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

This guide follows the OxfordAQA International AS and A-level Physics (9630) specification, version 4.4, for exams from May/June 2020 onwards. It covers sections 3.12.1 (Radius of the nucleus) and 3.12.2 (Mass and energy). Section 3.12 is International A-level only, so none of it is examined at International AS; the specification places it in Unit 4 of the International A2 assessments.

Links: revision notes, practice questions, the OxfordAQA Physics hub, the printable checklist, the free diagnostics, and the earlier topics atomic structure and radioactive decay.

What this unit covers

Section	What you must be able to do
3.12.1	Estimate a nuclear radius from the closest approach of alpha particles, using the Coulomb equation
3.12.1	Describe how electron diffraction gives a radius; know the intensity-angle graph
3.12.1	Know typical nuclear radii; use R = R₀A^(1/3); explain why it shows constant density; calculate nuclear density
3.12.2	Appreciate that E = mc² applies to all energy changes
3.12.2	Calculate mass difference, mass defect and binding energy; use u and 1 u = 931.5 MeV
3.12.2	Describe fission and fusion; calculate energy released from nuclear masses
3.12.2	Use the binding-energy-per-nucleon curve (plotted against A) to identify which regions give energy from fission or fusion

Constants used here (given in questions where needed): e = 1.60 × 10⁻¹⁹ C, 1/(4πε₀) = 8.99 × 10⁹ N m² C⁻², h = 6.63 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m s⁻¹, 1 u = 1.66 × 10⁻²⁷ kg, 1 MeV = 1.60 × 10⁻¹³ J.

1. Closest approach of alpha particles

An alpha particle fired straight at a nucleus slows down because both carry positive charge. At the point of closest approach it is momentarily at rest, so all its initial kinetic energy has become electric potential energy. The Coulomb equation for the potential energy of two point charges gives:

E_k = (Q₁Q₂)/(4πε₀r), with Q₁ = 2e for the alpha particle and Q₂ = Ze for the target nucleus.

Rearranged: r = (2Ze²)/(4πε₀E_k).

Worked example 1. A 9.0 MeV alpha particle approaches a copper nucleus (Z = 29) head-on. Estimate the closest approach.

• E_k = 9.0 × 1.60 × 10⁻¹³ = 1.44 × 10⁻¹² J
• r = (8.99 × 10⁹ × 2 × 29 × (1.60 × 10⁻¹⁹)²)/(1.44 × 10⁻¹²)
• r = 9.3 × 10⁻¹⁵ m (9.27 × 10⁻¹⁵ m before rounding)

Assumptions: the nucleus is much heavier than the alpha particle, so it stays still; the approach is head-on; only the electrostatic force acts. If the alpha particle reached the nucleus, the strong interaction would also act and the Coulomb equation would fail. So the result is an upper limit for the radius. Raising the alpha energy until measured scattering departs from the Coulomb prediction shows that the strong interaction has begun to act, and that fixes the nuclear size more tightly.

2. Electron diffraction

High-energy electrons feel the electric charge of the nucleus but not the strong interaction. A beam fired at a target is diffracted by the nuclei, because a particle has a de Broglie wavelength λ = h/p. For the size of a nucleus the wavelength must be comparable with, or smaller than, a few femtometres (1 fm = 10⁻¹⁵ m). That needs electrons with energies in the hundreds of MeV. At that energy the electron is moving at almost the speed of light, so p ≈ E/c and λ ≈ hc/E.

Treat the nucleus as a small circular obstacle of diameter D. The first diffraction minimum is at an angle θ where

sin θ ≈ 1.22λ/D, with D = 2R.

The intensity-angle graph. Plot detected electron intensity against scattering angle. The graph has:

• a high central maximum at 0°
• a fall to a first minimum at the angle above
• a much weaker secondary maximum, then a second minimum, and further maxima that keep getting lower

A nucleus has no sharp edge, so the minima are shallow rather than falling to zero. A larger nucleus gives a first minimum at a smaller angle; a higher electron energy (shorter λ) also moves it to a smaller angle.

Worked example 2. Electrons of energy 420 MeV give a first minimum at 32.0°. Find the nuclear radius and estimate the nucleon number, using R₀ = 1.2 fm.

• E = 420 × 1.60 × 10⁻¹³ = 6.72 × 10⁻¹¹ J
• λ = hc/E = (6.63 × 10⁻³⁴ × 3.00 × 10⁸)/(6.72 × 10⁻¹¹) = 2.96 × 10⁻¹⁵ m
• D = 1.22λ/sin 32.0° = 6.81 × 10⁻¹⁵ m
• R = D/2 = 3.4 × 10⁻¹⁵ m
• A = (R/R₀)³ = (3.41/1.2)³ = 22.9, so A ≈ 23

3. Size and density of nuclei

Typical nuclear radii are from about 1 fm for the lightest nuclei to about 7 fm for the heaviest. An atom has a radius of order 10⁻¹⁰ m, roughly 10⁴ to 10⁵ times larger.

Measurements for many nuclei fit

R = R₀A^(1/3), with R₀ ≈ 1.2 fm

where A is the nucleon number. This was found from experimental data: a graph of lg R against lg A is a straight line of gradient 1/3, and the intercept is lg R₀.

Why this means constant density. The volume of a spherical nucleus is (4/3)πR³ = (4/3)πR₀³A, so volume is proportional to A. The mass is almost A × (1 u), so also proportional to A. The ratio mass/volume does not depend on A:

ρ = (A × 1 u)/((4/3)πR₀³A) = (3 × 1 u)/(4πR₀³)

All nuclei have the same density, whatever their size. The nucleons are packed together in nuclear matter.

Worked example 3. (a) Calculate the nuclear density with R₀ = 1.2 × 10⁻¹⁵ m. (b) Find the radius of a nucleus with A = 64.

• (a) ρ = (3 × 1.66 × 10⁻²⁷)/(4π × (1.2 × 10⁻¹⁵)³) = 2.3 × 10¹⁷ kg m⁻³. A 5.0 cm³ sample of such material would have a mass of 1.1 × 10¹² kg.
• (b) R = 1.2 × 64^(1/3) = 1.2 × 4 = 4.8 fm

Compare this with a solid such as iron, density 7870 kg m⁻³: nuclear matter is about 3 × 10¹³ times denser. The reason is that almost all the mass of an atom sits in a nucleus that fills a tiny fraction of the atom’s volume.

4. Mass-energy and binding energy

Einstein’s relation E = mc² applies to every energy change, not only nuclear ones. A change in energy ΔE is linked to a change in mass Δm by:

ΔE = Δm c²

Burning 1.0 kg of coal releases roughly 3.0 × 10⁷ J, so the mass change is 3.0 × 10⁷/(3.00 × 10⁸)² = 3.3 × 10⁻¹⁰ kg. That is too small to measure. In nuclear reactions the fractional mass change is about a million times larger, and it can be measured.

Definitions.

• Mass defect: the total mass of the free nucleons minus the mass of the nucleus.
• Binding energy: the minimum energy needed to separate a nucleus into its individual protons and neutrons. It is the energy equivalent, Δm c², of the mass defect.
• Binding energy per nucleon: binding energy divided by A. A larger value means a more tightly bound, more stable nucleus.

The atomic mass unit. The unit u is defined as 1/12 of the mass of one carbon-12 atom. 1 u = 1.66 × 10⁻²⁷ kg, and the energy equivalent is 1 u = 931.5 MeV. A mass in u multiplied by 931.5 gives energy in MeV directly. (Using 1.66 × 10⁻²⁷ kg and c = 3.00 × 10⁸ m s⁻¹ gives about 934 MeV, because the rounded constants differ slightly from the accurate ones; use 931.5 when working in u.)

Worked example 4. The nuclear mass of neon-20 (10 protons, 10 neutrons) is 19.98695 u. Take m_p = 1.007276 u and m_n = 1.008665 u. Calculate the binding energy and the binding energy per nucleon.

• Mass of separate nucleons = 10 × 1.007276 + 10 × 1.008665 = 20.15941 u
• Mass defect = 20.15941 − 19.98695 = 0.17246 u
• Binding energy = 0.17246 × 931.5 = 160.6 MeV
• Per nucleon = 160.6/20 = 8.03 MeV

Take care with data. If you are given atomic masses, the electrons are included. Either subtract Z electron masses to get the nuclear mass, or use the hydrogen atom mass in place of the proton mass, so that the electrons cancel.

5. Binding energy per nucleon and the nuclear curve

The curve of mean binding energy per nucleon, plotted against A, rises steeply for the lightest nuclei: about 1.1 MeV for hydrogen-2, then 7.07 MeV for helium-4, a peak that stands above its neighbours. It reaches a broad maximum of about 8.8 MeV near iron-56, then falls slowly to about 7.6 MeV for uranium-235.

• Fusion region: light nuclei, with A below about 56. Joining two of them gives a nucleus with a higher binding energy per nucleon.
• Fission region: heavy nuclei, with A above about 56. Splitting one gives two nuclei in the middle of the graph, again with a higher binding energy per nucleon.

In both processes the products are more tightly bound than the reactants. The total binding energy rises, the total mass falls, and the difference appears as kinetic energy of the products and radiation.

Energy released = binding energy after the reaction − binding energy before it.

Worked example 5 (from the graph). Uranium-235 (7.6 MeV per nucleon) absorbs a neutron and splits into two fragments of A about 117, each at 8.5 MeV per nucleon. Estimate the energy released.

• Binding energy gained per nucleon = 8.5 − 7.6 = 0.9 MeV
• Nucleons involved = 235
• Energy ≈ 235 × 0.9 = 2.1 × 10² MeV, an estimate because graph values are read to two significant figures

6. Energy from nuclear masses

Worked example 6 (fusion). Two deuterium nuclei fuse: ²H + ²H → ³He + n. Nuclear masses: deuterium 2.013553 u, helium-3 3.014932 u, neutron 1.008665 u.

• Mass before = 2 × 2.013553 = 4.027106 u
• Mass after = 3.014932 + 1.008665 = 4.023597 u
• Mass difference = 0.003509 u
• Energy released = 0.003509 × 931.5 = 3.27 MeV

Worked example 7 (fission). Data supplied: atomic masses 235.043930 u (uranium-235), 93.915361 u (strontium-94), 139.921636 u (xenon-140); neutron 1.008665 u. Calculate the energy released by ²³⁵U + n → ⁹⁴Sr + ¹⁴⁰Xe + 2n.

The protons balance (92 = 38 + 54) so the electron masses cancel and atomic masses can be used.

• Before = 235.043930 + 1.008665 = 236.052595 u
• After = 93.915361 + 139.921636 + 2 × 1.008665 = 235.854327 u
• Difference = 0.198268 u
• Energy = 0.198268 × 931.5 = 184.7 MeV.

To convert to joules, multiply by 1.60 × 10⁻¹³: 184.7 MeV = 2.96 × 10⁻¹¹ J per fission.

Common errors

• Using R = R₀A instead of R = R₀A^(1/3), or forgetting that density does not depend on A.
• Taking the closest approach as the exact nuclear radius instead of an upper limit.
• Using the alpha charge as e instead of 2e in the Coulomb equation.
• Putting energies in MeV into a formula that needs joules (multiply by 1.60 × 10⁻¹³).
• Mixing nuclear and atomic masses when working out a mass defect.
• Subtracting the wrong way round; energy released needs mass before minus mass after.

Test yourself with the practice questions, then condense with the revision notes.

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.