High School Physics · Grades 10–12

32Radioactive Decay

In a museum basement, a Geiger counter clicks over a sliver of wood from a pharaoh’s coffin. Each click is lawless: nothing announced it, nothing predicts the next. Yet by Friday the lab will print a date good to a century. This chapter resolves the paradox — how events perfectly random one by one add up, by the trillion, to the steadiest clocks we own.

32.1 Lawless one by one, clockwork by the mole

Last year (Chapter 19) we met the unstable nuclei, balanced their $\alpha$ and $\beta$ equations, and saw half of any large sample vanish per half-life $T_{1/2}$ . What we could not say was when between the stairs — nor why a random process keeps a schedule at all. Both questions have one answer.

Definition 32.1 (Decay constant)

For each unstable nuclide there is a constant $\lambda$ , its decay constant (unit $\mathrm{s}^{-1}$ ), such that during any short interval $\dd t$ each nucleus has probability $p = \lambda\,\dd t$ of decaying — the same for every nucleus of the species, whatever its age, and untouched by temperature, pressure or chemistry.

Remark 32.2 (Lawless singly, exact in crowds)

A nucleus that has waited ten thousand years is exactly as likely to decay this second as one made this morning: no wearing out, no warning — which one goes next is genuinely unpredictable. Crowds are another matter. Toss $100$ coins: expect $50$ heads, but be unsurprised by $43$ — fluctuations of order $\sqrt n$ , here $10\%$ . Toss $10^{20}$ : the relative fluctuation $1/\sqrt n$ is $10^{-10}$ , the outcome certain to ten decimals. A gram of matter holds some $10^{22}$ nuclei: lawlessness averages into clockwork, an exact law for $N(t)$ , the number of nuclei still intact at time $t$ .

Two real samples of 20 nuclei stagger down unpredictably around the expected curve; a mole of nuclei would hug it to ten decimal places. — Two real samples of $20$ nuclei stagger down unpredictably around the expected curve; a mole of nuclei would hug it to ten decimal places.

32.2 The decay law

Proposition 32.3 (The evolution equation)

In a large sample of $N$ nuclei, during $\dd t$ the number changes by

\dd N = -\lambda N\,\dd t, \qquad\text{i.e.}\qquad \frac{\dd N}{\dd t} = -\lambda N :

the population shrinks at a rate proportional to itself.

Proof. Each of the $N$ nuclei decays during $\dd t$ with probability $\lambda\,\dd t$ : expected decays $N\lambda\,\dd t$ , and for large $N$ the actual count sticks to the expected one (Remark 32.2); each decay removes one nucleus. ∎

Theorem 32.4 (Law of radioactive decay)

A sample holding $N_0$ nuclei at $t = 0$ holds, at time $t$ ,

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

Proof. Check by substitution: $\dd N/\dd t = N_0(-\lambda)\,e^{-\lambda t} = -\lambda N(t)$ , and $N(0) = N_0 e^0 = N_0$ : equation and start both right. (That no other function fits is proved, with integrals, in the university volume.) ∎

Definition 32.5 (Exponential decay)

A quantity following $N_0\,e^{-\lambda t}$ undergoes exponential decay: in equal times, equal fractions are lost. It is exactly the discharging capacitor of Chapter 30, with $\lambda$ playing $1/(RC)$ : one differential equation, learned once, runs circuits and nuclei alike.

Proposition 32.6 (Half-life)

The half-life (Definition 19.14) of a nuclide is set by its decay constant: $T_{1/2} = \ln 2/\lambda$ , and after $n$ half-lives $N = N_0/2^{\,n}$ , whole $n$ or not.

Proof. $N(T_{1/2}) = N_0/2$ requires $e^{-\lambda T_{1/2}} = \tfrac12$ , i.e. $\lambda T_{1/2} = \ln 2$ ; and $N(n\,T_{1/2}) = N_0\,e^{-n\ln 2} = N_0/2^{\,n}$ for any $n$ : the exponential threads the staircase and fills in between the stairs. ∎

N(t) = N_0\,e- t passes through every stair of the staircase — and now says what happens in between. — $N(t) = N_0\,e^{-\lambda t}$ passes through every stair of the staircase — and now says what happens in between.

Example 32.7 (Twenty powers of ten)

With $1$ year $= 3.156 \times 10^{7}\,\mathrm{s}$ :

nuclide	$T_{1/2}$	$\lambda$ ( $\mathrm{s}^{-1}$ )
polonium-214	$1.6 \times 10^{-4}\,\mathrm{s}$	$4.2 \times 10^{3}$	uranium chain link
technetium-99m	$6.0$ hours	$3.2 \times 10^{-5}$	medical imaging
iodine-131	$8.0$ days	$1.0 \times 10^{-6}$	thyroid medicine
caesium-137	$30$ years	$7.3 \times 10^{-10}$	reactor fallout
carbon-14	$5730$ years	$3.8 \times 10^{-12}$	archaeology
potassium-40	$1.25 \times 10^{9}$ years	$1.8 \times 10^{-17}$	rock dating
uranium-238	$4.5 \times 10^{9}$ years	$4.9 \times 10^{-18}$	Earth’s inner heat

One law, twenty powers of ten of $\lambda$ .

32.3 Activity: counting the clicks

Proposition 32.8 (Activity)

The activity of a sample — its decays per second, in becquerels (Definition 19.18) — is

\mathcal A = \lambda N, \qquad\text{so}\qquad \mathcal A(t) = \mathcal A_0 \, e^{-\lambda t}:

proportional to the surviving population, and equally exponential.

Proof. $N$ nuclei, each with probability $\lambda$ per second: $\lambda N$ decays per second; multiplying Theorem 32.4 by $\lambda$ gives $\mathcal A(t) = \mathcal A_0 e^{-\lambda t}$ . ∎

Example 32.9 (Huge $N$ , tiny $\lambda$ )

A banana ticks at about $15\,\mathrm{Bq}$ , a human body at about $8000\,\mathrm{Bq}$ , a cubic metre of granite at some $10^{6}\,\mathrm{Bq}$ (Chapter 19). Behind the modest clicks sit astronomical populations: the body’s potassium-40 contributes $\mathcal A \approx 4.4 \times 10^{3}\,\mathrm{Bq}$ with $\lambda = 1.8 \times 10^{-17}\,\mathrm{s}^{-1}$ : $N = \mathcal A/\lambda \approx 2.5 \times 10^{20}$ nuclei, of which barely a few thousand fire each second.

Method 32.10 (Measuring a half-life)

Count decays with a Geiger counter over successive equal intervals: this samples $\mathcal A$ at successive times.
Plot $\ln \mathcal A$ against $t$ : since $\ln \mathcal A = \ln \mathcal A_0 - \lambda t$ , exponential decay shows as a straight line — crooked data means another law, or a mixture of nuclides.
Read $\lambda$ off as minus the slope; $T_{1/2} = \ln 2/\lambda$ .

Geiger counts of a short-lived nuclide: on semi-log axes the exponential becomes a straight line; here the slope gives 9.6 × 10-3\, s-1, so T_1/2 72\, s. — Geiger counts of a short-lived nuclide: on semi-log axes the exponential becomes a straight line; here the slope gives $\lambda \approx 9.6 \times 10^{-3}\,\mathrm{s}^{-1}$ , so $T_{1/2} \approx 72\,\mathrm{s}$ .

Remark 32.11 (Becquerels are not harm)

The becquerel counts decays, not damage. Harm is tracked by the dose in sieverts (Chapter 19), weighing the energy deposited per kilogram of tissue by each radiation’s destructiveness: a $5 \times 10^{8}\,\mathrm{Bq}$ technetium scan is routine, far fewer becquerels of inhaled $\alpha$ emitter are not; converting $\mathrm{Bq}$ to $\mathrm{Sv}$ is university material.

32.4 Dating: reading time off a ratio

The decay law run backwards is a clock: measure the surviving fraction, and the exponential names the elapsed time.

Method 32.12 (Radiocarbon dating)

Alive: living tissue exchanges carbon with the atmosphere and carries its carbon-14 proportion — one atom in $10^{12}$ , giving $\mathcal A_0 = 13.6$ decays per minute per gram of carbon.
Death starts the clock: intake stops, and the proportion decays with $T_{1/2} = 5730$ years.
Measure the sample’s activity $\mathcal A$ per gram of carbon; $\mathcal A = \mathcal A_0 e^{-\lambda t}$ inverts to $t = \frac{1}{\lambda}\ln(\mathcal A_0/\mathcal A) = T_{1/2}\,\ln(\mathcal A_0/\mathcal A)/\ln 2$ .

Example 32.13 (A hearth in a cave)

Charcoal from a buried hearth gives $5.0$ decays per minute per gram: $\mathcal A_0/\mathcal A = 13.6/5.0 = 2.72$ , so $t = 5730 \times \ln(2.72)/\ln 2 \approx 8.3 \times 10^{3}$ years. The fire went out eight thousand years ago — the wood itself is the archive.

The dating curve: enter at the measured ratio, read down to the age. The dot is the hearth of : ratio 0.37, age 8.3 × 103 years. — The dating curve: enter at the measured ratio, read down to the age. The dot is the hearth of Example 32.13: ratio $0.37$ , age $8.3 \times 10^{3}$ years.

Example 32.14 (Clocks for rocks)

Carbon-14 goes silent beyond some $50\,000$ years; geology keeps slower clocks. Potassium–argon: potassium-40 ( $T_{1/2} = 1.25 \times 10^{9}$ years) decays to argon, a gas that escapes molten lava but is trapped once rock crystallises — solidification zeroes the clock. Uranium–lead: uranium-238 heads a chain of decays ending at stable lead-206 (Chapter 19), and the lead-to-uranium ratio in a crystal dates it. Both clocks agree on the oldest minerals ( $\approx 4.4 \times 10^{9}$ years) and, via meteorites, on the Solar System’s age.

Remark 32.15 (Chains, and the gas in the cellar)

Between uranium and lead the chain passes through radium, then radon-222, a radioactive gas ( $T_{1/2} = 3.8$ days). Born in granite and soil, it seeps into basements and accumulates; its $\alpha$ -emitting daughters decay in the lungs — most people’s largest natural dose, and the cheapest to reduce: open the cellar window.

32.5 Exercises

Exercise 32.1 ★

Iodine-131 has $\lambda = 1.0 \times 10^{-6}\,\mathrm{s}^{-1}$ . Compute its half-life in seconds, then in days.

Solution

Solution of Exercise 32.1.

$T_{1/2} = \ln 2/\lambda = 0.693/1.0 \times 10^{-6} = 6.9 \times 10^{5}\,\mathrm{s} \approx 8.0$ days — iodine-131’s clock.

Exercise 32.2 ★

Radon-222 has $T_{1/2} = 3.8$ days. Compute $\lambda$ in $\mathrm{s}^{-1}$ , then the activity of a sample of $1.0 \times 10^{10}$ nuclei.

Solution

Solution of Exercise 32.2.

$T_{1/2} = 3.8 \times 86\,400 = 3.28 \times 10^{5}\,\mathrm{s}$ ; $\lambda = 0.693/3.28 \times 10^{5} = 2.1 \times 10^{-6}\,\mathrm{s}^{-1}$ ; $\mathcal A = \lambda N = 2.1 \times 10^{-6} \times 1.0 \times 10^{10} \approx 2.1 \times 10^{4}\,\mathrm{Bq}$ .

Exercise 32.3 ★

Verify by substitution that $N(t) = N_0\,e^{-\lambda t}$ satisfies $\dd N/\dd t = -\lambda N$ and $N(0) = N_0$ ; then compute $N(T_{1/2})/N_0$ .

Solution

Solution of Exercise 32.3.

$\dd N/\dd t = N_0(-\lambda)e^{-\lambda t} = -\lambda N$ — the equation holds; $N(0) = N_0 e^0 = N_0$ ; $N(T_{1/2})/N_0 = e^{-\lambda \ln 2/\lambda} = e^{-\ln 2} = \tfrac12$ .

Exercise 32.4 ★

A caesium-137 source ( $T_{1/2} = 30$ years) has activity $4.0 \times 10^{5}\,\mathrm{Bq}$ today. Compute its activity in $90$ years, then in $300$ years.

Solution

Solution of Exercise 32.4.

$90$ years $= 3\,T_{1/2}$ : $\mathcal A = 4.0 \times 10^{5}/2^3 = 5.0 \times 10^{4}\,\mathrm{Bq}$ . $300$ years $= 10\,T_{1/2}$ : $\mathcal A = 4.0 \times 10^{5}/1024 \approx 3.9 \times 10^{2}\,\mathrm{Bq}$ .

Exercise 32.5 ★

A semi-log plot of $\ln \mathcal A$ against $t$ for a technetium sample is a straight line through $(0, 20.00)$ and $(10\,\mathrm{hours}, 18.85)$ . Find $\lambda$ (in $\mathrm{h}^{-1}$ and $\mathrm{s}^{-1}$ ) and $T_{1/2}$ .

Solution

Solution of Exercise 32.5.

Slope $= (18.85 - 20.00)/10 = -0.115$ , so $\lambda = 0.115\,\mathrm{h}^{-1} = 0.115/3600 = 3.2 \times 10^{-5}\,\mathrm{s}^{-1}$ ; $T_{1/2} = 0.693/0.115 = 6.0$ hours — technetium-99m.

Exercise 32.6 ★★

One gram of radium-226 ( $T_{1/2} = 1600$ years; molar mass $226\,\mathrm{g}/\mathrm{mol}$ ; $N_A = 6.02 \times 10^{23}\,\mathrm{mol}^{-1}$ ): compute the number of nuclei, then $\lambda$ , then the activity — history’s first activity unit, the curie.

Solution

Solution of Exercise 32.6.

$N = (1.0/226) \times 6.02 \times 10^{23} = 2.7 \times 10^{21}$ ; $T_{1/2} = 1600 \times 3.156 \times 10^{7} = 5.05 \times 10^{10}\,\mathrm{s}$ , so $\lambda = 1.37 \times 10^{-11}\,\mathrm{s}^{-1}$ ; $\mathcal A = \lambda N \approx 3.7 \times 10^{10}\,\mathrm{Bq}$ — one curie, the activity of Marie Curie’s gram of radium.

Exercise 32.7 ★★

Two students seal samples of $20$ and $2.0 \times 10^{20}$ nuclei of one same nuclide. What does each expect after one half-life? Whose prediction is reliable? (Estimate the relative fluctuation $1/\sqrt n$ for each.)

Solution

Solution of Exercise 32.7.

Both expect half to remain: $10$ , and $1.0 \times 10^{20}$ . Fluctuations $\sim 1/\sqrt{n}$ of the count: with $10$ decays, about $30\%$ — $7$ or $13$ would be no surprise; with $1.0 \times 10^{20}$ , about $10^{-10}$ . Only the large sample obeys the law to measurable precision.

Exercise 32.8 ★★

A body holds about $140\,\mathrm{g}$ of potassium, of which $0.012\%$ is potassium-40 ( $T_{1/2} = 1.25 \times 10^{9}$ years, molar mass $40\,\mathrm{g}/\mathrm{mol}$ ). Compute the number of potassium-40 nuclei, then their activity; compare with the body’s total $8000\,\mathrm{Bq}$ .

Solution

Solution of Exercise 32.8.

$m = 140 \times 1.2 \times 10^{-4} = 1.7 \times 10^{-2}\,\mathrm{g}$ ; $N = (0.0168/40) \times 6.02 \times 10^{23} = 2.5 \times 10^{20}$ . $\lambda = 0.693/(1.25 \times 10^{9} \times 3.156 \times 10^{7}) = 1.76 \times 10^{-17}\,\mathrm{s}^{-1}$ ; $\mathcal A = \lambda N \approx 4.4 \times 10^{3}\,\mathrm{Bq}$ — about half the body’s total (most of the rest is carbon-14).

Exercise 32.9 ★★

A patient receives $500\,\mathrm{MBq}$ of technetium-99m ( $T_{1/2} = 6.0$ hours). Compute the activity $15$ hours later, then the time for it to fall below $1\,\mathrm{MBq}$ .

Solution

Solution of Exercise 32.9.

$\lambda = 0.693/6.0 = 0.116\,\mathrm{h}^{-1}$ . At $15$ hours: $\mathcal A = 500\,e^{-0.116 \times 15} = 500 \times 0.177 \approx 88\,\mathrm{MBq}$ . Below $1\,\mathrm{MBq}$ : $t = \ln(500)/0.116 \approx 54\,\mathrm{hours}$ — about $9$ half-lives.

Exercise 32.10 ★★

Charcoal from one pit shows $3.4$ decays per minute per gram of carbon; bone from another shows $30\%$ of the living rate ( $\mathcal A_0 = 13.6$ per minute per gram, $T_{1/2} = 5730$ years). Date both samples.

Solution

Solution of Exercise 32.10.

Charcoal: $3.4/13.6 = \tfrac14$ , exactly $2$ half-lives: $t = 1.15 \times 10^{4}$ years. Bone: $t = 5730 \times \ln(1/0.30)/\ln 2 = 5730 \times 1.74 \approx 1.0 \times 10^{4}$ years.

Exercise 32.11 ★★

A cellar is sealed with a batch of radon-222 inside ( $T_{1/2} = 3.8$ days). After how long has the batch’s activity fallen to $1\%$ ? Real cellars stay radioactive for decades: where does fresh radon come from, and why must ventilation be permanent rather than one-off?

Solution

Solution of Exercise 32.11.

$\lambda = 0.693/3.8 = 0.182\,\mathrm{d}^{-1}$ ; $t = \ln(100)/\lambda = 4.61/0.182 \approx 25$ days (about $6.6$ half-lives). But the uranium chain in the surrounding granite and soil breeds radon continuously: a one-off airing is undone within weeks, so ventilation must be permanent.

Exercise 32.12 ★★★

Show from $N = N_0 e^{-\lambda t}$ that $t = T_{1/2}\ln(N_0/N)/\ln 2$ . A bone retains $5.0\%$ of its living carbon-14: date it.

Solution

Solution of Exercise 32.12.

$N/N_0 = e^{-\lambda t}$ gives $\lambda t = \ln(N_0/N)$ , and $\lambda = \ln 2/T_{1/2}$ gives $t = T_{1/2}\ln(N_0/N)/\ln 2$ . Here $t = 5730 \times \ln(20)/\ln 2 = 5730 \times 4.32 \approx 2.5 \times 10^{4}$ years.

Exercise 32.13 ★★★

In a crystal that trapped no argon at solidification, each decayed potassium-40 nucleus leaves one argon atom in place. Show that $N_{\mathrm{Ar}}/N_{\mathrm K} = e^{\lambda t} - 1$ ; a rock shows a ratio of $3.0$ : find its age ( $T_{1/2} = 1.25 \times 10^{9}$ years).

Solution

Solution of Exercise 32.13.

$N_{\mathrm K} = N_0 e^{-\lambda t}$ and $N_{\mathrm{Ar}} = N_0 - N_{\mathrm K}$ , so $N_{\mathrm{Ar}}/N_{\mathrm K} = e^{\lambda t} - 1$ . Ratio $3.0$ : $e^{\lambda t} = 4$ , so $\lambda t = 2\ln 2$ and $t = 2\,T_{1/2} = 2.5 \times 10^{9}$ years.

Exercise 32.14 ★★★

A Geiger counter gives, at $t = 0$ , $60$ , $120$ , $180$ , $240$ seconds, rates of $400$ , $225$ , $127$ , $71$ , $40$ counts per second. Compute $\ln \mathcal A$ at each time, check alignment, and deduce $\lambda$ and $T_{1/2}$ .

Solution

Solution of Exercise 32.14.

$\ln \mathcal A = 5.99$ , $5.42$ , $4.84$ , $4.26$ , $3.69$ : drops of $0.57$ – $0.58$ per $60\,\mathrm{s}$ — aligned. Slope: $\lambda = 2.30/240 = 9.6 \times 10^{-3}\,\mathrm{s}^{-1}$ ; $T_{1/2} = 0.693/9.6 \times 10^{-3} \approx 72\,\mathrm{s}$ .

Exercise 32.15 ★★★

For a single nucleus, the probability of surviving $n$ half-lives is $1/2^{\,n}$ . Compute the probability that (a) one given nucleus survives $10$ half-lives; (b) all of $8$ watched nuclei survive one half-life. (c) Of $1000$ nuclei, how many are expected after $3$ half-lives — and can the law say which? Conclude on what the law predicts.

Solution

Solution of Exercise 32.15.

(a) $2^{-10} \approx 1 \times 10^{-3}$ . (b) $(1/2)^8 = 1/256 \approx 0.4\%$ . (c) $1000/8 = 125$ expected — but the law is silent on which: it predicts populations exactly and individuals not at all.

32.6 Problem: The age of things

Problem 32.1

Weekend problem — the age of things: a museum lab calibrates the carbon clock, dates a statue and a papyrus, learns why no clock reads past its dial, and pushes a volcanic rock back a billion years

A week in a museum’s dating laboratory. Data: living tissue shows $\mathcal A_0 = 13.6$ carbon-14 decays per minute per gram of carbon; $T_{1/2} = 5730$ years (carbon-14), $1.25 \times 10^{9}$ years (potassium-40); $1$ year $= 3.156 \times 10^{7}\,\mathrm{s}$ ; $N_A = 6.02 \times 10^{23}\,\mathrm{mol}^{-1}$ ; molar mass of carbon $12\,\mathrm{g}/\mathrm{mol}$ ; the counters resolve about $0.2$ decays per minute per gram.

Part I — Monday: calibrating the clock.

Why is the carbon-14 proportion constant in living tissue, and what changes at death?
Compute $\lambda$ for carbon-14 in $\mathrm{s}^{-1}$ .
Verify by substitution that $N_0 e^{-\lambda t}$ solves $\dd N/\dd t = -\lambda N$ .
From $\mathcal A_0$ , compute the number of carbon-14 atoms per gram of carbon in living tissue.
How many carbon atoms does one gram hold? Deduce the carbon-14 proportion — compare with “one in $10^{12}$ ”.
Show that a measured activity $\mathcal A$ dates a sample as $t = T_{1/2}\,\ln(\mathcal A_0/\mathcal A)/\ln 2$ .

Part II — Tuesday: the statue and the papyrus.

A wooden statue gives $9.1$ decays per minute per gram. Date the wood.
What event, exactly, does that date mark — the carving or something else? What caution follows?
An Egyptian papyrus gives $10.4$ decays per minute per gram. Date it. Is a scribe of twenty-two centuries ago plausible?
The activity is measured to $\pm 0.2$ decays per minute. Estimate the papyrus’s age uncertainty, using $\Delta t \approx \Delta\mathcal A/(\lambda \mathcal A)$ .
A dealer’s “ancient” parchment gives $13.4$ decays per minute per gram. Verdict?

Part III — Thursday: the limits of the clock.

After how many half-lives does a sample’s activity fall below the counters’ $0.2$ decays per minute per gram? What age is that?
The best laboratories reach about ten half-lives. What is the practical horizon of radiocarbon dating, in years?
A dinosaur bone is about $6.6 \times 10^{7}$ years old: how many carbon-14 half-lives? Using question 4, after how many half-lives is not even one atom of the gram’s carbon-14 left? Conclude about “carbon-dating dinosaurs”.
What kind of nuclide could date the dinosaur’s rock instead?

Part IV — Friday: the volcanic rock.

Argon is a gas, potassium a solid’s faithful resident: explain why the solidification of lava zeroes the potassium–argon clock.
Assuming each decayed potassium-40 nucleus leaves one trapped argon atom, show that $N_{\mathrm{Ar}}/N_{\mathrm K} = e^{\lambda t} - 1$ .
The rock under the museum’s fossil shows $N_{\mathrm{Ar}}/N_{\mathrm K} = 0.60$ . Date the rock.
Could carbon-14 have dated this rock, or potassium–argon the statue? Give each clock’s useful window in half-lives.
Friday report, one sentence: the three ages found this week, and the single law that read them all.

Solution

Solution of Problem 32.1.

1. Exchange with the atmosphere (eating, breathing) keeps the proportion topped up at the atmospheric value; death stops the intake and decay takes over.

2. $T_{1/2} = 5730 \times 3.156 \times 10^{7} = 1.81 \times 10^{11}\,\mathrm{s}$ , so $\lambda = 0.693/1.81 \times 10^{11} = 3.8 \times 10^{-12}\,\mathrm{s}^{-1}$ .

3. $\dd N/\dd t = -\lambda N_0 e^{-\lambda t} = -\lambda N$ , and $N(0) = N_0$ .

4. $\mathcal A_0 = 13.6/60 = 0.227\,\mathrm{Bq}$ per gram; $N = \mathcal A_0/\lambda = 0.227/3.8 \times 10^{-12} \approx 5.9 \times 10^{10}$ atoms per gram.

5. $(1/12) \times 6.02 \times 10^{23} = 5.0 \times 10^{22}$ carbon atoms; proportion $5.9 \times 10^{10}/5.0 \times 10^{22} \approx 1.2 \times 10^{-12}$ — one in $10^{12}$ , as promised.

6. $\mathcal A = \mathcal A_0 e^{-\lambda t}$ gives $\lambda t = \ln(\mathcal A_0/\mathcal A)$ , and $\lambda = \ln 2/T_{1/2}$ turns it into $t = T_{1/2}\ln(\mathcal A_0/\mathcal A)/\ln 2$ .

7. $t = 5730 \times \ln(13.6/9.1)/\ln 2 = 5730 \times 0.580 \approx 3.3 \times 10^{3}$ years.

8. The death of the wood — the tree’s felling, not the carving; a statue cut from old timber (or a fake carved from ancient wood) predates or postdates its material’s date.

9. $t = 5730 \times \ln(13.6/10.4)/\ln 2 = 5730 \times 0.387 \approx 2.2 \times 10^{3}$ years — fully consistent with a scribe of twenty-two centuries ago.

10. $\Delta t \approx \Delta\mathcal A/(\lambda\mathcal A)$ with $\lambda = \ln 2/5730 = 1.21 \times 10^{-4}$ per year: $\Delta t = (0.2/10.4)/1.21 \times 10^{-4} \approx 1.6 \times 10^{2}$ years — a date good to a century or two.

11. $t = 5730 \times \ln(13.6/13.4)/\ln 2 \approx 120$ years — indistinguishable from modern within the $\pm 160$ -year uncertainty: the parchment is recent, the “antique” a fake.

12. $13.6/2^n < 0.2$ needs $2^n > 68$ : $n = 7$ half-lives (since $2^7 = 128$ ), about $4.0 \times 10^{4}$ years.

13. Ten half-lives: about $5.7 \times 10^{4}$ years — the practical horizon, some sixty thousand years.

14. $6.6 \times 10^{7}/5730 \approx 11\,500$ half-lives. One gram’s $5.9 \times 10^{10}$ atoms are exhausted after $2^n > 5.9 \times 10^{10}$ , i.e. $n \approx 36$ half-lives ( $\approx 2 \times 10^{5}$ years): long before the dinosaurs’ age, not one carbon-14 atom is left — there is nothing to count.

15. A much slower clock: potassium-40 or uranium-238, with half-lives of billions of years.

16. Molten lava lets argon bubble away, so the freshly solidified crystal holds potassium but zero argon: the ratio starts at $0$ at solidification — the clock is zeroed.

17. As in the exercise: $N_{\mathrm K} = N_0 e^{-\lambda t}$ , $N_{\mathrm{Ar}} = N_0 - N_{\mathrm K}$ , so $N_{\mathrm{Ar}}/N_{\mathrm K} = e^{\lambda t} - 1$ .

18. $e^{\lambda t} = 1.60$ : $t = 1.25 \times 10^{9} \times \ln(1.60)/\ln 2 = 1.25 \times 10^{9} \times 0.678 \approx 8.5 \times 10^{8}$ years.

19. No, twice: at $8.5 \times 10^{8}$ years carbon-14 has run $10^5$ half-lives — silence; at $3.3 \times 10^{3}$ years potassium-40 has run $2.6 \times 10^{-6}$ of a half-life — the argon ratio $\approx \lambda t \sim 2 \times 10^{-6}$ is unmeasurably small. Each clock reads only from about a tenth of a half-life to about ten.

20. Statue $\approx 3.3 \times 10^{3}$ years, papyrus $\approx 2.2 \times 10^{3}$ years, rock $\approx 8.5 \times 10^{8}$ years — three clocks, one law: $N = N_0\,e^{-\lambda t}$ , read backwards.