---
title: "Radioactive Decay"
book: "High School Physics"
subject: physics
language: en
chapter: 32
exercises: 15
source: https://one-course.com/books/physics/2/en/chapter/32-radioactive-decay
---

# Chapter 32 — Radioactive 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](https://one-course.com/books/physics/2/en/chapter/19-the-nucleus-and-radioactivity#ch-g11-nucleus-radioactivity)) we met the [unstable](https://one-course.com/books/physics/2/en/chapter/29-work-and-mechanical-energy#prop-g12-work-and-energy-equilibria) nuclei, balanced their $\alpha$ and $\beta$ equations, and saw half of any large sample vanish per [half-life](https://one-course.com/books/physics/2/en/chapter/19-the-nucleus-and-radioactivity#def-g11-nucleus-radioactivity-halflife) $T_{1/2}$. 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](https://one-course.com/books/physics/2/en/chapter/29-work-and-mechanical-energy#prop-g12-work-and-energy-equilibria) [nuclide](https://one-course.com/books/physics/2/en/chapter/19-the-nucleus-and-radioactivity#def-g11-nucleus-radioactivity-notation) there is a constant $\lambda$, its *decay constant* ([unit](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit) $\mathrm{s}^{-1}$), such that during any short interval $\dd t$ each [nucleus](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder) has probability $p = \lambda\,\dd t$ of decaying — the same for every [nucleus](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder) of the species, whatever its age, and untouched by temperature, [pressure](https://one-course.com/books/physics/2/en/chapter/7-pressure-from-sport-to-diving#def-g10-pressure-pressure) or chemistry.

**Remark 32.2 (Lawless singly, exact in crowds).**

A [nucleus](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder) 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.](https://one-course.com/images/onecourse/chapters/physics-2/g12-radioactive-decay/fig-730e8fb206e0.svg)

*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](#rem-g12-radioactive-decay-crowds)); each decay removes one [nucleus](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder). ∎

**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](https://one-course.com/books/physics/2/en/chapter/30-rc-and-rl-circuits#def-g12-rc-rl-circuits-capacitor) of [Chapter 30](https://one-course.com/books/physics/2/en/chapter/30-rc-and-rl-circuits#ch-g12-rc-rl-circuits), with $\lambda$ playing $1/(RC)$: one differential equation, learned once, runs circuits and nuclei alike.

**Proposition 32.6 (Half-life).**

The [half-life](https://one-course.com/books/physics/2/en/chapter/19-the-nucleus-and-radioactivity#def-g11-nucleus-radioactivity-halflife) ([Definition 19.14](https://one-course.com/books/physics/2/en/chapter/19-the-nucleus-and-radioactivity#def-g11-nucleus-radioactivity-halflife)) of a [nuclide](https://one-course.com/books/physics/2/en/chapter/19-the-nucleus-and-radioactivity#def-g11-nucleus-radioactivity-notation) is set by its [decay constant](#def-g12-radioactive-decay-lambda): $T_{1/2} = \ln 2/\lambda$, 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.](https://one-course.com/images/onecourse/chapters/physics-2/g12-radioactive-decay/fig-abffc64a3a58.svg)

*$N(t) = N_0\,e^{-\lambda t}$ 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](https://one-course.com/books/physics/2/en/chapter/19-the-nucleus-and-radioactivity#def-g11-nucleus-radioactivity-notation) | $T_{1/2}$ | $\lambda$ ($\mathrm{s}^{-1}$) |  |
| --- | --- | --- | --- |
| 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](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-power) of ten of $\lambda$.

## 32.3 Activity: counting the clicks

**Proposition 32.8 (Activity).**

The [activity](https://one-course.com/books/physics/2/en/chapter/19-the-nucleus-and-radioactivity#def-g11-nucleus-radioactivity-activity) of a sample — its decays per second, in [becquerels](https://one-course.com/books/physics/2/en/chapter/19-the-nucleus-and-radioactivity#def-g11-nucleus-radioactivity-activity) ([Definition 19.18](https://one-course.com/books/physics/2/en/chapter/19-the-nucleus-and-radioactivity#def-g11-nucleus-radioactivity-activity)) — 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](#thm-g12-radioactive-decay-law) by $\lambda$ gives $\mathcal A(t) = \mathcal A_0 e^{-\lambda t}$. ∎

**Example 32.9 (Huge NNN, tiny λ\lambdaλ).**

A banana ticks at about $15\,\mathrm{Bq}$, a human body at about $8000\,\mathrm{Bq}$, a cubic [metre](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit) of granite at some $10^{6}\,\mathrm{Bq}$ ([Chapter 19](https://one-course.com/books/physics/2/en/chapter/19-the-nucleus-and-radioactivity#ch-g11-nucleus-radioactivity)). 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).**

1. Count decays with a Geiger counter over successive equal intervals: this samples $\mathcal A$ at successive times.
2. Plot $\ln \mathcal A$ against $t$ : since $\ln \mathcal A = \ln \mathcal A_0 - \lambda t$ , [exponential decay](#def-g12-radioactive-decay-exponential) shows as a *straight line* — crooked data means another law, or a mixture of [nuclides](https://one-course.com/books/physics/2/en/chapter/19-the-nucleus-and-radioactivity#def-g11-nucleus-radioactivity-notation) .
3. 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.](https://one-course.com/images/onecourse/chapters/physics-2/g12-radioactive-decay/fig-64d96d680aaa.svg)

*Geiger counts of a short-lived [nuclide](https://one-course.com/books/physics/2/en/chapter/19-the-nucleus-and-radioactivity#def-g11-nucleus-radioactivity-notation): 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](https://one-course.com/books/physics/2/en/chapter/19-the-nucleus-and-radioactivity#def-g11-nucleus-radioactivity-activity) counts decays, not damage. Harm is tracked by the [dose](https://one-course.com/books/physics/2/en/chapter/19-the-nucleus-and-radioactivity#rem-g11-nucleus-radioactivity-dose) in [sieverts](https://one-course.com/books/physics/2/en/chapter/19-the-nucleus-and-radioactivity#rem-g11-nucleus-radioactivity-dose) ([Chapter 19](https://one-course.com/books/physics/2/en/chapter/19-the-nucleus-and-radioactivity#ch-g11-nucleus-radioactivity)), weighing the [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) deposited per [kilogram](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit) of tissue by each radiation’s destructiveness: a $5 \times 10^{8}\,\mathrm{Bq}$ technetium scan is routine, far fewer [becquerels](https://one-course.com/books/physics/2/en/chapter/19-the-nucleus-and-radioactivity#def-g11-nucleus-radioactivity-activity) 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).**

1. *Alive:* living tissue exchanges carbon with the atmosphere and carries its carbon-14 proportion — one [atom](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder) in $10^{12}$ , giving $\mathcal A_0 = 13.6$ decays per minute per gram of carbon.
2. *Death starts the clock:* intake stops, and the proportion decays with $T_{1/2} = 5730$ years.
3. Measure the sample’s [activity](https://one-course.com/books/physics/2/en/chapter/19-the-nucleus-and-radioactivity#def-g11-nucleus-radioactivity-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.](https://one-course.com/images/onecourse/chapters/physics-2/g12-radioactive-decay/fig-aa355c6da75e.svg)

*The dating curve: enter at the measured ratio, read down to the age. The dot is the hearth of [Example 32.13](#ex-g12-radioactive-decay-hearth): 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](https://one-course.com/books/physics/2/en/chapter/29-work-and-mechanical-energy#rem-g12-work-and-energy-escape) molten lava but is trapped once rock crystallises — solidification zeroes the clock. *Uranium–lead*: uranium-238 heads a chain of decays ending at [stable](https://one-course.com/books/physics/2/en/chapter/29-work-and-mechanical-energy#prop-g12-work-and-energy-equilibria) lead-206 ([Chapter 19](https://one-course.com/books/physics/2/en/chapter/19-the-nucleus-and-radioactivity#ch-g11-nucleus-radioactivity)), 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](https://one-course.com/books/physics/2/en/chapter/19-the-nucleus-and-radioactivity#rem-g11-nucleus-radioactivity-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](https://one-course.com/books/physics/2/en/chapter/19-the-nucleus-and-radioactivity#def-g11-nucleus-radioactivity-halflife) in seconds, then in days.

**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](https://one-course.com/books/physics/2/en/chapter/19-the-nucleus-and-radioactivity#def-g11-nucleus-radioactivity-activity) of a sample of $1.0 \times 10^{10}$ nuclei.

**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 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](https://one-course.com/books/physics/2/en/chapter/19-the-nucleus-and-radioactivity#def-g11-nucleus-radioactivity-activity) $4.0 \times 10^{5}\,\mathrm{Bq}$ today. Compute its [activity](https://one-course.com/books/physics/2/en/chapter/19-the-nucleus-and-radioactivity#def-g11-nucleus-radioactivity-activity) in $90$ years, then in $300$ years.

**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 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](https://one-course.com/books/physics/2/en/chapter/19-the-nucleus-and-radioactivity#def-g11-nucleus-radioactivity-activity) — history’s first [activity](https://one-course.com/books/physics/2/en/chapter/19-the-nucleus-and-radioactivity#def-g11-nucleus-radioactivity-activity) [unit](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit), the curie.

**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](https://one-course.com/books/physics/2/en/chapter/19-the-nucleus-and-radioactivity#def-g11-nucleus-radioactivity-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](https://one-course.com/books/physics/2/en/chapter/19-the-nucleus-and-radioactivity#def-g11-nucleus-radioactivity-notation). What does each expect after one [half-life](https://one-course.com/books/physics/2/en/chapter/19-the-nucleus-and-radioactivity#def-g11-nucleus-radioactivity-halflife)? Whose prediction is reliable? (Estimate the relative fluctuation $1/\sqrt n$ for each.)

**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](https://one-course.com/books/physics/2/en/chapter/19-the-nucleus-and-radioactivity#def-g11-nucleus-radioactivity-activity); compare with the body’s total $8000\,\mathrm{Bq}$.

**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](https://one-course.com/books/physics/2/en/chapter/19-the-nucleus-and-radioactivity#def-g11-nucleus-radioactivity-activity) $15$ hours later, then the time for it to fall below $1\,\mathrm{MBq}$.

**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 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](https://one-course.com/books/physics/2/en/chapter/19-the-nucleus-and-radioactivity#def-g11-nucleus-radioactivity-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 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 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](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder) leaves one argon [atom](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder) 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 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 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](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder), the probability of surviving $n$ half-lives is $1/2^{\,n}$. Compute the probability that (a) one given [nucleus](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder) survives $10$ half-lives; (b) all of $8$ watched nuclei survive one [half-life](https://one-course.com/books/physics/2/en/chapter/19-the-nucleus-and-radioactivity#def-g11-nucleus-radioactivity-halflife). (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 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.**

1. Why is the carbon-14 proportion constant in living tissue, and what changes at death?
2. Compute $\lambda$ for carbon-14 in $\mathrm{s}^{-1}$ .
3. Verify by substitution that $N_0 e^{-\lambda t}$ solves $\dd N/\dd t = -\lambda N$ .
4. From $\mathcal A_0$ , compute the number of carbon-14 [atoms](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder) per gram of carbon in living tissue.
5. How many carbon [atoms](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder) does one gram hold? Deduce the carbon-14 proportion — compare with “one in $10^{12}$ ”.
6. Show that a measured [activity](https://one-course.com/books/physics/2/en/chapter/19-the-nucleus-and-radioactivity#def-g11-nucleus-radioactivity-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.**

7. A wooden statue gives $9.1$ decays per minute per gram. Date the wood.
8. What event, exactly, does that date mark — the carving or something else? What caution follows?
9. An Egyptian papyrus gives $10.4$ decays per minute per gram. Date it. Is a scribe of twenty-two centuries ago plausible?
10. The [activity](https://one-course.com/books/physics/2/en/chapter/19-the-nucleus-and-radioactivity#def-g11-nucleus-radioactivity-activity) is measured to $\pm 0.2$ decays per minute. Estimate the papyrus’s age uncertainty, using $\Delta t \approx \Delta\mathcal A/(\lambda \mathcal A)$ .
11. A dealer’s “ancient” parchment gives $13.4$ decays per minute per gram. Verdict?

**Part III — Thursday: the limits of the clock.**

12. After how many half-lives does a sample’s [activity](https://one-course.com/books/physics/2/en/chapter/19-the-nucleus-and-radioactivity#def-g11-nucleus-radioactivity-activity) fall below the counters’ $0.2$ decays per minute per gram? What age is that?
13. The best laboratories reach about ten half-lives. What is the practical horizon of radiocarbon dating, in years?
14. 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](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder) of the gram’s carbon-14 left? Conclude about “carbon-dating dinosaurs”.
15. What kind of [nuclide](https://one-course.com/books/physics/2/en/chapter/19-the-nucleus-and-radioactivity#def-g11-nucleus-radioactivity-notation) could date the dinosaur’s rock instead?

**Part IV — Friday: the volcanic rock.**

16. Argon is a gas, potassium a solid’s faithful resident: explain why the solidification of lava zeroes the potassium–argon clock.
17. Assuming each decayed potassium-40 [nucleus](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder) leaves one trapped argon [atom](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder) , show that $N_{\mathrm{Ar}}/N_{\mathrm K} = e^{\lambda t} - 1$ .
18. The rock under the museum’s fossil shows $N_{\mathrm{Ar}}/N_{\mathrm K} = 0.60$ . Date the rock.
19. Could carbon-14 have dated this rock, or potassium–argon the statue? Give each clock’s useful window in half-lives.
20. Friday report, one sentence: the three ages found this week, and the single law that read them all.

**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](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder) per gram.

**5.** $(1/12) \times 6.02 \times 10^{23} = 5.0 \times 10^{22}$ carbon [atoms](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder); 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](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder) 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](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder) 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](https://one-course.com/books/physics/2/en/chapter/19-the-nucleus-and-radioactivity#def-g11-nucleus-radioactivity-halflife) — 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](https://one-course.com/books/physics/2/en/chapter/19-the-nucleus-and-radioactivity#def-g11-nucleus-radioactivity-halflife) 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.