Half-Life and Radioactive Decay
The Clock Inside Every Atom
How do we know the Earth is 4.5 billion years old? How can archaeologists date a wooden artifact to 3000 BCE? The answer lies in radioactive decay, a process so predictable that it serves as nature’s most reliable stopwatch.
Each radioactive isotope decays at its own characteristic rate, measured by its half-life: the time for half the atoms to decay. This constant rate, independent of how much material you have, makes radioactive decay the perfect clock for measuring deep time.
Before You Start
Quick Self-Check
What is the solution to $\frac{dm}{dt} = km$ where $k < 0$? → Exponential Growth ODE
Solve $e^{-0.5t} = 0.25$ for $t$. → Natural Logarithm
If you struggled, review the linked prerequisite first.
Prerequisite Map
Quick Reference
| Property | Value |
|---|---|
| Concept | Differential Equations |
| Chapter | 6, Section 5 |
| Difficulty | Intermediate |
| Time | ~15 minutes |
Key Concepts
The Decay Model
Radioactive decay follows:
$$\frac{dm}{dt} = km \quad \text{where } k < 0$$
The solution is:
$$\boxed{m(t) = m_0 e^{kt}}$$
where $m_0$ is the initial mass and $k$ is the decay constant (negative).
Half-Life Definition
The half-life $h$ is the time for half the substance to decay:
$$m(h) = \frac{m_0}{2}$$
Relating $k$ and Half-Life
From $m_0 e^{kh} = \frac{m_0}{2}$:
$$e^{kh} = \frac{1}{2}$$ $$kh = \ln\left(\frac{1}{2}\right) = -\ln 2$$
$$\boxed{k = \frac{-\ln 2}{h} \approx \frac{-0.693}{h}}$$
Two Equivalent Forms
The decay formula can be written as:
| Form | Formula | When to Use |
|---|---|---|
| Exponential | $m(t) = m_0 e^{kt}$ | When $k$ is given |
| Half-life | $m(t) = m_0 \cdot 2^{-t/h}$ | When half-life $h$ is given |
These are equivalent since $e^{kt} = e^{(-\ln 2/h)t} = (e^{\ln 2})^{-t/h} = 2^{-t/h}$.
Common Half-Lives
| Isotope | Half-Life | Application |
|---|---|---|
| Carbon-14 | 5,730 years | Archaeological dating |
| Uranium-238 | 4.5 billion years | Geological dating |
| Iodine-131 | 8 days | Medical imaging |
| Radium-226 | 1,590 years | Historical interest |
| Cobalt-60 | 5.3 years | Cancer treatment |
Finding Time for Target Mass
To find when mass reaches $m$:
$$m_0 e^{kt} = m$$ $$t = \frac{\ln(m/m_0)}{k} = \frac{-h \ln(m/m_0)}{\ln 2}$$
Or using the half-life form: $$2^{-t/h} = \frac{m}{m_0}$$ $$t = -h \log_2\left(\frac{m}{m_0}\right) = h \log_2\left(\frac{m_0}{m}\right)$$
Common Pitfalls
| Mistake | Why It’s Wrong | Correct Approach |
|---|---|---|
| Forgetting $k$ is negative | Decay means $k < 0$; using positive $k$ gives growth | $k = -\ln 2/h$ (note the minus sign) |
| Confusing half-life with $k$ | Half-life is time; $k$ is rate (per time) | $k = -\ln 2/h$, different units |
| Thinking mass reaches zero | $e^{kt} \to 0$ but never equals 0 | Decay is asymptotic |
| Wrong base conversion | $2^{-t/h} \neq e^{-t/h}$ | Use $e^{-(\ln 2/h)t} = 2^{-t/h}$ |
| Using percentage remaining incorrectly | “25% remaining” means $m/m_0 = 0.25$ | Set up $2^{-t/h} = 0.25$ and solve |
Worked Example
Problem: Strontium-90 has a half-life of 28 years. A sample has initial mass 50 mg.
(a) Find the decay constant $k$. (b) Write the mass function $m(t)$. (c) Find the mass after 40 years. (d) When will only 10 mg remain?
Solution:
(a) Decay constant: $$k = \frac{-\ln 2}{28} \approx -0.02476 \text{ per year}$$
(b) Mass function: $$m(t) = 50e^{-0.02476t}$$
Or equivalently: $m(t) = 50 \cdot 2^{-t/28}$
(c) After 40 years: $$m(40) = 50e^{-0.02476 \times 40} = 50e^{-0.990} \approx 50 \times 0.372 = 18.6 \text{ mg}$$
(d) When $m = 10$: $$50 \cdot 2^{-t/28} = 10$$ $$2^{-t/28} = 0.2$$ $$-\frac{t}{28} = \log_2(0.2) = \frac{\ln 0.2}{\ln 2} \approx -2.322$$ $$t = 28 \times 2.322 \approx 65 \text{ years}$$
Practice Problems
A substance has a half-life of 10 days. If you start with 80 grams, how much remains after 30 days?
Cesium-137 has a half-life of 30 years. Find the decay constant $k$ and write the decay formula for an initial mass of 100 mg.
A radioactive sample decays from 200 mg to 125 mg in 4 hours. Find the half-life.
An ancient wooden artifact contains 35% of the carbon-14 found in living trees. Carbon-14 has a half-life of 5730 years. How old is the artifact?
Two isotopes decay simultaneously. Isotope A has half-life 100 years; isotope B has half-life 400 years. Initially there is twice as much A as B.
- Write formulas for both quantities over time.
- When will the amounts be equal?
- Prove that for any two decaying isotopes, if one starts with more material but has a shorter half-life, the amounts will eventually be equal.
CCI-Style Conceptual Questions
Question 1: A sample decays to 25% of its original amount. How many half-lives have passed?
Answer
Two half-lives. After 1 half-life: 50%. After 2 half-lives: 25%.
In general, after $n$ half-lives, the fraction remaining is $(1/2)^n$.
Question 2: If you double the initial amount of a radioactive substance, what happens to its half-life?
Answer
Nothing; the half-life stays the same. Half-life is a property of the isotope, not the sample size. Doubling the initial amount just means you’ll have twice as much at every point in time.
Question 3: Isotope X has half-life 10 years and isotope Y has half-life 20 years. After 20 years, what fraction of each original sample remains?
Answer
- X: 20 years = 2 half-lives, so $(1/2)^2 = 1/4 = 25\%$ remains
- Y: 20 years = 1 half-life, so $1/2 = 50\%$ remains
Common Misconceptions
after two half-lives, a radioactive sample has zero mass remaining because each half-life removes half and two halves equal one whole.
This is the rate-as-fixed-number error. After each half-life, the remaining mass is halved relative to the amount at the start of that period, not relative to the original sample. After one half-life, $\frac{1}{2}$ of the original remains. After two half-lives, $\frac{1}{2}$ of $\frac{1}{2}$ remains, giving $\frac{1}{4}$ of the original. In the model $m(t) = m_0 e^{kt}$ with $k < 0$, the mass approaches zero asymptotically and never actually reaches zero.
Mastery Checklist
Mental Model
The “Coin Flip” Analogy:
Imagine each atom flipping a coin every time period. If it lands heads, the atom decays; tails, it survives. After one half-life, about half the atoms have decayed. The surviving atoms flip again; half of them decay. This random process, applied to trillions of atoms, produces the smooth exponential curve we observe.
Still Confused?
| If you’re struggling with... | Review this |
|---|---|
| The decay equation $m' = km$ | Exponential Growth ODE |
| Solving for $t$ in $e^{kt} = c$ | Natural Logarithm |
| Converting between bases | The key identity: $e^{-(\ln 2/h)t} = 2^{-t/h}$ |
Connections
Looking back:
- Exponential Growth ODE gives the mathematical foundation
- Half-life is just the decay analog of doubling time
Looking ahead:
- Carbon-14 dating uses half-life for archaeological chronology
- Pharmacokinetics uses similar models for drug clearance
- Nuclear physics uses decay chains for complex isotope analysis
Real-world connections:
- Determining age of fossils, rocks, and artifacts
- Medical dosimetry (radiation treatment planning)
- Nuclear waste management and storage
- Smoke detectors (americium-241 decay)
| Previous | Up | Next |
|---|---|---|
| Population Growth | Chapter 6 Skills | Newton’s Cooling |
Last updated: 2026-01-22