3 higher-level sections hidden.
Educerie · IB Diploma · Physics
Theme E Nuclear and quantum physics · E.3 Radioactive decay
What you must be able to do
| You must be able to | Level | What it looks like in the exam |
|---|---|---|
| Define isotopes and write them in nuclear notation | SL, HL | Paper 1A, or 1 mark inside a longer question |
| Define mass defect and binding energy, and calculate them using E = mc² with masses in kg, MeV c⁻² or u | SL, HL | "Calculate the binding energy per nucleon of…" (3 marks) |
| Sketch and interpret the binding energy per nucleon curve | SL, HL | "Explain, with reference to the graph, why energy is released in…" (2 to 3 marks) |
| Calculate the energy released in a nuclear reaction from masses | SL, HL | "Determine the energy released in this decay" (2 to 3 marks) |
| Describe the strong nuclear force: short-range, attractive, between nucleons | SL, HL | "State two properties of the strong nuclear force" (2 marks) |
| Describe decay as random and spontaneous | SL, HL | "Outline what is meant by random" (1 to 2 marks) |
| Write α, β⁻, β⁺ and γ decay equations, including neutrinos and antineutrinos | SL, HL | "Complete the decay equation" (2 marks) |
| Compare the penetration and ionising ability of α, β and γ | SL, HL | "Explain why an alpha source is used in…" (2 to 3 marks) |
| Use activity, count rate and half-life with a whole number of half-lives, and correct for background | SL, HL | Paper 1B: "Determine the half-life from the data" (3 marks) |
| Choose an isotope for a medical use, pipe leaks, thickness control or dating, and justify it | SL, HL | "Suggest why this isotope is suitable" (3 marks) |
| Give the evidence for the strong force, and use N/Z to explain stability | HL only | "Explain why heavy stable nuclei have more neutrons than protons" (2 to 3 marks) |
| Explain why the binding energy per nucleon is roughly constant above A ≈ 60 | HL only | "Explain the shape of the graph for large A" (2 marks) |
| Explain how α and γ spectra show nuclear energy levels, and how the β spectrum shows the neutrino | HL only | "Explain how the beta energy spectrum is evidence for the neutrino" (3 marks) |
| Use N = N₀ exp(−λt), A = λN and T½ = ln 2/λ for any time interval, and explain λ as a probability | HL only | "Determine the age of the sample" (3 to 4 marks) |
Before you start
You need nuclear notation and the nuclear model from E.1: Z protons, A nucleons, A − Z neutrons. You need the electronvolt (1 MeV = 1.60 × 10⁻¹³ J) and the electric force between charges. HL students need the natural logarithm and the exponential function; if ln and e are rusty, look at them before section 11.
1The idea in one paragraph
Put protons and neutrons together into a nucleus and the result weighs less than the parts did apart. The missing mass has become energy, released when the nucleus formed, and the same energy would be needed to pull it apart again: the binding energy, E = mc². Nuclei in the middle of the periodic table, around iron, are the most tightly bound. Some combinations of protons and neutrons are unstable, and those nuclei change on their own, at random, by emitting an alpha particle, a beta particle or a gamma ray. Which nucleus decays next cannot be predicted, but a sample of billions follows a precise rule: in each half-life, half of what is left decays. That rule is what makes radioactive isotopes useful in hospitals, in industry and for dating the past.
2Isotopes and the force that holds a nucleus together
Isotopes are nuclei of the same element with different numbers of neutrons: the same proton number Z, a different nucleon number A. Carbon-12 (¹²₆C) and carbon-14 (¹⁴₆C) both have 6 protons; carbon-14 has 8 neutrons instead of 6. Isotopes are chemically identical, because chemistry depends on the electrons and so on Z, but their nuclei can behave very differently: carbon-12 is stable and carbon-14 is radioactive.
A nucleus poses an obvious question. Protons are positive, and at a separation of about 10⁻¹⁵ m they repel each other with a force of hundreds of newtons. Gravity between them is far too weak to help: the electric force between two protons is about 10³⁶ times their gravitational attraction. If only those two forces existed, no nucleus with more than one proton could hold together. Something else must pull nucleons together, and the guide asks you to know it exists.
The strong nuclear force acts between nucleons (proton–proton, proton–neutron, neutron–neutron). It is attractive and very short-range, about 10⁻¹⁵ m, and inside that range it is much stronger than the electric force.
Its short range explains a lot. Each nucleon is held only by its nearest neighbours, while every proton repels every other proton in the nucleus. That imbalance is why very large nuclei are unstable, and why stable nuclei need extra neutrons, which add strong-force attraction without adding repulsion (section 9).
3Mass defect and binding energy
Weigh a helium-4 nucleus and weigh its parts, two protons and two neutrons, separately. The nucleus is lighter.
Mass defect Δm = (total mass of the separate nucleons) − (mass of the nucleus)
Binding energy EB = Δm c², the energy needed to separate a nucleus completely into its individual nucleons (and the energy released when it forms from them)
This is mass–energy equivalence, E = mc²: mass is a form of energy, and in nuclear reactions the changes of mass are large enough to measure. Nuclear masses are given in one of three units, and you must be able to use all of them:
| Unit | Value | When to use it |
|---|---|---|
| kilogram | the SI unit | use E = mc² with c = 3.00 × 10⁸ m s⁻¹, answer in J |
| unified atomic mass unit, u | 1 u = 1.661 × 10⁻²⁷ kg | masses of particles and nuclei in tables |
| MeV c⁻² | 1 u = 931.5 MeV c⁻² | multiply a mass in u by 931.5 to get its energy in MeV directly |
The unit MeV c⁻² is a mass: a mass of 1 MeV c⁻² has an energy of 1 MeV. It turns E = mc² into a multiplication by 931.5.
Worked example. Find the binding energy per nucleon of helium-4. Mass of the ⁴₂He nucleus = 4.001506 u; mp = 1.007276 u; mn = 1.008665 u (data booklet).
The same in SI units: Δm = 0.030376 × 1.661 × 10⁻²⁷ = 5.045 × 10⁻²⁹ kg, and EB = Δmc² = 5.045 × 10⁻²⁹ × (3.00 × 10⁸)² = 4.54 × 10⁻¹² J, which is 28.4 MeV. The last figure differs only because the constants are rounded.
Keep enough decimal places. The mass defect is a small difference between two large numbers, and rounding the masses to three figures first would make it vanish.
4The binding energy per nucleon curve
Total binding energy grows with the size of the nucleus, so it says little on its own. What measures how tightly a nucleus is held is the binding energy per nucleon: the average energy needed to remove one nucleon. Figure 1 plots it against nucleon number.
Read four features from it.
- It rises steeply for light nuclei: hydrogen-2 has about 1.1 MeV per nucleon, carbon-12 about 7.7 MeV.
- Helium-4 sits above the curve around it (7.07 MeV). It is an unusually tightly bound nucleus, which is one reason it is the nucleus thrown out in alpha decay.
- It peaks around A = 56 to 62, near iron and nickel, at about 8.8 MeV per nucleon. These are the most stable nuclei.
- It falls slowly for heavier nuclei, to about 7.6 MeV per nucleon for uranium.
Why energy is released. A nuclear reaction releases energy whenever the products are more tightly bound than what went in: higher binding energy per nucleon means the nucleons have fallen into a deeper hole, and the difference comes out as kinetic energy of the products and as gamma photons. So there are two routes up the curve:
- Fusion: joining light nuclei (left of the peak) into a heavier one, as in stars (E.5).
- Fission: splitting a very heavy nucleus (right of the peak) into two middle-sized ones, as in reactors (E.4).
- Alpha decay of a heavy nucleus is also a step up the curve, which is why it happens by itself.
Nothing is gained by fusing iron or splitting it: iron is already at the top.
Worked example: energy from a decay. Radium-226 decays by alpha emission to radon-222. Nuclear masses: ²²⁶₈₈Ra 225.977135 u, ²²²₈₆Rn 221.970400 u, ⁴₂He 4.001506 u.
The mass after is less than the mass before, so energy is released, mostly as the kinetic energy of the alpha particle. If a question gives atomic masses instead of nuclear ones, check that the electrons balance on both sides; in alpha decay they do (88 = 86 + 2).
5Radioactive decay: what a nucleus can do
An unstable nucleus changes into a more stable one by emitting radiation. This is radioactive decay, and it has two properties the guide names.
- Spontaneous: nothing outside triggers it. Heating, pressure or chemical reactions do not change the rate, because they affect electrons and not the nucleus.
- Random: it is impossible to predict which nucleus will decay next, or when a given nucleus will decay. Each nucleus has a fixed chance of decaying in the next second, and that chance does not change with the nucleus's age.
The randomness shows directly in a counter: the clicks from a source are irregular, and repeated one-minute counts vary around an average.
There are three kinds of decay. In each, nucleon number and charge are both conserved: the top numbers balance, and the bottom numbers balance, on both sides of the equation. Figure 2 shows what each one does to a nucleus's numbers of protons and neutrons.
Alpha decay. The nucleus emits an alpha particle, a helium-4 nucleus ⁴₂α, which takes 2 protons and 2 neutrons with it. A falls by 4 and Z falls by 2.
parent (A, Z) → daughter (A − 4, Z − 2) + ⁴₂α, for example ²²⁶₈₈Ra → ²²²₈₆Rn + ⁴₂α
Beta-minus decay. A neutron in the nucleus changes into a proton, and an electron and an antineutrino are created and emitted. The electron is the beta-minus particle, ⁰₋₁β. A is unchanged; Z rises by 1.
n → p + ⁰₋₁β + ν̄, for example ¹⁴₆C → ¹⁴₇N + ⁰₋₁β + ν̄
Beta-plus decay. A proton changes into a neutron, and a positron (the electron's antiparticle, same mass, charge +e) and a neutrino are emitted. A is unchanged; Z falls by 1.
p → n + ⁰₊₁β + ν, for example ¹⁸₉F → ¹⁸₈O + ⁰₊₁β + ν
Gamma decay. After an alpha or beta decay the new nucleus is often left in an excited state, with extra energy. It drops to a lower state by emitting a gamma photon, a high-energy electromagnetic wave. Neither A nor Z changes; the nucleus simply loses energy. An asterisk marks the excited nucleus.
⁶⁰₂₈Ni* → ⁶⁰₂₈Ni + γ
Neutrinos and antineutrinos (ν and ν̄) have no charge and almost no mass, and they interact so weakly with matter that almost all of them pass straight through the Earth. They are there because energy and momentum must be conserved in beta decay (section 10). In an equation they carry 0 for both numbers; you must still write them. The rule to remember: β⁻ comes with an antineutrino, β⁺ with a neutrino.
Worked example. Complete: ²³⁸₉₂U → ? + ⁴₂α. Nucleon numbers: 238 = A + 4, so A = 234. Proton numbers: 92 = Z + 2, so Z = 90. The product is ²³⁴₉₀Th (thorium; the question gives the symbol). Then ²³⁴₉₀Th decays by β⁻: A stays 234, Z becomes 91, giving ²³⁴₉₁Pa + ⁰₋₁β + ν̄.
6Penetration and ionisation
As alpha, beta and gamma radiation pass through matter they knock electrons off atoms, creating ions. That is ionisation, and it is how radiation both damages living cells and is detected. Each ionisation takes a little of the radiation's energy, so the more strongly a radiation ionises, the sooner it runs out of energy, and the less far it penetrates. Figure 3 shows the result.
| Alpha α | Beta β | Gamma γ | |
|---|---|---|---|
| What it is | helium nucleus, charge +2e | electron (β⁻) or positron (β⁺), charge ∓e | high-energy photon, no charge |
| Ionising ability | strong | moderate | weak |
| Range in air | a few cm | tens of cm to about a metre | very large; intensity falls gradually |
| Stopped by | a sheet of paper, or the outer layer of skin | a few mm of aluminium | never completely; reduced by several cm of lead or thick concrete |
Alpha particles are slow and doubly charged, so they interact with almost every atom they pass. Gamma photons are uncharged and interact only occasionally. That is why an alpha source is harmless outside the body but very dangerous if swallowed or breathed in, while gamma is the reverse: it reaches you from a distance, but does less damage per centimetre.
7Activity, count rate and half-life
The activity A of a source is the number of nuclei that decay per second. Its unit is the becquerel: 1 Bq = 1 decay per second. A detector such as a Geiger–Müller tube records a count rate, the number of decays it detects per second or per minute. Count rate is always less than the activity, because radiation goes out in all directions and only some of it enters the detector, and not every particle that enters is counted. But count rate is proportional to activity, so it follows the same pattern.
Half-life T½ is the time taken for half of the radioactive nuclei in a sample to decay, which is also the time for the activity (and the count rate) to halve. It is the same for every sample of a given isotope, whatever its size or starting activity. After n half-lives, the fraction remaining is (½)ⁿ:
| Half-lives passed | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| Fraction of nuclei (or activity) left | 1 | ½ | ¼ | ⅛ | 1/16 |
| Fraction decayed | 0 | ½ | ¾ | ⅞ | 15/16 |
Worked example. Iodine-131 has a half-life of 8.0 days. A hospital receives a sample with an activity of 800 kBq. What is its activity 24 days later, and what fraction of the nuclei have decayed?
Background radiation. A detector counts even with no source present, because radiation is everywhere: radon gas from rocks and soil, cosmic rays, rocks and building materials, food (potassium-40), and medical sources. Before using any count rate, measure the background with the source removed and subtract it from every reading. The background does not halve with the source, so leaving it in makes the count rate level off above zero and makes the half-life look longer. Figure 4 shows both curves.
Determining a half-life. For an isotope whose half-life is minutes to days, measure the count rate at regular intervals, subtract the background, plot corrected count rate against time, and read off the time for the count rate to halve. Do this from several different starting points and average them, because the random scatter in the data makes any one reading uncertain. Because decay is random, counts should also be taken over long enough intervals that each count is large: the larger the count, the smaller its fractional scatter. HL adds a better method using a log graph (section 11). For an isotope with a very long half-life, the count rate hardly changes during an experiment, so instead you measure the activity and the number of nuclei and use A = λN (HL).
8Choosing an isotope for the job
The guide names four uses. In each, the choice comes down to two questions: which radiation (from its penetration and ionisation) and what half-life (long enough to do the job, short enough not to linger or, for dating, matched to the age being measured).
Medical tracers. A patient is given a small amount of a radioactive isotope that collects in the organ being studied, and a gamma camera outside the body images it. Technetium-99m is the most widely used. It emits gamma only, so the radiation escapes the body to reach the camera and, being weakly ionising, does little damage on the way. Its half-life is about 6 hours: long enough to prepare the dose and complete the scan, short enough that the activity in the patient falls quickly afterwards. An alpha emitter would be useless here, since nothing would leave the body, and dangerous, since all its ionisation would happen inside.
Leaks in underground pipes. A small amount of a gamma emitter (sometimes a beta emitter for shallow pipes) is added to the liquid in the pipe. A detector is carried along the ground above. Where the pipe leaks, the tracer collects in the soil and the count rate rises. Gamma is needed to get through the soil. The half-life should be hours to a few days, so the source lasts through the search but does not stay in the ground or the water supply. Sodium-24, with a half-life of about 15 hours, is a standard choice.
Thickness control. In a factory making paper, plastic film or aluminium foil, a source sits on one side of the sheet and a detector on the other, as in Figure 5. If the sheet becomes thicker it absorbs more radiation and the count rate falls; the controller then presses the rollers together. For paper the source is beta: alpha would be stopped completely by any paper, and gamma would pass straight through with almost no change, so neither would respond to small changes in thickness. For thick steel sheet, gamma is used. The half-life must be long (years to decades, such as strontium-90 at about 29 years), so that a falling count rate means a thicker sheet and not a weakening source.
Radioactive dating. Living things take in carbon, including a small, steady proportion of carbon-14. When they die, no new carbon-14 comes in and what is there decays with a half-life of 5730 years. Comparing the activity per gram of an old sample with that of living material gives its age. If a wooden tool has one quarter of the carbon-14 activity per gram of new wood, two half-lives have passed: it is about 11 500 years old. Carbon-14 dating works for ages up to a few tens of thousands of years; beyond that too little is left to measure. For rocks, isotopes with half-lives of millions or billions of years are used, such as potassium-40 (1.25 × 10⁹ years). Match the half-life to the age: an isotope with a short half-life has vanished from an old sample, and one with a very long half-life has barely changed in a young one.
9HLWhy some nuclei are stable: the strong force and N/Z
SL students can skip to section 12.
The evidence for the strong nuclear force. The guide asks for evidence, not just the claim.
- Nuclei exist. Protons 10⁻¹⁵ m apart repel strongly; gravity is 10³⁶ times too weak to balance that. A third, attractive force must act.
- It is short-range. Nuclear density is the same for all nuclei (E.1), and the binding energy per nucleon levels off for large A (below): each nucleon is bound only to its nearest neighbours. And Rutherford scattering matches a purely electric force until the alpha particle reaches the nuclear surface (E.1): the extra force switches on only at about 10⁻¹⁵ m.
- It becomes repulsive at very short range. Nucleons do not collapse into each other; the constant density shows they keep a minimum spacing.
- It acts on neutrons too. Neutrons are uncharged, yet they are bound in nuclei; the deuteron, one proton and one neutron, is held together by something other than electricity.
The ratio of neutrons to protons. Figure 6 plots every known stable nucleus by its number of neutrons N against its number of protons Z. The stable nuclei lie in a narrow band of stability.
For light nuclei the band follows N = Z: helium-4, carbon-12 and oxygen-16 all have equal numbers. For heavier nuclei it bends upwards, and lead-208 has N/Z = 126/82 = 1.54. The reason is the difference in range between the two forces. The strong force pulls each nucleon only towards its near neighbours, but the electric repulsion acts between every pair of protons in the nucleus. As Z grows, the total repulsion grows faster than the attraction, and extra neutrons are needed to add attraction without adding charge. Beyond Z = 82 no amount of extra neutrons is enough, and every nucleus is unstable.
The figure also predicts how an unstable nucleus decays: it moves towards the band.
- Above the band (too many neutrons): β⁻ decay turns a neutron into a proton, moving one step down and to the right.
- Below the band (too many protons): β⁺ decay turns a proton into a neutron, moving up and to the left.
- Very heavy nuclei (A above about 200): α decay removes two protons and two neutrons at once, the quickest way to lose size.
Why the binding energy curve is nearly flat above A ≈ 60. Beyond about 60 nucleons, the binding energy per nucleon stays close to 8 to 9 MeV, falling only slowly (Figure 1). If every nucleon attracted every other one, adding nucleons would keep adding binding energy per nucleon. It does not, because the strong force saturates: once a nucleon is surrounded by its nearest neighbours, adding more nucleons further away gives it no extra binding. So each nucleon contributes roughly the same amount, and the total binding energy is roughly proportional to A. The slow fall at large A is the electric repulsion, which, acting between every pair of protons, grows faster than A.
10HLNuclear energy levels and the neutrino
Alpha and gamma spectra show discrete nuclear energy levels. Measure the energies of the alpha particles from one isotope and they are not spread out: they come at one or a few sharp values. The gamma photons emitted by the daughter nucleus also have sharp energies. Just as line spectra showed discrete atomic levels in E.1, these show that a nucleus has discrete energy levels. Figure 7 draws an example with invented values.
A parent nucleus decays by alpha emission either straight to the daughter's ground state, giving alpha particles of 5.30 MeV, or to an excited state 0.20 MeV above it, giving alpha particles of 5.10 MeV. The excited daughter then emits a 0.20 MeV gamma photon. The alpha energies differ by exactly the gamma energy, which is the signature of a level. (A small share of the energy goes to the recoiling daughter nucleus; here it is ignored.) Gamma photons from the nucleus are the same kind of photon as those from atomic transitions, but the nuclear levels are MeV apart instead of eV, so the photons are around a million times more energetic.
The beta spectrum is continuous, and that is evidence for the neutrino. Beta particles from one isotope come out with every energy from nearly zero up to a maximum, as in Figure 8. That was a crisis in the 1920s. If a nucleus with a fixed energy turned into a daughter with a fixed energy by emitting only an electron, the electron would always carry the same energy, as alpha particles do. Some energy seemed to disappear, and some momentum too.
In 1930 Wolfgang Pauli proposed that conservation of energy and momentum are right, and that a third, unseen particle carries off the missing share. That is the (anti)neutrino. The energy released in each decay is fixed, but it is shared between the electron and the antineutrino in different proportions each time, so the electron's energy varies. The maximum on the graph is the case where the antineutrino takes almost nothing. Because the neutrino interacts so weakly, it was not detected directly until 1956, by Cowan and Reines, using the huge flux from a nuclear reactor. Conservation laws predicted a particle a quarter of a century before anyone saw one.
11HLThe decay law
At SL you worked in whole half-lives. HL needs any time at all. Because each nucleus has the same fixed probability of decaying per unit time, the number decaying per second is proportional to the number still there. That gives exponential decay:
N = N₀ exp(−λt), A = λN = λN₀ exp(−λt), T½ = ln 2/λ
exp(−λt) means e raised to the power −λt; the data booklet prints it as a superscript.
N is the number of undecayed nuclei at time t and N₀ the number at t = 0. λ is the decay constant, in s⁻¹ (or h⁻¹, y⁻¹: any time unit, used consistently). A = λN says that activity is proportional to the number of nuclei, which is why activity and count rate decay with the same half-life as N. Putting N = N₀/2 at t = T½ gives exp(−λT½) = ½, so λT½ = ln 2.
λ and probability. The decay constant is the probability that a given nucleus decays in unit time, but only approximately, and only when λt is small. The exact probability of decay within time t is 1 − exp(−λt). For λ = 0.10 s⁻¹, the probability of decay within 1 s is 1 − exp(−0.10) = 0.095, close to λ × 1 s = 0.10. Within 10 s the exact probability is 1 − exp(−1) = 0.63, nowhere near λ × 10 s = 1.0. λt is a good estimate of the probability only while λt ≪ 1.
Worked example: activity from mass. Find the activity of 1.0 μg of iodine-131 (T½ = 8.02 days, molar mass 131 g mol⁻¹).
Worked example: any time interval. What fraction of an iodine-131 sample remains after 20 days?
About 18% remains. Check: 20 days is 2.5 half-lives, and ½ raised to the power 2.5 is 0.177.
Worked example: dating. A piece of charcoal has 35% of the carbon-14 activity per gram of living wood. T½ = 5730 years. How old is it?
Finding the half-life with a log graph. Take natural logs of A = A₀ exp(−λt):
ln A = ln A₀ − λt
A graph of ln(corrected count rate) against t is a straight line with gradient −λ, as in Figure 9. It uses every data point at once, shows up any point that does not fit, and gives T½ = ln 2/λ from the gradient. This is the Paper 1B method. The same exponential shape appears wherever a rate of change is proportional to the amount present, as in the absorption of gamma rays in a thickness of lead.
12Where marks are lost
- Adding the masses the wrong way round. Mass defect is (parts) − (nucleus), and it is positive. A negative mass defect means the subtraction is reversed.
- Rounding masses before subtracting. Keep all the given decimal places until Δm is found.
- Confusing binding energy with energy stored in the nucleus. Binding energy is the energy needed to pull the nucleus apart. More binding energy per nucleon means more stable, not "more energy to release".
- Leaving out the neutrino, or putting the wrong one in. β⁻ decay emits an antineutrino; β⁺ decay emits a neutrino. A decay equation without it loses the mark.
- Saying gamma decay changes the element. Gamma emission changes neither A nor Z; the nucleus only loses energy.
- Treating half-life as half of the lifetime. After two half-lives a quarter is left, not none.
- Forgetting background. Subtract it from every reading before halving anything.
- Using λ = probability of decay for a long interval (HL). It is only a good approximation when λt is small.
13Draw it right
- Binding energy per nucleon: axes "nucleon number A" and "binding energy per nucleon / MeV"; a steep rise, helium-4 above the trend, a broad peak near A = 56 at about 8.8 MeV, a gentle fall to about 7.6 MeV at A = 238. Label fusion to the left of the peak and fission to the right.
- Decay curve: start at N₀ (or A₀), mark N₀/2, N₀/4, N₀/8 with dashed lines to the time axis at equal intervals T½, 2T½, 3T½. The curve approaches the axis and never reaches it.
- Background: the uncorrected curve levels off at the background line, not at zero.
- Penetration diagram: α stopped by paper, β by a few mm of aluminium, γ reduced by lead but still continuing.
- HL, N against Z: N on the vertical axis, the N = Z line dashed, the band bending above it; β⁻ arrows down-right from above the band, β⁺ arrows up-left from below, α arrows down-left at the top end.
- HL, beta spectrum: a smooth curve from zero to a sharp maximum energy, falling to zero at Emax; alpha spectrum as one or a few sharp lines.
- HL, ln A against t: a straight line with negative gradient −λ; the intercept is ln A₀.
14Try it
Marks in brackets. mp = 1.007276 u, mn = 1.008665 u, 1 u = 931.5 MeV c⁻². Answers and marker's notes follow.
Q1. The mass of an iron-56 nucleus (⁵⁶₂₆Fe) is 55.9207 u. Calculate its binding energy per nucleon, in MeV. 3 marks
Q2. Complete each decay equation, including any neutrino or antineutrino. 3 marks
(a) ²³⁸₉₂U → Th + ⁴₂α 1 mark
(b) ¹⁴₆C → N + ⁰₋₁β + … 1 mark
(c) ²²₁₁Na → Ne + ⁰₊₁β + … 1 mark
Q3. (Data-based, Paper 1B style. Invented data.) A student measures the count rate from a source. With the source removed, the background count rate is 30 counts per minute. 6 marks
| Time / min | 0 | 2 | 4 | 6 | 8 | 10 |
|---|---|---|---|---|---|---|
| Count rate / counts per minute | 830 | 534 | 347 | 230 | 156 | 109 |
(a) Determine the half-life of the source. 3 marks
(b) Explain why the background must be subtracted. 1 mark
(c) Repeating the experiment gives slightly different count rates. Suggest why. 2 marks
Q4. A hospital chooses an isotope to image a patient's kidneys with a camera outside the body. Explain why the isotope should emit gamma radiation only and have a half-life of a few hours. 4 marks
Q5 (HL). Carbon-14 has a half-life of 5730 years. A bone has a carbon-14 activity per gram equal to 22% of that of living bone. 4 marks
(a) Calculate the decay constant of carbon-14, in y⁻¹. 1 mark
(b) Determine the age of the bone. 3 marks
Q6 (HL). Explain how the energy spectrum of beta particles provides evidence for the existence of the neutrino. 3 marks
15In one breath
Isotopes share a proton number but not a nucleon number. A nucleus weighs less than its separate nucleons; the mass defect times c² is the binding energy, the energy to pull the nucleus apart, with 1 u = 931.5 MeV c⁻² turning masses straight into MeV. Binding energy per nucleon rises steeply, peaks near iron at about 8.8 MeV and falls slowly, so fusing light nuclei and splitting heavy ones both release energy. Nucleons are held by the strong nuclear force: attractive, short-range, between all nucleons. Decay is random and spontaneous: α removes 2 protons and 2 neutrons, β⁻ turns a neutron into a proton with an electron and an antineutrino, β⁺ turns a proton into a neutron with a positron and a neutrino, γ removes only energy. Alpha ionises most and is stopped by paper, beta by millimetres of aluminium, gamma is only reduced by lead. In each half-life the activity halves; subtract background first. Choose the radiation by its penetration and the half-life by how long the job lasts. HL: nuclei need extra neutrons as Z grows because repulsion acts between all protons and the strong force only between neighbours; the strong force saturates, so binding energy per nucleon is nearly constant above A ≈ 60; sharp α and γ energies show nuclear levels, the continuous β spectrum demands the neutrino; N = N₀ exp(−λt), A = λN, T½ = ln 2/λ, and λ is the probability of decay per unit time only when λt is small.
Answers
Q1. Mass of parts = 26 × 1.007276 + 30 × 1.008665 = 26.189176 + 30.259950 = 56.449126 u. Δm = 56.449126 − 55.9207 = 0.528426 u. EB = 0.528426 × 931.5 = 492.2 MeV. Per nucleon = 492.2 ÷ 56 = 8.79 MeV. M1 for the mass of 26 protons and 30 neutrons, M1 for Δm × 931.5, A1 for 8.79 MeV (accept 8.8). Dividing by 26 instead of 56 scores M1 M1 A0.
Q2. (a) ²³⁸₉₂U → ²³⁴₉₀Th + ⁴₂α. (b) ¹⁴₆C → ¹⁴₇N + ⁰₋₁β + ν̄. (c) ²²₁₁Na → ²²₁₀Ne + ⁰₊₁β + ν. A1 for each complete line; (b) needs the antineutrino and (c) the neutrino. Swapping ν and ν̄ loses that mark.
Q3. (a) Corrected count rates: 800, 504, 317, 200, 126, 79 counts per minute. The count rate falls from 800 to 400 in about 3 minutes, and from 400 to 200 between about 3 and 6 minutes (and 200 is reached at exactly 6 min, a quarter of 800). T½ = 3.0 min. (b) The background is not from the source and does not decay, so leaving it in would make the count rate fall more slowly and give too long a half-life. (c) Radioactive decay is random, so the number of decays in any interval varies about an average; the background also fluctuates. (a) M1 for subtracting 30 from every reading, M1 for finding the time to halve from the corrected values or a graph, A1 for 3.0 min (accept 2.8 to 3.2). (b) R1. (c) R1 for random decay, R1 for variation about a mean (or background fluctuating). Halving the uncorrected 830 gives about 3.2 min and scores M0 M1 A0.
Q4. Gamma radiation is weakly ionising and very penetrating, so it passes out of the body to reach the camera, and it causes little ionisation in the patient's cells along the way. Alpha or beta radiation would be absorbed inside the body: none would reach the camera, and all its ionising energy would be deposited in the tissue. A half-life of a few hours is long enough for the isotope to be prepared, given and imaged before the activity has fallen too far, but short enough that the activity in the patient becomes small within a day or so, limiting the dose. R1 for gamma escaping the body to the detector, R1 for low ionisation so less damage (or alpha/beta absorbed and damaging), R1 for the half-life long enough for the procedure, R1 for short enough to limit the patient's exposure.
Q5 (HL). (a) λ = ln 2 ÷ 5730 = 1.21 × 10⁻⁴ y⁻¹. (b) A/A₀ = exp(−λt) = 0.22, so t = ln(1/0.22) ÷ λ = 1.514 ÷ 1.21 × 10⁻⁴ = 1.25 × 10⁴ years. (a) A1. (b) M1 for A/A₀ = 0.22 = exp(−λt), M1 for taking logs correctly, A1 for 1.25 × 10⁴ years (accept 1.2 to 1.3 × 10⁴). Using 0.78 instead of 0.22 gives 2.1 × 10³ years and scores M0 M1 A0.
Q6 (HL). In a given beta decay the energy released is fixed, because the parent and daughter nuclei have fixed energies. If only the electron were emitted, every beta particle would have the same energy. Instead, beta particles have a continuous range of energies up to a maximum. So the energy must be shared with another particle, emitted at the same time, that is not detected: the (anti)neutrino. The maximum beta energy corresponds to the neutrino taking almost none. R1 for the fixed energy released, R1 for the observed continuous spectrum contradicting a two-body decay, R1 for a third particle sharing the energy (conservation of energy, and of momentum). A statement that "the neutrino exists" with no reasoning scores 0.
Educerie · written from the published IB Diploma Programme Physics guide, first assessment 2025, section E.3 Radioactive decay. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.