The Strong Nuclear Force

The force that binds the nucleus

A gold nucleus (Au-197) contains 79 protons packed into a sphere of diameter ~14 fm. The Coulomb repulsion between 79 positive charges at this distance is roughly 10⁻⁸ N, enormous on the nuclear scale. Yet the gold nucleus is stable, with a half-life longer than the age of the universe. Something overcomes that repulsion. In 1935, Hideki Yukawa at Osaka University predicted a new force carrier, the π meson, mass ~200 MeV, that mediates the attraction between nucleons. The strong nuclear force has four defining properties that distinguish it from all other forces:

• Short range: Effective only over distances of ~1–3 fm ($10^{-15}$ m). Beyond this, it drops to zero rapidly. This is why nuclei have sizes of a few femtometres, adding more nucleons doesn't increase the range of attraction.
• Attractive and charge-independent: It attracts proton-proton, neutron-neutron, and proton-neutron pairs with roughly equal strength. It does not depend on electric charge.
• Saturated: Each nucleon interacts only with its immediate neighbours, not with all other nucleons. This explains why nuclear density is approximately constant throughout the periodic table.
• Repulsive at very short distances: Below ~0.5 fm, the strong force becomes repulsive, preventing nucleons from collapsing into each other.

The strong force is mediated by mesons (pions) in the older Yukawa theory, or by gluons in quantum chromodynamics (QCD), the modern theory. Gluons bind quarks together inside protons and neutrons, and the residual strong force between nucleons is a secondary effect of this quark-gluon interaction.

Because the strong force is short-range, it cannot affect electrons, which orbit at ~$10^5$ times the nuclear radius. Electrons experience only electromagnetic and gravitational forces (the latter being negligible).

Strong nuclear force properties: (1) short range ~1–3 fm, (2) attractive and charge-independent (p-p = n-n = p-n), (3) saturated (each nucleon feels only neighbours), (4) repulsive below ~0.5 fm. Electrons orbit at ~10⁻¹⁰ m, 10⁵× the nuclear radius, so the strong force is negligible there. Mediated by pions (Yukawa) or gluons (QCD).

Write the four properties, charge-independence and saturation are the ones students most often miss.

The energy that holds nuclei together

We just saw that the strong nuclear force binds nucleons together over short distances of ~1–3 fm. That raises a question: how do we actually measure how tightly bound a nucleus is? This card answers it → by calculating the mass defect Δm = (free nucleon masses) − (nuclear mass) and using E = mc² to convert it to binding energy per nucleon.

The mass defect is the difference between the mass of the separated nucleons and the actual mass of the nucleus:

The binding energy is the energy equivalent of this mass defect (Einstein's mass-energy equivalence):

This is the energy required to completely separate the nucleus into individual protons and neutrons. Conversely, it is the energy released when nucleons combine to form a nucleus.

The binding energy per nucleon ($E_b/A$) is a key indicator of nuclear stability:

• It increases rapidly for light nuclei, peaks near iron-56 (~8.8 MeV/nucleon), then gradually decreases for heavier nuclei.
• Iron-56 is the most tightly bound nucleus, the peak of the stability curve.
• Light nuclei can release energy by fusion (combining toward iron on the curve).
• Heavy nuclei can release energy by fission (splitting toward iron on the curve).

For very heavy nuclei ($A > 200$), the cumulative Coulomb repulsion between many protons overwhelms the short-range strong force. These nuclei are unstable and undergo radioactive decay (alpha, beta, or spontaneous fission).

Mass defect: Δm = Zm_p + Nm_n − m_nucleus. Binding energy: E_b = Δm × 931.5 MeV (with Δm in u). E_b/A peaks at Fe-56 (~8.8 MeV/nucleon), the most stable nucleus. Fusion of light nuclei and fission of heavy nuclei both release energy by moving products toward the Fe-56 peak.

Write the formula chain: Δm → × 931.5 → E_b → ÷ A → E_b/A. You will need this in every nuclear calculation question.

The competition between strong force and Coulomb repulsion

We just saw that binding energy per nucleon peaks at iron-56, with heavy nuclei having lower E_b/A and less stability. That raises a question: why exactly do heavy nuclei become unstable, what's physically changing? This card answers it → Coulomb repulsion is long-range and cumulative (every proton repels every other proton), while the strong force saturates; beyond A ≈ 200, repulsion wins.

Nuclear stability depends on the balance between two competing effects:

• Strong nuclear force: Short-range (~1–3 fm), attractive between all nucleons. Each nucleon only interacts with its immediate neighbours (saturation). Adding more nucleons adds more strong force, but only locally.
• Electromagnetic repulsion: Long-range ($1/r^2$). Every proton repels every other proton regardless of distance. Adding more protons increases total Coulomb repulsion across the entire nucleus.

For light nuclei ($A \lesssim 40$), the strong force dominates and $N \approx Z$ gives stability. For heavier nuclei, extra neutrons are needed to "dilute" the proton-proton repulsion, the neutron-to-proton ratio $N/Z$ increases above 1 for stable heavy nuclei (e.g. lead-208 has $N/Z \approx 1.54$).

Beyond about $A \approx 200$, no combination of protons and neutrons can stabilise the nucleus. These nuclei are unstable and decay by:

• Alpha decay: Emits $^4_2\text{He}$, reduces both $Z$ and $A$, most common for very heavy nuclei.
• Beta decay ($\beta^-$): Converts a neutron to a proton, shifts the nucleus toward the stability valley when there are too many neutrons.
• Beta decay ($\beta^+$): Converts a proton to a neutron, used when $N/Z$ is too low.
• Spontaneous fission: The nucleus splits into two fragments, occurs for the very heaviest nuclides.

Nuclear stability: strong force (short-range, saturated) vs. EM repulsion (long-range, cumulative with every proton). Light nuclei: N ≈ Z. Heavy nuclei: N/Z > 1 needed. A > 200: all unstable → alpha decay (−2Z, −4A), β⁻ (n→p, too many neutrons), β⁺ (p→n, too few neutrons).

Draw the two competing forces in your notes, the contrast between long-range EM and short-range strong is the key to all stability questions.

Splitting a heavy nucleus to release nuclear energy

We just saw that heavy nuclei sit on the right of the binding-energy-per-nucleon curve, well below the iron-56 peak, so they are only loosely bound per nucleon. That raises a question: if a heavy nucleus could be broken into mid-mass fragments that lie closer to the peak, where would the lost binding energy go? This card answers it → nuclear fission splits a heavy nucleus into two more tightly bound fragments, and the gain in binding energy is released as kinetic energy and radiation.

Nuclear fission is the process in which a heavy nucleus, such as uranium-235, absorbs a slow (thermal) neutron and becomes so unstable that it splits into two lighter daughter fragments, releasing two or three free neutrons and a large amount of energy. A representative reaction is:

Notice that mass number is conserved (1 + 235 = 141 + 92 + 3) and atomic number is conserved (0 + 92 = 56 + 36 + 0). The exact fragments vary from event to event, but the pattern is always: one heavy nucleus in, two mid-mass fragments plus a few neutrons out.

Why is energy released? Uranium-235 has a binding energy per nucleon of only about 7.6 MeV, while the mid-mass fragments (around mass number 90 to 140) lie much closer to the iron-56 peak at about 8.5 MeV per nucleon. The products are therefore more tightly bound than the original nucleus. The increase of roughly 0.9 MeV per nucleon over 235 nucleons gives:

This binding energy gain corresponds to a tiny loss of mass (the mass defect of the reaction), released by $E = \Delta m c^2$ mostly as kinetic energy of the fast-moving fragments. About 200 MeV per fission is roughly a million times the energy of a chemical reaction such as burning a molecule of fuel.

Because each fission also frees two or three neutrons, those neutrons can trigger further fissions. This is a chain reaction, and it can be controlled or uncontrolled:

• Controlled chain reaction (nuclear reactor): on average exactly one neutron from each fission goes on to cause the next fission, so the reaction is steady and self-sustaining. A moderator (water or graphite) slows the fast neutrons to thermal speeds, where U-235 is far more likely to absorb them. Control rods (boron or cadmium) absorb surplus neutrons and are raised or lowered to hold the rate constant. The steady heat boils water to drive turbines.
• Uncontrolled chain reaction (nuclear weapon): on average more than one neutron from each fission causes a further fission, so the number of fissions grows exponentially. The energy of a huge number of nuclei is released in a fraction of a second.

A chain reaction can only be sustained if enough fissile material is present so that neutrons strike another nucleus before escaping the surface. The minimum amount needed for a self-sustaining reaction is the critical mass.

Fission: a heavy nucleus (e.g. U-235) absorbs a neutron and splits into two mid-mass fragments + 2 to 3 neutrons + ~200 MeV. Energy is released because the fragments lie closer to the Fe-56 peak (higher E_b/A than U-235), so the products are more tightly bound. Chain reaction: controlled (reactor, avg 1 neutron per fission causes next, moderator slows neutrons, control rods absorb them) vs uncontrolled (weapon, >1 per fission, exponential). Critical mass = minimum fissile mass to sustain the chain.

Write the U-235 fission equation and check both 235 = 141 + 92 + 3(1) and 92 = 56 + 36, conservation of nucleon number and charge is a common exam mark.

Combining light nuclei to release nuclear energy

We just saw that fission releases energy by moving heavy nuclei down toward the iron-56 peak from the right. That raises a question: the curve also rises steeply on the left, so could energy be released by moving light nuclei up toward the peak instead? This card answers it → nuclear fusion joins light nuclei into a more tightly bound product, climbing the curve from the light side, and stars run on exactly this.

Nuclear fusion is the process in which two light nuclei combine to form a single heavier nucleus, releasing energy. In the cores of stars like the Sun, hydrogen nuclei fuse through the proton-proton chain, with the overall result that four protons become a helium-4 nucleus, two positrons and two neutrinos, releasing about 26.7 MeV. The reaction targeted for fusion power on Earth is the deuterium-tritium (D-T) reaction:

Mass number is conserved (2 + 3 = 4 + 1) and atomic number is conserved (1 + 1 = 2 + 0).

Why is energy released? Deuterium and tritium are light nuclei with low binding energy per nucleon (~1.1 and ~2.8 MeV per nucleon). The helium-4 product is far more tightly bound, at about 7.1 MeV per nucleon, much closer to the iron-56 peak. The reaction therefore moves up the binding-energy-per-nucleon curve from the light side, so the product is more tightly bound than the reactants. The increase in binding energy appears as a mass defect, and that lost mass is released as energy by $E = \Delta m c^2$. Per nucleon, fusion of the lightest elements releases even more energy than fission.

Conditions required. Both nuclei are positively charged, so they repel each other electrically (the Coulomb barrier). To get close enough for the short-range strong force to bind them, the nuclei must collide at enormous speed. This demands:

• Very high temperature (millions of kelvin) so that nuclei have enough kinetic energy to overcome the Coulomb repulsion;
• Very high pressure or density so that collisions are frequent enough to sustain the reaction.

In the Sun these conditions are produced by the immense gravity of its huge mass, which compresses and heats the core to about 15 million kelvin. Fusion powers all main-sequence stars. On Earth it is extremely hard to achieve because no material can contain plasma at such temperatures, so experimental reactors use magnetic confinement (tokamaks) or inertial confinement (lasers) to hold the fuel together long enough for fusion to release more energy than was supplied.

Fusion: light nuclei combine into a heavier nucleus + energy (e.g. proton-proton chain in stars; D + T → He-4 + n + 17.6 MeV). Energy is released because the product (He-4) has a higher binding energy per nucleon than the light reactants, moving UP toward the Fe-56 peak from the light side, so mass is lost (E = Δmc²). Conditions: very high temperature and pressure to overcome Coulomb repulsion between the positive nuclei. Gravity supplies these in stars; on Earth confinement is the key difficulty.

Link both processes to Figure 2: fission slides down from the right, fusion climbs up from the left, and both end nearer the Fe-56 peak where E_b/A is highest.