Apply Radioactive Decay — Apply E.3 Radioactive decay in problem-solving and explanations - IB Physics

SECTION 1

Historical Context & Motivation

At the end of the nineteenth century, scientists believed that atoms were permanent, indivisible building blocks of matter. That view shattered in a single decade when researchers discovered that certain elements spontaneously emit invisible radiation. The phenomenon — radioactive decay — revealed that atomic nuclei can be unstable and will transform into entirely different elements over time. Understanding this process became essential not only for physics but also for medicine, archaeology, energy production, and environmental science.

1896 • Becquerel's Discovery

Henri Becquerel observed that uranium salts emitted penetrating rays that fogged photographic plates, even in the dark. This was the first recorded evidence of natural radioactivity.

1898 • The Curies Isolate Radium

Marie and Pierre Curie isolated two new radioactive elements — polonium and radium — proving that radioactivity was an atomic property, not a chemical reaction.

1903 • Rutherford Classifies Radiation

Ernest Rutherford identified three types of nuclear radiation: alpha (α), beta (β), and gamma (γ), each with different penetrating powers and charges.

1907 • Half-Life Concept Formalised

Rutherford introduced the half-life as a constant characteristic of each radioactive isotope, enabling quantitative predictions of decay.

1949 • Carbon-14 Dating

Willard Libby developed radiocarbon dating, applying half-life calculations to determine the age of organic materials — a landmark application of radioactive decay.

The central question that emerged from these discoveries was this: if nuclei decay at random, how can we make precise, reliable predictions about the amount of material remaining after a given time? The answer lies in the mathematics of exponential decay and the statistical nature of large populations of atoms. In IB Physics section E.3, you will learn to apply these tools to solve problems involving half-life, decay constant, and activity.

SECTION 2

Core Principles of Radioactive Decay

Radioactive decay is governed by a handful of powerful ideas. Each one helps explain why decay is predictable for large samples even though the behaviour of any single nucleus is completely random. These principles form the foundation for every calculation in this topic.

Random & Spontaneous

Each nuclear decay event is random — you cannot predict when a specific nucleus will decay. It is also spontaneous, meaning it occurs without any external trigger.

Half-Life (t₁/₂)

The half-life is the time required for half of the undecayed nuclei in a sample to decay. It is constant for a given isotope regardless of the amount of material.

Decay Constant (λ)

The decay constant λ represents the probability of decay per nucleus per unit time. A large λ means the isotope decays quickly; a small λ means it is long-lived.

Activity (A)

The activity of a sample is the number of decays per second, measured in becquerels (Bq). Activity decreases exponentially as the number of undecayed nuclei decreases.

Exponential Decay Law

The number of remaining nuclei follows an exponential decay pattern: N = N₀ ×

e^{−λt}

. This law emerges because the rate of decay is proportional to the number of nuclei present.

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KEY TAKEAWAY

Think of radioactive decay like popcorn popping in a microwave. You cannot predict which kernel will pop next, but you can reliably predict that after a certain time, about half the unpopped kernels will have popped. The half-life is like the time it takes for half the remaining kernels to pop — it stays roughly constant even as fewer and fewer kernels remain.

SECTION 3

Visualising Exponential Decay

The graph below shows how the number of undecayed nuclei (N) changes over time (t) for a radioactive sample. Notice how the curve never actually reaches zero — it just keeps halving. Each successive half-life reduces the remaining sample by another 50%.

The exponential decay curve shows how the fraction of undecayed nuclei N/N₀ decreases over successive half-lives. After one half-life, 50% remains; after two, 25%; after three, 12.5%. The dashed lines mark the half-life points on both axes.

The key feature of this graph is its shape: a smooth curve that drops steeply at first and then flattens out. This is characteristic of all exponential decay processes. The rate of decay is fastest when there are the most undecayed nuclei, and it slows down as fewer remain. Importantly, no matter how long you wait, the curve never reaches exactly zero — there is always some fraction remaining, though it becomes negligibly small after many half-lives.

SECTION 4

Mathematical Framework

The IB E.3 syllabus requires you to apply several interconnected equations. Each connects the number of undecayed nuclei, time, decay constant, and activity. Let's build them step by step.

EXPONENTIAL DECAY LAW

N = N₀ × e^(−λt)

N = number of undecayed nuclei at time t; N₀ = initial number of nuclei; λ = decay constant (s⁻¹); t = elapsed time (s); e ≈ 2.718 (Euler's number).

HALF-LIFE AND DECAY CONSTANT

t₁/₂ = ln 2 / λ ≈ 0.693 / λ

t₁/₂ = half-life; ln 2 ≈ 0.693. This equation lets you convert between the half-life (easy to measure) and the decay constant (useful in calculations).

ACTIVITY

A = λN = A₀ × e^(−λt)

A = activity at time t (Bq); A₀ = initial activity; λ = decay constant; N = number of undecayed nuclei at time t. Since A is proportional to N, activity also decays exponentially.

HALF-LIFE SHORTCUT

N = N₀ × (½)^(t / t₁/₂)

This equivalent form avoids the exponential function. It is especially handy when the elapsed time is a whole number of half-lives, because you simply halve repeatedly.

💡

IB Exam Tip

The IB data booklet provides the equations N = N₀ ×

e^{−λt}

and t₁/₂ = ln 2 / λ. You do not need to memorise them, but you must be able to select the correct one and rearrange it for any unknown. Practice rearranging for λ, t, N₀, and t₁/₂.

SECTION 5

Types of Radioactive Decay

When a nucleus decays, it emits one of several types of radiation. Understanding the differences matters for writing balanced nuclear equations and for predicting the resulting daughter nucleus. The diagram below summarises the three main types of decay and their effects on the nucleus.

Comparison of alpha, beta-minus, and gamma decay. Alpha decay reduces both mass number and atomic number. Beta-minus decay converts a neutron to a proton, increasing atomic number by 1. Gamma emission releases energy without changing the nucleus's composition.

Property	Alpha (α)	Beta-minus (β⁻)	Gamma (γ)
Particle emitted	⁴₂He nucleus	Electron (e⁻) + antineutrino	High-energy photon
Change in A (mass no.)	−4	0	0
Change in Z (atomic no.)	−2	+1	0
Ionising power	Highest	Moderate	Lowest
Penetrating power	Lowest (paper)	Moderate (aluminium)	Highest (thick lead)

Summary of the three main types of radioactive decay

SECTION 6

Worked Example — Half-Life Calculation

A sample of iodine-131 (used in thyroid treatment) has an initial activity of 8.0 × 10⁵ Bq. The half-life of iodine-131 is 8.02 days. Determine: (a) the decay constant λ, (b) the activity after 24 days, and (c) the time for the activity to drop to 1.0 × 10⁵ Bq.

Iodine-131 Decay Calculation

1
Step 1 — Identify Given Values

A₀ = 8.0 × 10⁵ Bq, t₁/₂ = 8.02 days. Convert the half-life to seconds for SI consistency if needed, but since the IB often works in convenient time units, we'll keep days here.
2
Step 2 — Calculate the Decay Constant λ

Using the relation λ = ln 2 / t₁/₂: λ = 0.693 / 8.02 days

λ ≈ 0.0864 day⁻¹
3
Step 3 — Find Activity After 24 Days (Part b)

Apply A = A₀ × $e^{−λt}$ : A = 8.0 × 10⁵ × $e^{−0.0864 × 24}$ A = 8.0 × 10⁵ × $e^{−2.074}$ A = 8.0 × 10⁵ × 0.1257

A ≈ 1.0 × 10⁵ Bq
4
Step 3 — Alternative Shortcut Check

24 days ÷ 8.02 days ≈ 3 half-lives. After 3 half-lives, the fraction remaining is (½)³ = ⅛. So A = 8.0 × 10⁵ × ⅛ = 1.0 × 10⁵ Bq. This confirms our exponential result.
5
Step 4 — Find Time to Reach 1.0 × 10⁵ Bq (Part c)

Rearrange A = A₀ × $e^{−λt}$ for t: t = −ln(A / A₀) / λ t = −ln(1.0 × 10⁵ / 8.0 × 10⁵) / 0.0864 t = −ln(0.125) / 0.0864 t = 2.079 / 0.0864

t ≈ 24.1 days (≈ 3 half-lives)

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Strategy Note

Whenever the elapsed time is an exact multiple of the half-life, use the shortcut N = N₀ × (½)ⁿ — it's faster and avoids rounding errors. Use the exponential form when the time is not a neat multiple of t₁/₂.

SECTION 7

Strengths & Limitations of Decay Models

The exponential decay model is remarkably powerful, but like all models in physics it has both strengths and limitations. Understanding these will help you interpret exam questions correctly and avoid common mistakes.

Strengths	Limitations
Accurately predicts average behaviour for large samples (billions of nuclei).	Cannot predict when a single specific nucleus will decay.
Half-life is constant and independent of external conditions (temperature, pressure, chemical bonding).	The continuous exponential model breaks down for very small numbers of nuclei, where statistical fluctuations dominate.
Simple mathematics (exponentials and logarithms) allow precise calculations.	Does not account for decay chains — the daughter nucleus may also be radioactive and decay further.
Widely applicable: medical imaging, carbon dating, nuclear power, geology.	Assumes a pure sample of one isotope; mixtures require summing multiple exponentials.

Strengths and limitations of the simple exponential decay model

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KEY TAKEAWAY

The exponential model is like a weather forecast: it works brilliantly for large-scale trends ("50% chance of rain over the whole city") but fails at predicting individual events ("will this specific raindrop hit my window?"). With trillions of nuclei, the law of large numbers makes the model extremely reliable.

SECTION 8

Connections to Advanced Theory

The E.3 treatment of radioactive decay uses a semi-classical, statistical approach. At university level and in the IB HL extensions, these ideas connect to deeper physics. The table below maps what you know now to where these ideas lead.

E.3 Concept (What You Know)	Advanced Extension
Half-life is a fixed property of each isotope.	Quantum tunnelling probability determines the half-life; heavier alpha particles must tunnel through a Coulomb barrier whose width depends on nuclear structure.
N = N₀ × $e^{−λt}$ assumes a single-step decay.	Decay chains (e.g., uranium-238 → radon-222 → polonium-218 → …) require solving coupled differential equations (Bateman equations).
Activity measured in becquerels (Bq).	Dose concepts (Gray, Sievert) weight activity by radiation type and tissue sensitivity for biological risk assessment.
Beta decay emits an electron and antineutrino.	The weak nuclear force mediates beta decay via W boson exchange — a key prediction of the electroweak theory.

From E.3 to advanced nuclear physics

Even though these advanced topics are beyond the scope of the IB exam, knowing they exist can deepen your understanding. For instance, if a question asks why some nuclei have very long half-lives while others decay almost instantly, the answer lies in the quantum tunnelling probability — the thicker or higher the energy barrier, the less likely a particle is to escape the nucleus, and the longer the half-life.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A student claims that after two half-lives, a radioactive sample will have completely decayed. Explain why this statement is incorrect, and state what fraction of the original sample actually remains after two half-lives.

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After one half-life, half of the original nuclei remain (½ of N₀). After a second half-life, half of that remaining amount decays, leaving ½ × ½ = ¼ of the original sample. The sample never fully reaches zero because each half-life only removes half of whatever is currently present. The fraction remaining after n half-lives is (½)ⁿ, which approaches but never equals zero.

Question 2: BASIC CALCULATION

Phosphorus-32 has a half-life of 14.3 days. Calculate the decay constant λ in day⁻¹ and in s⁻¹.

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Using λ = ln 2 / t₁/₂: λ = 0.693 / 14.3 = 0.0485 day⁻¹. To convert to s⁻¹, note that 14.3 days = 14.3 × 24 × 3600 = 1.236 × 10⁶ s. Therefore λ = 0.693 / (1.236 × 10⁶) = 5.61 × 10⁻⁷ s⁻¹. Always check that the units of λ match the units of time in your decay equation to keep calculations consistent.

Question 3: INTERMEDIATE

A sample of strontium-90 (t₁/₂ = 28.8 years) initially contains 5.0 × 10²⁰ atoms. How many atoms remain after 100 years? What is the activity at that time?

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First, find λ = 0.693 / 28.8 = 0.02406 yr⁻¹. Then use N = N₀ ×

e^{−λt}

= 5.0 × 10²⁰ ×

e^{−0.02406 × 100}

= 5.0 × 10²⁰ ×

e^{−2.406}

= 5.0 × 10²⁰ × 0.09005 = 4.5 × 10¹⁹ atoms. For activity, convert λ to s⁻¹: λ = 0.693 / (28.8 × 3.156 × 10⁷ s) = 7.63 × 10⁻¹⁰ s⁻¹. Then A = λN = 7.63 × 10⁻¹⁰ × 4.5 × 10¹⁹ ≈ 3.4 × 10¹⁰ Bq. This remains a very high activity, illustrating why strontium-90 is a significant environmental concern.

Question 4: APPLIED

An archaeologist finds a wooden artefact in which the carbon-14 activity is 1.9 Bq per gram of carbon. Living wood has a carbon-14 activity of 15.2 Bq per gram. The half-life of carbon-14 is 5730 years. Estimate the age of the artefact.

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The ratio of current to original activity is A/A₀ = 1.9 / 15.2 = 0.125. Using A = A₀ ×

e^{−λt}

, we get 0.125 =

e^{−λt}

. Taking the natural logarithm: ln(0.125) = −λt, so −2.079 = −λt. Now λ = 0.693 / 5730 = 1.209 × 10⁻⁴ yr⁻¹. Therefore t = 2.079 / (1.209 × 10⁻⁴) ≈ 17,200 years. Alternatively, note that 0.125 = (½)³, so the artefact is about 3 half-lives old: 3 × 5730 = 17,190 years. Both methods agree, confirming the estimate.

Question 5: CRITICAL THINKING

A hospital receives two radioactive sources: Source X has a half-life of 6 hours and initial activity 2.0 × 10⁸ Bq; Source Y has a half-life of 60 hours and initial activity 1.0 × 10⁷ Bq. After how many hours will both sources have the same activity? Discuss which source poses a greater long-term radiation hazard.

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Set

A_X

A_Y

: 2.0 × 10⁸ ×

e^{−λ_X × t}

= 1.0 × 10⁷ ×

e^{−λ_Y × t}

. Here λ_X = 0.693/6 = 0.1155 h⁻¹ and λ_Y = 0.693/60 = 0.01155 h⁻¹. Dividing both sides: 20 =

e^{(λ_X − λ_Y) × t}

e^{0.104 × t}

. Taking ln: ln 20 = 0.104t, so t = 2.996 / 0.104 ≈ 28.8 hours. After this time, Source Y is the greater long-term hazard because its longer half-life means it remains significantly active for weeks, whereas Source X decays to negligible levels within days. In waste management, long-lived isotopes are generally a bigger concern even if their initial activity is lower.

SUMMARY

Summary — Radioactive Decay (E.3)

Summary

Radioactive decay is a random, spontaneous process in which unstable nuclei transform by emitting alpha particles, beta particles, or gamma rays. The number of undecayed nuclei follows the exponential decay law N = N₀ × $e^{−λt}$ , where λ (decay constant) is linked to the half-life by t₁/₂ = ln 2 / λ. The activity A = λN also decays exponentially and is measured in becquerels.

To solve IB problems, choose the appropriate equation, substitute known values, and rearrange for the unknown. Use the half-life shortcut N = N₀ × (½)ⁿ when time is a whole number of half-lives. Remember that the model applies to large populations of nuclei and cannot predict the behaviour of any single nucleus. Applications range from carbon dating and medical imaging to nuclear waste management.