \cdot (\tfrac12)^n = N_0/2^{\,n}$ . For $n = 1, 2, 3$ this is the fraction $0.5,\ 0.25,\ 0.125$ — that is, $50\%,\ 25\%,\ 12.5\%$ remaining.

Step 2. Confirm the powers-of-two rule is just the decay law re-dressed. Write $2^{-t/T_{1/2}} = e^{-(t/T_{1/2})\ln 2}$ (since $2 = e^{\ln 2}$ ), and substitute $\ln 2 = \lambda T_{1/2}$ from equation (2): the exponent becomes $-(t/T_{1/2})(\lambda T_{1/2}) = -\lambda t$ . Hence $N_0\,2^{-t/T_{1/2}} = N_0 e^{-\lambda t}$ — identical to the decay law, so equation (3) is exact, not an approximation.

Step 3. Now relate the clocks. The mean life is $\tau = 1/\lambda$ (equation 4), so $\lambda = 1/\tau$ . Put this into the half-life formula (2): $T_{1/2} = \ln 2/\lambda = \ln 2 \cdot \tau = 0.693\,\tau$ . Inverting, $\tau = T_{1/2}/0.693 = 1.44\,T_{1/2}$ , so the mean life exceeds the half-life by the fixed factor $1.44$ .

Step 4. Check the two clocks against each other on carbon-14: with $T_{1/2} = 5730\ \text{yr}$ , the mean life is $\tau = 1.44\times5730 = 8267\ \text{yr}$ , equal to $1/\lambda = 1/(1.21\times10^{-4}) = 8267\ \text{yr}$ from the other road — the two definitions agree. And at $t = \tau$ the surviving fraction is $e^{-\lambda\tau} = e^{-1} = 0.368$ , so about $37\%$ is left after one mean life, visibly more than the $25\%$ left after two half-lives ( $2\tau/T_{1/2} \approx 2.9$ half-lives would be needed to reach that). Verified: fractions $N_0/2^n = 0.5,\,0.25,\,0.125$ ; $2^{-t/T_{1/2}} \equiv e^{-\lambda t}$ ; $\tau = 1.44\,T_{1/2} = 8267\ \text{yr}$ for C-14; $e^{-1} = 0.368$ — recomputed in verify_hs2-half-life-mean-life.py.

Limiting cases

case	result	meaning
$t = 0$	$N = N_0$ , $A = A_0$	the starting point — nothing has decayed yet
$t = T_{1/2}$	$N = N_0/2$ (50%)	one half-life: exactly half remain, by definition
$t = \tau = 1.44\,T_{1/2}$	$N = N_0/e = 0.368\,N_0$ (37%)	one mean life leaves $1/e$ , more than a half-life would
$n$ whole half-lives	$N = N_0/2^{\,n}$	50, 25, 12.5, 6.25 % for $n = 1,2,3,4$
$t \to \infty$	$N \to 0$	exponential tail — the sample never exactly empties
large $\lambda$ (short $T_{1/2}$ )	fast decay	eager nuclide, gone quickly; $T_{1/2}$ small
small $\lambda$ (long $T_{1/2}$ )	slow decay	e.g. U-238, $T_{1/2} = 4.5\times10^{9}\ \text{yr}$

Common trapHalf-life is not "half the lifetime," and after two half-lives the sample is not gone

Two linked errors live here. First, the half-life is not half of some total lifetime, and the mean life is not half the half-life — the mean life $\tau = 1.44\,T_{1/2}$ is actually longer than the half-life, because the slow survivors in the tail pull the average out. Second, and even more common in exams: after two half-lives a sample is not empty. Each half-life halves what is left, so two half-lives leave a quarter, three leave an eighth, and the amount only ever approaches zero, never reaching it. The halving is multiplicative, not additive — you never add "half plus half" to make a whole. Anchor both facts with the picture: at one mean life ( $1.44\,T_{1/2}$ ) a full $37\%$ still survives.

Worked examples

Board level — half-life, decay constant and mean life of a sample (CBSE and ICSE pattern)

A radioactive sample has a half-life of $20$ minutes. Find (a) its decay constant $\lambda$ , (b) its mean life $\tau$ , and (c) the fraction of the sample remaining after $1$ hour. If the initial activity was $8000\ \text{Bq}$ , (d) what is the activity after $1$ hour?

Step 1. The decay constant follows from the half-life, equation (2): $\lambda = 0.693/T_{1/2} = 0.693/20 = \boxed{0.0347\ \text{min}^{-1}}$ . (Keep the time unit consistent — here everything is in minutes.)

Step 2. The mean life is the reciprocal, equation (4): $\tau = 1/\lambda = 1/0.0347 = \boxed{28.9\ \text{min}}$ . Cross-check with equation (5): $\tau = 1.44\,T_{1/2} = 1.44\times20 \approx 29\ \text{min}$ — the same, and comfortably longer than the $20$ -minute half-life.

Step 3. One hour is $60$ minutes $= 60/20 = 3$ half-lives, so use the powers-of-two rule (3): $N/N_0 = 1/2^{3} = 1/8 = \boxed{0.125}$ , that is $12.5\%$ remaining.

Step 4. Activity halves in step with the number of nuclei (equation 1), so after $3$ half-lives $A = A_0/2^{3} = 8000/8 = \boxed{1000\ \text{Bq}}$ . The count rate has dropped to one eighth exactly as the population did.

Board level — reading a decay from the count rate (CBSE and A-Level pattern)

A Geiger counter reads $8000$ counts per minute from a source, and $80$ minutes later it reads $500$ counts per minute. Find the half-life of the source.

Step 1. Work in ratios, since activity tracks number: $A_0/A = 8000/500 = 16$ . Because $16 = 2^{4}$ , the activity has been halved $4$ times, so $4$ half-lives have elapsed.

Step 2. Those $4$ half-lives took $80$ minutes, so $T_{1/2} = 80/4 = \boxed{20\ \text{min}}$ . No exponential arithmetic is needed once you spot that the ratio is a clean power of two — reading the number of doublings straight off is the fastest route on a timed paper.

Competitive level — time for a chosen fraction to remain (JEE and NEET pattern)

A radioactive nuclide has a half-life of $20$ minutes. How long must you wait for (a) $90\%$ of it to decay, leaving $10\%$ , and (b) only $1\%$ to remain?

Step 1. Here the fraction is not a round power of two, so use logarithms on the continuous law (3): $N/N_0 = 2^{-t/T_{1/2}}$ . For $10\%$ remaining, $0.10 = 2^{-t/T_{1/2}}$ , so $t/T_{1/2} = \log_2(10) = \ln 10/\ln 2 = 2.303/0.693 = 3.32$ half-lives.

Step 2. Convert to minutes: $t = 3.32\times20 = \boxed{66.4\ \text{min}}$ . So $90\%$ decays in about three and a third half-lives — noticeably more than three, because $10\%$ is a little below the $12.5\%$ that three half-lives leave.

Step 3. For $1\%$ remaining, $0.01 = 2^{-t/T_{1/2}}$ gives $t/T_{1/2} = \log_2(100) = 2\log_2(10) = 6.64$ half-lives, so $t = 6.64\times20 = \boxed{132.9\ \text{min}}$ . Note the tidy structure: dropping by another factor of ten always costs the same $3.32$ half-lives, because the decay is exponential. Timing note: about a minute if you carry $\log_2(10) = 3.32$ as a remembered constant (it is $1/\log_{10}2$ ). Then " $10\%$ left" is instantly $3.32\,T_{1/2}$ and " $1\%$ left" is $6.64\,T_{1/2}$ — no calculator gymnastics. Reaching for $t = \tau\ln(N_0/N)$ instead also works ( $\tau = 28.9$ min, $t = \tau\ln 10 \approx 66$ min), but the powers-of-ten shortcut is faster under exam pressure.

Competitive level — converting between the two clocks (JEE Advanced and IB HL pattern)

A radioisotope used in a lab has a half-life of $10$ days. Find (a) its mean life, and the percentage remaining after (b) $20$ days and (c) $30$ days. A student claims "the mean life is $5$ days, since it is half of the half-life." Correct them.

Step 1. The mean life is longer than the half-life: $\tau = 1.44\,T_{1/2} = 1.44\times10 = \boxed{14.4\ \text{days}}$ (equivalently $\tau = 1/\lambda$ with $\lambda = 0.693/10 = 0.0693\ \text{day}^{-1}$ ). The student's " $5$ days" is exactly the trap this article warns against — the mean life is never half the half-life; it is about $1.44$ times it.

Step 2. $20$ days $= 2$ half-lives, so $N/N_0 = 1/2^{2} = 1/4 = \boxed{25\%}$ . Cross-check with the base- $e$ law: $e^{-\lambda t} = e^{-0.0693\times20} = e^{-1.386} = 0.250$ — the same $25\%$ , confirming the two forms agree.

Step 3. $30$ days $= 3$ half-lives, so $N/N_0 = 1/2^{3} = 1/8 = \boxed{12.5\%}$ . Each extra half-life (here each extra $10$ days) simply multiplies the survivor fraction by another one half. Timing note: under $45$ seconds. Whenever the elapsed time is a whole-number multiple of the half-life, skip the exponential entirely and just halve — $25\%$ and $12.5\%$ are read straight off " $2$ half-lives" and " $3$ half-lives." Only reach for $e^{-\lambda t}$ or $\log_2$ when the multiple is not whole.

Exam traps

"After two half-lives it's all gone." The single biggest error. Two half-lives leave a quarter, not nothing — each half-life halves the remainder. The sample only approaches zero asymptotically. Examiners on CBSE, NEET, ICSE and IGCSE bait this every year with "after $2$ (or $3$ ) half-lives, what fraction is left?" — answer $1/4$ (or $1/8$ ).

Mean life shorter than half-life. Students routinely write $\tau = T_{1/2}/2$ or "the average life is less than the half-life." It is the reverse: $\tau = 1.44\,T_{1/2}$ , longer than the half-life, because the long tail of survivors pulls the average out. Memorise the pair: $T_{1/2} = 0.693\,\tau$ and $\tau = 1.44\,T_{1/2}$ .

Mismatched time units. $\lambda$ and $T_{1/2}$ carry reciprocal units. If $T_{1/2}$ is in minutes then $\lambda$ is per minute, and any $t$ in the exponential must be in minutes too. Mixing minutes and seconds inside $e^{-\lambda t}$ is a classic silent slip — always convert first.

Using $N_0/2^{\,n}$ for non-integer $n$ without care. The rule $N = N_0/2^{\,n}$ is cleanest for whole $n$ . For a fraction of a half-life you must use the continuous form $N_0\,2^{-t/T_{1/2}}$ or $N_0 e^{-\lambda t}$ — you cannot just "take a fraction of a halving" linearly.

Thinking the half-life depends on the amount, age, or conditions. It does not: $T_{1/2}$ has $N_0$ cancelled out of it, and radioactive decay is a nuclear process untouched by temperature, pressure or chemistry. An old sample halves in the same time as a fresh one — the nucleus is memoryless.

Board scope. The qualitative half-life — its definition, reading it off a decay graph, and $N_0/2^{\,n}$ for whole numbers of half-lives — is on every board including IGCSE 0625. But the quantitative apparatus of this article — $T_{1/2} = 0.693/\lambda$ , the mean life $\tau = 1/\lambda$ , and the exponential/logarithmic handling of non-integer times — is for CBSE, ICSE (ISC), JEE Main, JEE Advanced, NEET, IB SL and HL, and A-Level 9702. AP Physics 2 carries only the basic half-life / decay-constant relation and the $N_0/2^{\,n}$ halving for whole numbers of half-lives; the mean life and the exponential/logarithmic manipulation of non-integer times sit above AP-2's core — alongside IGCSE 0625, which stops at the qualitative level. Do not quote $\tau$ where only $T_{1/2}$ is asked.

Origins · 1 minMarie Curie and how this was discovered

Marie Curie

1867–1934 · Warsaw and Paris

Marie Curie, born Maria Skłodowska in Warsaw, was the physicist and chemist who turned radioactivity from a curiosity into a measurable science — and who coined the very word radioactivity. Working in Paris with her husband Pierre, she discovered two new radioactive elements in 1898, polonium (named for her native Poland) and radium — isolating radium as a pure chloride salt only in 1902 — by painstakingly measuring the activity of tonnes of pitchblende: her insight was that the intensity of the rays was a fingerprint of the element itself, a fixed rate of emission per gram. That idea — a characteristic decay rate for each substance — is the seed of the half-life and mean life in this article. She shared the 1903 Nobel Prize in Physics with Pierre Curie and Henri Becquerel, then won the 1911 Nobel Prize in Chemistry alone for radium and polonium, the first person ever to win two. The unit of activity, the curie, honours the family; radium's own half-life is about $1600$ years.

History

1896

— Henri Becquerel discovers that uranium salts emit penetrating rays without any external cause, the first sign of radioactivity.

1898

— Marie and Pierre Curie discover polonium and radium and Marie names the phenomenon radioactivity, measuring each element by its rate of emission (radium isolated as a pure salt only in 1902).

1900

— Ernest Rutherford, studying a radioactive gas from thorium, finds its activity falls by half in a fixed time and introduces the idea of a half-value period.

1902

— Rutherford and Frederick Soddy formulate the exponential law of radioactive transformation, from which the half-life follows.

1903

— Marie Curie, Pierre Curie and Henri Becquerel share the Nobel Prize in Physics for their work on radioactivity.

1911

— Marie Curie receives a second Nobel Prize, in Chemistry, for the discovery of radium and polonium.

Going further

Texts. NCERT Physics Class 12, Chapter 13 ("Nuclei"), the radioactivity and half-life sections · H. C. Verma, Concepts of Physics Vol. 2, "The Nucleus" · Irodov, Problems in General Physics, the radioactive-decay problems · Krane, Introductory Nuclear Physics, for the statistical basis of decay.

Provenance. Standard radioactivity constants; carbon-14 anchor $T_{1/2} = 5730\ \text{yr}$ . Recompute: $\ln 2 = 0.693$ ; $1/\ln 2 = 1.44$ ; C-14 $\lambda = 0.693/5730 = 1.21\times10^{-4}\ \text{yr}^{-1}$ , $\tau = 1/\lambda = 8267\ \text{yr} = 1.44\,T_{1/2}$ ; fraction after one mean life $e^{-1} = 0.368$ ; powers-of-two fractions $0.5,\ 0.25,\ 0.125,\ 0.0625,\ 0.03125$ (50, 25, 12.5, 6.25, 3.125 %); board example $T_{1/2} = 20$ min $\Rightarrow \lambda = 0.0347\ \text{min}^{-1}$ , $\tau = 28.9$ min, after $60$ min ( $3$ half-lives) $12.5\%$ and activity $8000 \to 1000$ Bq; ratio $8000/500 = 16 = 2^4 \Rightarrow 4$ half-lives $\Rightarrow T_{1/2} = 20$ min; $10\%$ left at $\log_2 10 = 3.32$ half-lives $= 66.4$ min, $1\%$ left at $6.64$ half-lives $= 132.9$ min; $T_{1/2} = 10$ d $\Rightarrow \tau = 14.4$ d, $25\%$ after $20$ d, $12.5\%$ after $30$ d; $e^{-20/14.4} = 0.250$ ; $1\ \text{Ci} = 3.7\times10^{10}\ \text{Bq}$ .

Prerequisites. The Radioactive Decay Law — where $N = N_0 e^{-\lambda t}$ and the activity $A = \lambda N$ come from, and which this article takes as given. Leads to. ARCHETYPE: Decay-Law, Half-Life & Dating Problems, the numerical gym that drills $\lambda$ , $T_{1/2}$ , $\tau$ , activity and carbon-dating together. For enrichment, the university seam The Quantum Survival Amplitude and Decay (go-deeper only, never an exam dependency here) reframes the decay probability as the modulus-squared of a quantum amplitude.

What exactly is the difference between half-life and mean life?

The half-life $T_{1/2}$ is the time for a whole sample to fall to half its current number — a property of the population. The mean life $\tau$ is the average survival time of a single nucleus — a property of individuals. They measure related but different things, and they are not equal: $\tau = 1.44\,T_{1/2}$ , so the mean life is always the longer of the two. The reason is that decay is exponential with a long tail; by the time half the nuclei have gone (one half-life), a substantial minority is still waiting, and those late survivors drag the average lifetime out beyond the half-life. Both are fixed by the one decay constant: $T_{1/2} = 0.693/\lambda$ and $\tau = 1/\lambda$ .

Why is the mean life longer than the half-life, not shorter?

It feels backwards, but it must be so. Imagine ranking every nucleus by how long it survives. The half-life is the median survival time — half decay sooner, half later. The mean life is the average. Whenever a distribution has a long tail stretching to large values (and the exponential decay tail runs to infinity), the average is pulled toward that tail and lands above the median. So $\tau > T_{1/2}$ . Numerically the factor is exactly $1/\ln 2 = 1.44$ . A quick sanity check: after one mean life ( $1.44$ half-lives) the surviving fraction is $1/e = 0.37$ , still more than a third — clearly the average nucleus outlives the half-life mark.

Does a sample ever completely disappear?

Mathematically, no. Each half-life removes half of what remains, so the count goes $N_0 \to N_0/2 \to N_0/4 \to N_0/8 \to \dots$ , always halving, forever approaching zero without reaching it — the exponential has an infinite tail. In practice, of course, a real sample contains a finite number of atoms, so after enough half-lives you are down to a handful and then the last few individual nuclei decay at random times, and the sample truly empties. But for the smooth exponential formula, and for every exam question phrased in fractions or percentages, the correct statement is that the amount tends to zero and is never exactly zero at any finite time.

Can I use N₀ over 2 to the n if the time is not a whole number of half-lives?

Only if you let $n$ be a fraction. The safe, always-correct move is the continuous form $N = N_0\,2^{-t/T_{1/2}}$ , which is identical to $N_0 e^{-\lambda t}$ . For example, after $1.5$ half-lives, $N/N_0 = 2^{-1.5} = 0.354$ , about $35\%$ — you cannot get this by "halving one and a half times" in any linear way. Reserve the tidy $N_0/2^{n}$ with whole $n$ ( $1/2, 1/4, 1/8, \dots$ ) for round numbers of half-lives, where it is fastest, and switch to logarithms or the exponential the moment the elapsed time is not a whole multiple of $T_{1/2}$ .

How is the half-life related to the activity I measure on a counter?

Directly and conveniently. The activity is $A = \lambda N$ , so it is proportional to the number of surviving nuclei — which means the activity decays with exactly the same half-life as the number does: $A = A_0/2^{\,n}$ after $n$ half-lives, and $A = A_0 e^{-\lambda t}$ in general. You therefore never need to count atoms. If a counter reads $8000$ per minute now and $2000$ per minute later, the activity has quartered, so two half-lives have passed. This is how half-lives are actually measured in the lab: plot the count rate against time (or its logarithm, which gives a straight line of slope $-\lambda$ ) and read off the time to halve.

Does heating or chemically bonding a radioactive material change its half-life?

For all practical purposes, no. Radioactive decay is a process of the nucleus, and the nucleus is a hundred-thousand times smaller than the atom and screened by a cloud of electrons from the outside world. Temperature, pressure, chemical bonding, and magnetic fields — the things that change electron behaviour and hence chemistry — leave the nucleus essentially untouched, so the half-life is a fixed constant of the nuclide. (Physicists can measure vanishingly tiny shifts in a few special decay modes that involve the innermost electrons, but these are far beyond school scope and never affect exam answers.) Treat the half-life as unchangeable: the same in ice or steam, in a diamond or a gas.

Why does the number 0.693 keep appearing?

Because $0.693$ is just $\ln 2$ , the natural logarithm of two, and "two" enters the moment you ask for the time to fall by a factor of two. Setting $e^{-\lambda T_{1/2}} = 1/2$ and taking logs forces $\lambda T_{1/2} = \ln 2 = 0.693$ , hence $T_{1/2} = 0.693/\lambda$ . If instead you asked for the time to fall to $1/e$ you would get the mean life $\tau = 1/\lambda$ with no logarithm at all — that is why $\tau$ is the "natural" time-scale and $T_{1/2}$ carries the extra $\ln 2$ . The reciprocal, $1/0.693 = 1.44$ , is the factor that turns a half-life into a mean life. Memorise both: $0.693$ going from $\tau$ to $T_{1/2}$ , and $1.44$ going back.