IB DP Physics: Nuclear structure, binding energy and types of radioactive decay (DP Physics)

This guide covers the first part of sub-topic E.3, Radioactive decay, in the IB Physics guide (first assessment 2025): nuclear structure, binding energy and the types of decay. The core understandings apply to SL and HL. Five further understandings are additional higher level, and each is marked “HL only”. Activity, count rate, half-life and the decay law are the second part of E.3 and are not taught here. Use the revision notes for recall and the practice questions for testing. Find your weak spots first with the free 10-minute diagnostics. See also the course hub and the printable checklist.

What this unit covers

Part	What you must be able to do	Tier
E.3	Describe isotopes	SL and HL
E.3	Use nuclear binding energy and mass defect	SL and HL
E.3	Interpret how binding energy per nucleon varies with nucleon number	SL and HL
E.3	Apply mass-energy equivalence to nuclear reactions	SL and HL
E.3	Describe the strong nuclear force	SL and HL
E.3	Explain why decay is random and spontaneous	SL and HL
E.3	Write and interpret alpha, beta-minus, beta-plus and gamma changes and equations	SL and HL
E.3	Recognise neutrinos and antineutrinos	SL and HL
E.3	Compare penetration and ionizing ability of the three radiations	SL and HL
E.3	Use masses in kg, MeV/c² and u	SL and HL
E.3	Evidence for the strong force	HL only
E.3	Neutron-to-proton ratio and stability	HL only
E.3	Approximate constancy of binding energy per nucleon above A = 60	HL only
E.3	Discrete nuclear energy levels from alpha and gamma spectra	HL only
E.3	Continuous beta spectrum as evidence for the neutrino	HL only

The guide leaves the weak nuclear force out of this course.

Isotopes and nuclide notation

A nuclide is written with nucleon number A (protons plus neutrons) and proton number Z. The neutron number is N = A − Z. Isotopes of an element have the same Z and different N, so different A. They have the same electron arrangement, so the same chemistry, but different nuclear mass and often different stability.

Worked example 1. Magnesium has Z = 12. Find N for magnesium-24, -25 and -26.

N = A - Z
Mg-24: 24 - 12 = 12
Mg-25: 25 - 12 = 13
Mg-26: 26 - 12 = 14

Same Z, three values of N: three isotopes of one element.

Mass units

Nuclear masses are given in three ways, and you must convert between them.

Unit	Meaning	Conversion
kg	SI unit	1 u = 1.661 × 10⁻²⁷ kg
u	unified atomic mass unit, one twelfth of the mass of a carbon-12 atom	1 u = 931.5 MeV/c²
MeV/c²	mass whose energy equivalent is stated in MeV	1 MeV/c² = 1.78 × 10⁻³⁰ kg

The third line comes from m = E/c² with 1 MeV = 1.602 × 10⁻¹³ J and c = 3.00 × 10⁸ m s⁻¹. Rounded constants give a slightly different figure for the energy of 1 u (about 933 MeV). Use the conversion you are given and do not switch routes mid-question.

Mass defect and binding energy

A bound nucleus has less mass than its separate nucleons. The mass defect Δm is the mass of the separated protons and neutrons minus the mass of the nucleus. The binding energy is the energy equivalent of that missing mass, ΔE = Δmc². It is the energy needed to separate the nucleus completely into free nucleons. A larger binding energy per nucleon means a more tightly bound nucleus.

To use tabulated atomic masses, count a hydrogen atom (mass 1.007825 u) for each proton. The electrons then cancel.

Worked example 2. Calcium-40 (Z = 20, N = 20) has atomic mass 39.962591 u. Take m(hydrogen atom) = 1.007825 u and m(neutron) = 1.008665 u. Find the mass defect, the binding energy and the binding energy per nucleon.

separated masses = 20 x 1.007825 + 20 x 1.008665
                 = 20.15650 + 20.17330 = 40.32980 u
mass defect      = 40.32980 - 39.962591 = 0.367209 u
binding energy   = 0.367209 x 931.5 = 342 MeV
per nucleon      = 342 / 40 = 8.55 MeV

In kg the mass defect is 0.367209 × 1.661 × 10⁻²⁷ = 6.10 × 10⁻²⁸ kg. As a mass it is 342 MeV/c². The three numbers describe one quantity.

The binding energy per nucleon curve

Plot binding energy per nucleon against nucleon number A. You need to interpret it, not just recall it.

Region	What the curve does	Meaning
Very light (A below about 20)	Rises steeply, with helium-4 standing above its neighbours	Adding nucleons helps a great deal
A about 50 to 65	Broad maximum, about 8.8 MeV per nucleon, near iron and nickel	Most tightly bound nuclei
A above about 65	Falls slowly, to about 7.6 MeV per nucleon for uranium	Heavy nuclei are less tightly bound

Energy is released in any change that moves nucleons to higher binding energy per nucleon. Joining light nuclei (to the left of the peak) does this, and so does splitting a very heavy nucleus (to the right of it). Afterwards the nucleons are more tightly bound overall and the system has less mass; the missing mass appears as kinetic energy and radiation.

Worked example 3. Compare iron-56 (atomic mass 55.934936 u, Z = 26) with calcium-40 from example 2.

separated = 26 x 1.007825 + 30 x 1.008665 = 26.20345 + 30.25995 = 56.46340 u
defect    = 56.46340 - 55.934936 = 0.528464 u
BE        = 0.528464 x 931.5 = 492 MeV
per nucleon = 492 / 56 = 8.79 MeV

Iron-56 has 8.79 MeV per nucleon against 8.55 MeV for calcium-40, so it is more tightly bound. A larger total binding energy alone does not decide stability. Compare per nucleon.

Mass-energy equivalence in nuclear reactions

In a reaction, the energy released equals the mass difference (reactants minus products) multiplied by c². If the products have less mass, kinetic energy appears. If they have more, that energy must be supplied.

Worked example 4. Polonium-210 decays by alpha emission to lead-206. Atomic masses: Po-210 209.982874 u, Pb-206 205.974465 u, He-4 4.002603 u. Find the energy released, and the kinetic energy of the alpha particle if the nuclei start at rest.

mass lost = 209.982874 - 205.974465 - 4.002603 = 0.005806 u
energy    = 0.005806 x 931.5 = 5.41 MeV

Momentum is conserved and starts at zero, so the alpha particle and the lead nucleus have equal and opposite momenta. For equal momentum p, kinetic energy is p²/2m, so the lighter particle takes the larger share, in the ratio 206 : 4 by mass.

alpha energy = 5.41 x 206/210 = 5.31 MeV   (lead recoil: 0.10 MeV)

The strong nuclear force

Protons in a nucleus repel each other electrically. Gravity between nucleons is far too weak to hold them together. So a third interaction must act: the strong nuclear force. It is attractive between nucleons, acts over only a few femtometres (1 fm = 10⁻¹⁵ m) and is much stronger than the electric repulsion at that range. Because its range is so short, each nucleon is bound only by its close neighbours.

Random and spontaneous decay

Decay is spontaneous: it needs no outside trigger, and temperature, pressure or chemical state do not change it. It is random: you cannot say when one particular nucleus will decay. You can only state a probability, and large samples then behave predictably. This is why a decay count shows scatter about the average rather than a smooth line.

Alpha, beta and gamma changes

Type	Particle emitted	Change in nucleus
Alpha (α)	helium-4 nucleus	A falls by 4, Z falls by 2
Beta-minus (β⁻)	electron and an antineutrino	A same, Z rises by 1 (a neutron becomes a proton)
Beta-plus (β⁺)	positron and a neutrino	A same, Z falls by 1 (a proton becomes a neutron)
Gamma (γ)	photon	A and Z unchanged; nucleus loses energy

Both A and total charge balance in every equation. The neutrino ν goes with β⁺ and the antineutrino with β⁻, which keeps the count of leptons balanced: the electron counts as +1 and its antineutrino as −1.

Worked example 5. Write equations for the decay of radium-226 (Z = 88) by alpha emission, carbon-14 (Z = 6) by β⁻ to nitrogen, and sodium-22 (Z = 11) by β⁺ to neon.

²²⁶₈₈Ra → ²²²₈₆Rn + ⁴₂He
¹⁴₆C   → ¹⁴₇N  + ⁰₋₁e + antineutrino
²²₁₁Na → ²²₁₀Ne + ⁰₊₁e + neutrino

Check the top numbers: 226 = 222 + 4; 14 = 14 + 0; 22 = 22 + 0. Check the bottom: 88 = 86 + 2; 6 = 7 − 1; 11 = 10 + 1.

Gamma emission usually follows another decay, because the daughter nucleus is formed in an excited state. For example cobalt-60 decays by β⁻ to an excited nickel-60 nucleus, which then emits gamma photons.

Penetration and ionizing ability

Radiation	Ionizing ability	Penetration
Alpha	Strongest	Stopped by a few centimetres of air or a sheet of paper
Beta	Moderate	Stopped by a few millimetres of aluminium
Gamma	Weakest	Reduced, not fully stopped, by thick lead or concrete

The more strongly a radiation ionizes, the faster it loses energy along its path, so it travels less far. Alpha particles are heavy and doubly charged, so they ionize densely. Uses follow from this. A thickness gauge for sheet metal needs a radiation that is partly absorbed, so the count changes with thickness. A tracer for a leak in an underground pipe needs radiation that escapes the ground and the pipe. For a medical tracer inside the body you want a gamma emitter, which can leave the body, with a short half-life (from the second part of E.3) so that exposure is low.

HL only: further understandings

Evidence for the strong force. Stable nuclei exist at all, despite the electric repulsion between their protons. Scattering experiments in which energetic alpha particles approach light nuclei also depart from the pure electric prediction once the particles get close enough, which shows a second interaction acting at very short range.

Neutron-to-proton ratio. For light stable nuclides N is about equal to Z. Heavier stable nuclides need more neutrons than protons, up to about 1.5 neutrons per proton. Extra neutrons add strong-force attraction without adding electric repulsion. A nuclide above the stable band has too many neutrons and tends to β⁻ decay (example 5, carbon-14); one below it has too few and tends to β⁺ decay (sodium-22). Very heavy nuclides shed mass by alpha decay.

Constancy above A = 60. From about A = 60 the curve is nearly flat at roughly 8 MeV per nucleon. The strong force is short range, so a nucleon bonds only to its close neighbours and the binding energy grows in proportion to A, not faster. The slow fall at high A reflects the electric repulsion, which acts across the whole nucleus.

Discrete nuclear levels. Alpha particles from a given nuclide, and gamma photons, are found at a few definite energies, as line spectra. This means a nucleus has discrete energy levels, like an atom. An alpha line of lower energy goes with a daughter formed in an excited level, which then emits a gamma photon equal to the energy gap.

Continuous beta spectrum. Beta particles come out with every energy from zero up to a maximum. If only an electron and a daughter nucleus were emitted, energy and momentum conservation would fix the electron energy, as in alpha decay. A continuous spread, with energy and momentum apparently missing, is evidence for a third particle, the neutrino, sharing the energy.

Common errors

• Using total binding energy, not binding energy per nucleon, to judge stability.
• Reversing the subtraction: mass defect is separated nucleons minus nucleus.
• Forgetting that an excited nucleus emits a gamma photon without changing A or Z.
• Giving the neutrino with β⁻ instead of the antineutrino.
• Saying gamma rays are “stopped by lead”, when thickness only reduces the intensity.
• Mixing kg and MeV/c² in one calculation.

Official syllabus

IB Physics guide, first assessment 2025, International Baccalaureate Organization: sub-topic E.3, Radioactive decay (understandings on nuclear structure, binding energy and radioactive decay types, including the additional higher level understandings).