Year 12 Physics Module 8 ⏱ ~45 min 5 MC · 2 Short Answer Lesson 13 of 17

Radioactive Decay

In 1898, Marie Curie in Paris isolated polonium and radium from 10 tonnes of pitchblende ore. She measured the half-life of Ra-226 as 1,600 years (modern value: 1,599 ± 4 years) using an electroscope to measure the ionisation rate of the emitted radiation. She coined the term "radioactivity" and became the first scientist to win two Nobel Prizes, Physics in 1903 and Chemistry in 1911. Her quantitative electroscope method established the experimental framework for measuring radioactive decay that is still used today.

Today's hook: In 1898, Marie Curie in Paris processed 10 tonnes of uranium ore (pitchblende) to isolate 0.1 g of radium-226. Using an electroscope to measure the ionisation rate, the number of charged particles emitted per second, she determined that Ra-226 has a half-life of 1,600 years: exactly half of any sample decays in that time, then half of the remainder in another 1,600 years. Yet she could not predict which individual atom would decay next. How can a process be simultaneously perfectly predictable in bulk and completely unpredictable for each individual atom?
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Worksheets

Practise this lesson

Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.

Build Foundations & guided practice Apply Application & data response Master Mastery tasks Exam Exam-style questions
Before you read, predict

A radioactive sample has a half-life of 10 minutes. You start with 1000 atoms.

Before reading on, answer:

  1. How many atoms remain after 10 minutes? After 20 minutes?
  2. Why can't we predict which specific atom will decay next?
  3. What does the decay rate depend on?

Warm-up: In alpha decay, an alpha particle is identical to:

Learning Intentions
goals

Know, Decay Modes

  • Alpha, beta, gamma
  • Conservation laws
  • Decay equations
  • Spontaneous vs artificial transmutation

Understand, Exponential Decay

  • Half-life and decay constant
  • Activity and count rate
  • Carbon dating
  • Conservation of mass-energy in transmutations

Can Do, Calculate Decay

  • Use $N = N_0 e^{-\lambda t}$
  • Calculate half-life
  • Analyse decay chains
  • Find energy released using $E = \Delta m\,c^2$
Scan these before reading
vocab
Half-life ($t_{1/2}$)Time for half of radioactive nuclei to decay; $t_{1/2} = \ln(2)/\lambda$.
Decay constant ($\lambda$)Probability of decay per unit time; $\lambda = \ln(2)/t_{1/2}$.
Activity ($A$)Decays per second; $A = \lambda N$; measured in becquerels (Bq).
Alpha particle$^4_2\text{He}$ nucleus; high ionisation, low penetration.
TransmutationChange of one element into another by a nuclear reaction; spontaneous (radioactive decay) or artificial (induced by bombardment, e.g. Rutherford's $\alpha + {}^{14}_{7}\text{N} \to {}^{17}_{8}\text{O} + p$).
Mass-energy equivalence$E = \Delta m\,c^2$; the small mass defect between reactants and products of a nuclear reaction appears as released or absorbed energy.
Q-valueEnergy released in a transmutation. A positive Q-value is exothermic (can occur spontaneously); a negative Q-value is endothermic (needs energy input).
Cross-lesson links: L14 showed what holds nuclei together. L15 examines what happens when that stability fails, radioactive decay releases the energy stored in nuclear binding. The decay types (α, β, γ) each involve different physics: strong force (α), weak force (β), EM radiation (γ); and each has direct applications from medical imaging to carbon-14 dating.

Core Content

1
Types of Radioactive Decay
+5 XP

Alpha, beta, and gamma

Place a small sample of radium-226 in front of a Geiger counter. The counter clicks spontaneously, at random intervals, yet if you measure the total count rate for 10 minutes and again in 100 years, the average rate will decrease by exactly the factor predicted by t₁/₂ = 1,600 years, the value Marie Curie first determined in Paris in 1898. This spontaneous, probabilistic, yet statistically precise behaviour is the signature of radioactive decay. The three main decay modes differ in what the nucleus emits. Alpha decay ($\alpha$): A heavy nucleus emits an alpha particle ($^4_2\text{He}$), reducing its mass number by 4 and atomic number by 2:

$$^A_Z\text{X} \rightarrow \; ^{A-4}_{Z-2}\text{Y} + \; ^4_2\text{He}$$

Alpha particles are massive and doubly charged, so they cause intense ionisation but have low penetration (stopped by paper or skin). They occur in very heavy nuclei ($Z > 82$) where the strong force is overwhelmed by Coulomb repulsion.

Beta-minus decay ($\beta^-$): A neutron converts to a proton, emitting an electron and an antineutrino:

$$n \rightarrow p + e^- + \bar{\nu}_e$$ $$^A_Z\text{X} \rightarrow \; ^A_{Z+1}\text{Y} + e^- + \bar{\nu}_e$$

Beta particles are fast electrons with moderate penetration (stopped by a few mm of aluminium). Beta decay occurs when a nucleus has too many neutrons relative to protons.

Beta-plus decay ($\beta^+$) / Electron capture: A proton converts to a neutron, emitting a positron and neutrino (or capturing an orbital electron):

$$p \rightarrow n + e^+ + \nu_e$$ $$^A_Z\text{X} \rightarrow \; ^A_{Z-1}\text{Y} + e^+ + \nu_e$$

This occurs when a nucleus has too many protons relative to neutrons.

Gamma decay ($\gamma$): An excited nucleus releases excess energy as a high-energy photon:

$$^A_Z\text{X}^* \rightarrow \; ^A_Z\text{X} + \gamma$$

Gamma rays are highly penetrating (require several cm of lead or thick concrete) but cause less ionisation per unit path length than alpha or beta.

Alpha (α) Beta (β) Gamma (γ) paper STOPPED Al STOPPED Pb through Paper Aluminium Lead α: charge +2, mass 4 u, highest ionisation, lowest penetration β⁻: charge −1, mass ~0 u, moderate ionisation and penetration γ: charge 0, massless photon, lowest ionisation, highest penetration

Figure 1, Penetration comparison of alpha, beta and gamma radiation. Alpha is stopped by paper; beta by a few mm of aluminium; gamma requires several cm of lead.

Stop and check

Complete the decay equation: $^{238}_{92}\text{U} \rightarrow \; ^{234}_{90}\text{Th} + \; ?$ What type of decay is this?

Alpha (α): emits ⁴He, A−4, Z−2, stopped by paper. Beta-minus (β⁻): n→p+e⁻+ν̄_e, A unchanged, Z+1, stopped by Al. Beta-plus (β⁺): p→n+e⁺+ν_e, A unchanged, Z−1. Gamma (γ): excited nucleus emits photon, A and Z unchanged, requires lead. A and Z are conserved in every decay equation.

Write the table of decay types with changes to A and Z, you need to complete decay equations from memory in the HSC.

In alpha decay, the mass number of the daughter nucleus decreases by 4.

In beta-minus decay, the atomic number of the daughter nucleus decreases by 1.

Gamma decay does not change the mass number or atomic number of the nucleus.

2
The Mathematics of Decay
+5 XP

Exponential decay and half-life

We just saw the three decay modes and what they each emit. That raises a question: how quickly does a sample decay, and can we predict how many nuclei remain at any future time? This card answers it → the exponential law N = N₀e^(−λt) exactly predicts population at any time, and the half-life t₁/₂ = ln2/λ is the key measurable quantity.

Radioactive decay is a random process at the individual nucleus level, but the statistical behaviour of a large sample follows a precise exponential law:

$$N = N_0 e^{-\lambda t}$$

where $N$ is the number of remaining nuclei, $N_0$ is the initial number, $\lambda$ is the decay constant, and $t$ is time.

The half-life ($t_{1/2}$) is the time for half the nuclei to decay:

$$t_{1/2} = \dfrac{\ln 2}{\lambda} = \dfrac{0.693}{\lambda}$$

The activity ($A$) is the number of decays per second:

$$A = \lambda N = \lambda N_0 e^{-\lambda t} = A_0 e^{-\lambda t}$$

Activity is measured in becquerels (Bq), where 1 Bq = 1 decay per second. The older unit is the curie (Ci), where 1 Ci = $3.7\times10^{10}$ Bq.

After $n$ half-lives, the remaining fraction is $(1/2)^n$. For example, after 3 half-lives, $1/8$ of the original sample remains.

Carbon-14 dating: Living organisms maintain a constant $^{14}$C/$^{12}$C ratio. When they die, $^{14}$C decays with $t_{1/2} = 5{,}730$ years. Measuring the remaining $^{14}$C activity allows dating of organic materials up to ~50,000 years old.

Number of half-lives (n) N / N₀ 1 1/2 1/4 1/8 0 1 2 3 4 ½ ¼ ⅛ N = N₀ e^(−λt) = N₀(½)^n

Figure 2, Exponential decay curve: after each half-life the number of remaining nuclei halves. The curve approaches zero but never reaches it, there is always a non-zero probability of a nucleus surviving.

Radioactive Decay Formulae

$N = N_0 e^{-\lambda t}$, number of remaining nuclei

$t_{1/2} = \ln(2)/\lambda = 0.693/\lambda$, half-life

$A = \lambda N = A_0 e^{-\lambda t}$, activity (Bq)

$N = N_0(1/2)^{t/t_{1/2}}$, alternative half-life form

Stop and check

A sample has activity 800 Bq and half-life 6 hours. What is its activity after 24 hours? How many half-lives have passed?

Exponential decay: N = N₀e^(−λt), λ = ln2/t₁/₂. Half-life: t₁/₂ = 0.693/λ. Activity: A = λN (Bq). After n half-lives: fraction = (½)ⁿ. Carbon-14 dating: t₁/₂ = 5,730 yr; ¹⁴C/¹²C ratio falls after death; key assumption is atmospheric ratio constant.

Record the three forms of the decay equation and the half-life formula, switching between them cleanly is what separates Band 5 from Band 4 answers.

A sample has initial activity 400 Bq and half-life 2 hours. After 6 hours (3 half-lives), its activity in Bq is _____ Bq. Enter the numerical value.

HSC Tip and Misconceptions, Final Check
+5 XP
HSC Tip, Decay Calculations

A common exam trap: confusing activity with number of nuclei. Activity $A = \lambda N$ decreases exponentially just like $N$, but they are different quantities. When using $N = N_0(1/2)^{t/t_{1/2}}$, make sure $t$ and $t_{1/2}$ are in the same units. For decay equations, always check conservation of mass number ($A$) and atomic number ($Z$). In beta decay, an antineutrino ($\bar{\nu}_e$) is also emitted, don't forget it in complete equations. Gamma decay does not change $A$ or $Z$; it only releases energy. For carbon dating, the key assumption is that the atmospheric $^{14}$C/$^{12}$C ratio has been constant over time, this is approximately true but requires calibration.

Wrong: "Activity and number of nuclei are the same thing."
Right: $A = \lambda N$, activity equals the decay constant multiplied by the number of remaining nuclei. Both decrease exponentially, but they are different quantities with different units (Bq vs. nuclei).
Wrong: "After two half-lives, all atoms have decayed."
Right: After $n$ half-lives, the fraction remaining is $(1/2)^n$. It never reaches zero, radioactive decay is asymptotic. After 2 half-lives, 25% of nuclei remain.

Three of these statements about radioactive decay are correct. Pick the odd one out.

4
Conservation of Mass-Energy in Nuclear Transmutations
+5 XP

Where the released energy comes from: the mass defect

We just saw how to balance a decay equation and predict how many nuclei remain. That raises a question: a Geiger counter shows alpha particles flying out with kinetic energy, so where does that energy come from? This card answers it → the products of a nuclear reaction are very slightly lighter than the reactants, and that missing mass reappears as energy through $E = mc^2$.

Weigh the reactants of any nuclear reaction on an imaginary ultra-precise balance, then weigh the products. They do not match. A tiny amount of mass, the mass defect $\Delta m$, has vanished, and in its place an exactly equivalent amount of energy has appeared as kinetic energy of the products and gamma photons. Mass alone is not conserved, but mass-energy together is. This is the deeper meaning of every decay equation you have written.

A nuclear transmutation is any process that changes one nuclide into another. There are two kinds:

  • Spontaneous transmutation is radioactive decay: an unstable nucleus changes on its own, with no trigger, for example alpha decay $^{210}_{84}\text{Po} \rightarrow \; ^{206}_{82}\text{Pb} + \; ^4_2\text{He}$ or beta-minus decay $^{14}_6\text{C} \rightarrow \; ^{14}_7\text{N} + e^- + \bar{\nu}_e$.
  • Artificial (induced) transmutation happens when a nucleus is deliberately bombarded by a particle. Rutherford produced the first artificial transmutation in 1919 by firing alpha particles at nitrogen gas: $^4_2\text{He} + \; ^{14}_7\text{N} \rightarrow \; ^{17}_8\text{O} + \; ^1_1\text{H}$. Nuclear fission and fusion are also transmutations.

Whichever kind it is, two bookkeeping rules always hold. The total nucleon number $A$ (sum of top numbers) is conserved, and the total charge $Z$ (sum of bottom numbers) is conserved. On top of these, the total mass-energy is conserved. Writing the mass defect as the reactant mass minus the product mass:

$$\Delta m = m_{\text{reactants}} - m_{\text{products}}$$

the energy released (the Q-value) is found from Einstein's mass-energy equivalence:

$$E = \Delta m \, c^2 \qquad \text{or, using nuclear units,} \qquad E = \Delta m \times 931.5 \text{ MeV/u}$$

where $1\text{ u} = 1.66\times10^{-27}\text{ kg}$ is the atomic mass unit and $1\text{ u}$ of mass is equivalent to $931.5\text{ MeV}$ of energy.

Worked Example, Alpha Decay of Polonium-210

Reaction: $^{210}_{84}\text{Po} \rightarrow \; ^{206}_{82}\text{Pb} + \; ^4_2\text{He}$

Step 1, check conservation of A and Z. $A$: $210 = 206 + 4$ ✓. $Z$: $84 = 82 + 2$ ✓.

Step 2, masses (in u). Reactant: Po-210 $= 209.982874$. Products: Pb-206 $= 205.974465$, He-4 $= 4.002602$, so $m_{\text{products}} = 205.974465 + 4.002602 = 209.977067$.

Step 3, mass defect. $\Delta m = 209.982874 - 209.977067 = 0.005807\text{ u}$. The products are lighter than the parent, so mass is NOT conserved on its own.

Step 4, energy released. $E = 0.005807 \times 931.5 = 5.41\text{ MeV}$.

Check in joules: $E = \Delta m\, c^2 = (0.005807 \times 1.66\times10^{-27}) \times (3.0\times10^8)^2 = 8.7\times10^{-13}\text{ J}$, which matches $5.41\text{ MeV}$ (since $1\text{ MeV} = 1.6\times10^{-13}\text{ J}$).

The missing $0.005807\text{ u}$ of mass reappears as $5.41\text{ MeV}$ of kinetic energy shared by the recoiling Pb-206 nucleus and the alpha particle. Mass-energy is conserved.

Mass-energy balance: Po-210 alpha decay BEFORE (reactant) Po-210 nucleus 209.982874 u at rest AFTER (products) Pb-206 + He-4 209.977067 u released energy 5.41 MeV (KE + recoil) mass defect Δm = 209.982874 - 209.977067 = 0.005807 u E = Δm c² = 0.005807 × 931.5 = 5.41 MeV

Figure 3, The reactant is slightly heavier than the products. The missing mass (the mass defect) is converted to the kinetic energy of the products via $E = \Delta m\, c^2$, so mass-energy is conserved even though mass alone is not.

Stop and check

In beta-minus decay of carbon-14, $^{14}_6\text{C} \rightarrow \; ^{14}_7\text{N} + e^- + \bar{\nu}_e$, verify that both $A$ and $Z$ balance. If the products are lighter than the parent by $\Delta m$, where does the energy $\Delta m\, c^2$ go?

Transmutation = any change of nuclide. Spontaneous = radioactive decay (alpha, beta); artificial = nucleus bombarded by a particle (Rutherford: ⁴He + ¹⁴N → ¹⁷O + ¹H). In every reaction A and Z are conserved AND mass-energy is conserved. Mass alone is NOT conserved: the mass defect Δm = m(reactants) − m(products) reappears as energy. E = Δm c² = Δm × 931.5 MeV/u. Example: Po-210 alpha decay, Δm = 0.005807 u → 5.41 MeV.

Memorise the four-step method (balance A and Z, sum masses, find Δm, multiply by 931.5), and the conversion 1 u = 931.5 MeV. The HSC asks you to apply E = mc² to a transmutation almost every year.

In a nuclear transmutation the products are found to be slightly less massive than the reactants. This "missing" mass is best described as:

5
Exothermic and Endothermic Transmutations
+5 XP

The sign of the Q-value: when energy is released and when it must be supplied

We just saw a spontaneous alpha decay that released 5.41 MeV because the products were lighter than the parent. That raises a question: do all transmutations release energy? This card answers it → not necessarily. If the products are heavier than the reactants the reaction is endothermic and energy must be supplied, which is exactly why Rutherford needed fast alpha particles.

The released energy is called the Q-value, and its sign tells you everything. Define it from the mass defect, $Q = (m_{\text{reactants}} - m_{\text{products}})c^2$. A positive $Q$ means the products are lighter, mass-energy is released, and the reaction is exothermic, it can happen spontaneously (as in radioactive decay). A negative $Q$ means the products are heavier, so energy must be put in, and the reaction is endothermic, it cannot happen unless the bombarding particle supplies enough kinetic energy.

  • Exothermic ($Q > 0$): spontaneous alpha and beta decays, nuclear fission of heavy nuclei, and fusion of light nuclei. Energy emerges as kinetic energy of products and radiation.
  • Endothermic ($Q < 0$): many artificial transmutations. The kinetic energy of the incoming particle pays the mass-energy debt.

Rutherford's 1919 reaction is a clear endothermic case. Using atomic masses $^4_2\text{He} = 4.002602\text{ u}$, $^{14}_7\text{N} = 14.003074\text{ u}$, $^{17}_8\text{O} = 16.999132\text{ u}$ and $^1_1\text{H} = 1.007825\text{ u}$:

$$\Delta m = (4.002602 + 14.003074) - (16.999132 + 1.007825) = -0.001281 \text{ u}$$ $$Q = \Delta m \times 931.5 = -1.19 \text{ MeV}$$

The negative sign tells us the products are heavier, so this transmutation absorbs about $1.19\text{ MeV}$. The alpha particle must arrive with at least this much kinetic energy (more, once recoil is accounted for) for the reaction to proceed. This is precisely why a spontaneous source of fast alpha particles was needed, and why particle accelerators were later built to drive endothermic transmutations.

The sign of the Q-value Exothermic, Q > 0 products lighter than reactants reactant mass product mass energy released (decay, fission, fusion) Endothermic, Q < 0 products heavier than reactants reactant mass product mass energy supplied (e.g. Rutherford 1.19 MeV)

Figure 4, Sign convention for $Q = \Delta m\,c^2$. When products are lighter ($Q > 0$) energy is released and the reaction is exothermic; when products are heavier ($Q < 0$) energy must be supplied and the reaction is endothermic.

Stop and check

A proposed transmutation has $Q = -2.4\text{ MeV}$. Can it occur spontaneously? What is the minimum kinetic energy the bombarding particle must supply (ignoring recoil)?

Q-value Q = (m(reactants) − m(products))c². Q > 0: products lighter, energy released, EXOTHERMIC, can be spontaneous (decay, fission, fusion). Q < 0: products heavier, energy absorbed, ENDOTHERMIC, needs energy input. Rutherford ⁴He + ¹⁴N → ¹⁷O + ¹H has Δm = −0.001281 u, Q = −1.19 MeV (endothermic), so the alpha must be fast.

Always state the sign of Q and interpret it: positive means released and spontaneous-possible, negative means must be supplied. Markers reward the interpretation, not just the number.

For a transmutation, the calculated Q-value is negative ($Q < 0$). This tells you that:

Activities

Activity 1, Decay Equations and Half-life Calculations
ApplyBand 4

Practice writing decay equations and using $N = N_0(1/2)^n$

  1. Write the complete decay equation for $^{226}_{88}\text{Ra}$ undergoing alpha decay. Identify the daughter nucleus.
  2. Write the complete decay equation for $^{14}_6\text{C}$ undergoing beta-minus decay. Verify conservation of $A$ and $Z$.
  3. A sample of $^{32}\text{P}$ has a half-life of 14.3 days. If you start with $8.0 \times 10^{10}$ atoms, how many remain after 42.9 days?
  4. The activity of a sample drops from 3,200 Bq to 200 Bq in 4 hours. Calculate the half-life of the isotope.
Activity 2, Carbon Dating and Radiation Types
AnalyseBand 5

Apply decay concepts to real-world contexts

  1. A wooden artefact has $^{14}$C activity of 6.25 Bq/g. Fresh wood has activity 50 Bq/g. Given $t_{1/2}(^{14}\text{C}) = 5{,}730$ years, estimate the age of the artefact. Show your working clearly.
  2. Describe one medical application of radioactive isotopes. State the type of radiation emitted and explain why that type is chosen for this application.
  3. Compare the biological danger of alpha radiation from an external source versus an internal source. Use ionisation and penetration properties in your answer.
  4. A student claims that after 10 half-lives, "all the radioactive atoms will have decayed." Evaluate this claim and correct any error.