OxfordAQA International AS and A-level Physics: Nuclear radius, mass-energy and binding energy (9630) – Revision Notes

Condensed notes for sections 3.12.1 (Radius of the nucleus) and 3.12.2 (Mass and energy) of the OxfordAQA International AS and A-level Physics (9630) specification, version 4.4, exams from May/June 2020 onwards. Section 3.12 is International A-level only. For full explanations, read the study guide; test yourself with the practice questions.

Also useful: the OxfordAQA Physics hub, the printable checklist, the free diagnostics and the earlier note on radioactive decay.

Key definitions

Term	Meaning
Closest approach	Distance at which an alpha particle stops momentarily; all its kinetic energy has become electric potential energy
Mass defect	Mass of the free nucleons minus the nucleus mass
Binding energy	Minimum energy needed to separate a nucleus into its protons and neutrons; equals Δm c²
Binding energy per nucleon	Binding energy ÷ A; larger means more tightly bound
Atomic mass unit, u	Defined as 1/12 of the mass of one carbon-12 atom; 1.66 × 10⁻²⁷ kg; 1 u = 931.5 MeV
Fission	A heavy nucleus splits into two medium-mass nuclei, usually after absorbing a neutron
Fusion	Two light nuclei combine into one heavier nucleus

Formulas

Quantity	Relationship
Closest approach (head-on, Z = target proton number)	E_k = (2e × Ze)/(4πε₀r), so r = 2Ze²/(4πε₀E_k)
Electron wavelength (de Broglie)	λ = h/p; for very high energies p ≈ E/c, so λ ≈ hc/E
First diffraction minimum (circular obstacle)	sin θ ≈ 1.22λ/D, with D = 2R
Nuclear radius	R = R₀A^(1/3), R₀ ≈ 1.2 fm
Nuclear density	ρ = 3 × (1 u)/(4πR₀³), independent of A
Mass-energy	ΔE = Δm c²
Energy released	Δm (in u) × 931.5 = energy in MeV
Mass defect	Δm = Zm_p + (A − Z)m_n − m_nucleus

Unit conversions: 1 MeV = 1.60 × 10⁻¹³ J; 1 fm = 10⁻¹⁵ m.

3.12.1 Radius of the nucleus

Typical sizes. Nuclear radii run from about 1 fm to about 7 fm. Atoms are about 10⁻¹⁰ m across.

Method for closest approach.

1. Convert the alpha energy from MeV to J.
2. Put kinetic energy equal to the electric potential energy: E_k = (2e)(Ze)/(4πε₀r).
3. Rearrange for r and substitute.
4. State the result as an upper limit for the radius, since it assumes a head-on approach, a stationary nucleus and only the Coulomb force.

Reminder. A 6.0 MeV alpha particle approaches a calcium nucleus (Z = 20). E_k = 9.6 × 10⁻¹³ J, so r = 8.99 × 10⁹ × 2 × 20 × (1.60 × 10⁻¹⁹)²/(9.6 × 10⁻¹³) = 9.6 × 10⁻¹⁵ m.

Electron diffraction.

• Electrons are used because they interact with the nuclear charge but not through the strong interaction.
• Their energy must be high enough (hundreds of MeV) for the wavelength to be comparable with a nucleus.
• The intensity-angle graph: strong central maximum at 0°, first minimum, a weaker secondary maximum, further minima and maxima, each maximum lower than the one before. Minima are shallow because the nucleus has no sharp edge.
• A bigger nucleus gives the first minimum at a smaller angle. Higher electron energy also gives a smaller angle.

Method in steps. Find λ from energy; find D from sin θ = 1.22λ/D; halve D to get R; if asked for A, use A = (R/R₀)³.

Constant density. Volume ∝ R³ ∝ A and mass ∝ A, so mass/volume is the same for all nuclei. About 2.3 × 10¹⁷ kg m⁻³ with R₀ = 1.2 fm. A graph of lg R against lg A is a straight line with gradient 1/3 and intercept lg R₀.

Radius ratio reminder. A = 216 against A = 8: R is in the ratio (216/8)^(1/3) = 3.0.

3.12.2 Mass and energy

Mass-energy. E = mc² applies to every energy change. Chemical changes alter mass too, but by far too little to measure. For nuclear changes the mass change is measurable.

Method for binding energy.

1. Work out the mass of the separate nucleons: Z protons plus (A − Z) neutrons.
2. Subtract the mass of the nucleus to get the mass defect.
3. Multiply the defect in u by 931.5 to get the binding energy in MeV.
4. Divide by A for the binding energy per nucleon.

Reminder. Nitrogen-14 has a nuclear mass of 13.999231 u. Mass of separate nucleons = 7 × 1.007276 + 7 × 1.008665 = 14.111587 u. Mass defect = 14.111587 − 13.999231 = 0.112356 u. Binding energy = 0.112356 × 931.5 = 104.7 MeV. Per nucleon = 104.7/14 = 7.48 MeV.

Method for energy released in a reaction. Add up the masses before; add up the masses after; the difference in u times 931.5 is the energy released in MeV. If the mass after is smaller, energy is released. Use nuclear masses throughout, or atomic masses throughout when the numbers of electrons before and after match.

The binding energy per nucleon graph

Region	Nucleon number	What happens
Rises steeply	A below about 20	Fusion releases large energy per nucleon; helium-4 stands above neighbours
Peak	Around 56 (iron)	Most tightly bound nuclei; neither fission nor fusion releases energy
Slow fall	A above about 56	Fission releases energy; uranium-235 is about 7.6 MeV per nucleon

• Fusion region: left of the peak. Fission region: right of the peak.
• Energy released = Σ(A × BE per nucleon) after − Σ(A × BE per nucleon) before.
• Typical values: hydrogen-2 about 1.1, helium-4 7.07, carbon-12 7.68, iron-56 8.79, uranium-235 7.59 (MeV per nucleon).

Must-know distinctions

• Mass defect vs binding energy: a mass (u or kg) vs an energy (MeV or J); linked by c².
• Binding energy vs binding energy per nucleon: total vs average. Compare stability using the per-nucleon value.
• Fission vs fusion: heavy splits vs light joins; both move nuclei toward the peak.
• Closest approach vs radius: the closest approach is only an upper limit.
• Atomic mass vs nuclear mass: atomic mass includes Z electrons.

Quick self-test

1. A 5.0 MeV alpha particle is fired head-on at a silver nucleus (Z = 47). Find the closest approach.
2. Calculate the radius of a nucleus with A = 125, using R₀ = 1.2 fm.
3. Find the nuclear density if R₀ = 1.4 fm.
4. Why is the density of all nuclei the same?
5. Convert a mass of 0.50 u into MeV and joules.
6. Which mass in u is equivalent to 8.0 MeV?
7. The nuclear mass of lithium-7 is 7.014357 u. Find the binding energy per nucleon (m_p = 1.007276 u, m_n = 1.008665 u).
8. A reaction has a mass difference of 0.0042 u. How much energy is released?
9. Electrons of energy 300 MeV are diffracted by a nucleus of diameter 6.0 fm. Find the angle of the first minimum.
10. A nucleus has A = 150 and binding energy per nucleon 8.3 MeV. It splits into two with A = 75, each with 8.7 MeV per nucleon. Is energy released, and how much?
11. State what is meant by binding energy.

Answers

1. r = 8.99 × 10⁹ × 2 × 47 × (1.60 × 10⁻¹⁹)²/(8.0 × 10⁻¹³) = 2.7 × 10⁻¹⁴ m.
2. R = 1.2 × 125^(1/3) = 1.2 × 5 = 6.0 fm.
3. ρ = 3 × 1.66 × 10⁻²⁷/(4π × (1.4 × 10⁻¹⁵)³) = 1.4 × 10¹⁷ kg m⁻³.
4. Volume and mass are both proportional to A, so mass/volume does not depend on A.
5. 0.50 × 931.5 = 466 MeV = 466 × 1.60 × 10⁻¹³ = 7.5 × 10⁻¹¹ J.
6. 8.0/931.5 = 0.0086 u.
7. Separate nucleons: 3 × 1.007276 + 4 × 1.008665 = 7.056488 u; defect 0.042131 u; binding energy 39.2 MeV; 5.61 MeV per nucleon.
8. 0.0042 × 931.5 = 3.9 MeV.
9. λ = hc/E = 4.14 fm; sin θ = 1.22 × 4.14/6.0 = 0.842; θ = 57°.
10. Binding energy before = 150 × 8.3 = 1245 MeV; after = 150 × 8.7 = 1305 MeV; 60 MeV released (products more tightly bound).
11. The minimum energy needed to separate a nucleus completely into its individual protons and neutrons.

Where marks are usually lost

• Using e instead of 2e for the alpha particle in the Coulomb equation.
• Leaving the alpha energy in MeV when the equation needs joules.
• Presenting the closest approach as the nuclear radius with no comment on the assumptions.
• Using R = R₀A, or cubing instead of cube-rooting.
• Claiming density depends on A, or giving no reasoning for constant density.
• Subtracting in the wrong order, so that energy released comes out negative.
• Using atomic masses for the products and nuclear masses for the reactants.
• Mixing up mass defect (u) with binding energy (MeV).
• Saying only “iron is stable” instead of comparing binding energy per nucleon before and after.

Official syllabus

OxfordAQA International AS and A-level Physics (9630), version 4.4, for International AS and A-level exams from May/June 2020 onwards, published by OxfordAQA. Sections 3.12.1 and 3.12.2.