The Strong Nuclear Force
In 1935, Hideki Yukawa at Osaka University predicted the existence of a new massive particle that would mediate the strong nuclear force. Using the uncertainty principle, he estimated its mass as ~200 MeV from the nuclear force range of ~10⁻¹⁵ m. The π meson (pion) was discovered in 1947 by Cecil Powell using cosmic ray photographic plates from Mount Chacaltaya in Bolivia at 5,220 m altitude. Yukawa was awarded the Nobel Prize in Physics in 1949, Powell in 1950.
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
The nucleus contains protons that repel each other electrically, yet most nuclei are stable.
Before reading on, answer:
- What must be true about the force between nucleons if the nucleus is stable?
- Why doesn't this strong force pull electrons into the nucleus?
- Why do very heavy nuclei tend to be unstable despite the strong force?
Warm-up: The strong nuclear force acts between:
Know — Strong Force Properties
- Short range (~1-3 fm)
- Attractive between all nucleons
- Independent of charge
Understand — Nuclear Stability
- Binding energy per nucleon
- Mass defect
- Fusion and fission energy release
Can Do — Analyse Nuclear Reactions
- Calculate binding energy
- Determine stability
- Predict decay modes
Core Content
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).
Figure 1 — Strong nuclear force vs. nucleon separation. Repulsive below ~0.5 fm, strongly attractive at ~1–3 fm, rapidly falling to zero beyond ~3 fm.
Why is the strong nuclear force called "strong"? Compare its strength to the electromagnetic force at nuclear distances (~1 fm). Why does its short range mean electrons are unaffected?
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 strong nuclear force is approximately charge-independent — it attracts proton-proton pairs with roughly the same strength as proton-neutron pairs.
The strong nuclear force has infinite range, just like gravity and electromagnetism.
Because the strong force saturates, nuclear density is approximately constant across the periodic table.
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:
$$\Delta m = [Zm_p + Nm_n] - m_{\text{nucleus}}$$The binding energy is the energy equivalent of this mass defect (Einstein's mass-energy equivalence):
$$E_b = \Delta m \cdot c^2$$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).
$\Delta m = Zm_p + Nm_n - m_{\text{nucleus}}$ — mass defect
$E_b = \Delta m \cdot c^2$ — binding energy
$E_b / A$ — binding energy per nucleon (stability indicator)
1 u = 931.5 MeV/$c^2$ — atomic mass unit conversion
Figure 2 — Binding energy per nucleon vs mass number. The curve peaks at iron-56. Both fusion (light nuclei) and fission (heavy nuclei) release energy by moving toward the peak.
Calculate the binding energy per nucleon for helium-4.
Given: $m_p = 1.007276$ u, $m_n = 1.008665$ u, $m_{\text{He}} = 4.002602$ u, 1 u = 931.5 MeV/$c^2$. Helium-4 has $Z = 2$, $N = 2$.
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.
Carbon-12 has 6 protons and 6 neutrons. Using $m_p = 1.007276$ u, $m_n = 1.008665$ u, and $m_C = 12.000000$ u, the mass defect $\Delta m$ in atomic mass units (to 4 decimal places) is: _____ u.
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.
A common exam trap: using atomic mass instead of nuclear mass in binding energy calculations. Atomic mass includes electrons, so you must either subtract electron masses or use atomic masses consistently (the electron masses cancel if you use atomic masses for both sides). For binding energy per nucleon, always divide by $A$ (mass number), not by $Z$ or $N$. Remember: fusion of light nuclei and fission of heavy nuclei both release energy because they move toward the peak of the binding energy curve at iron-56. When calculating mass defect in u, multiply by 931.5 MeV/u to get binding energy in MeV.
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.
Very heavy nuclei (A > 200) tend to be unstable because:
Three of these statements about the strong nuclear force are correct. Pick the odd one out.
Activities
Practice mass defect and binding energy per nucleon
- Calculate the binding energy per nucleon for helium-4 ($Z=2$, $N=2$, $m_{\text{He}}=4.002602$ u, $m_p=1.007276$ u, $m_n=1.008665$ u, 1 u = 931.5 MeV/$c^2$).
- The binding energy per nucleon of Fe-56 is 8.79 MeV. Calculate the total binding energy of Fe-56 in MeV.
- Uranium-238 has binding energy per nucleon ~7.57 MeV. Is more energy released by fissioning U-238 into two medium-mass fragments (assume 8.4 MeV/nucleon in products) or by fusing two helium-4 nuclei into beryllium-8? Explain using the binding energy curve.
- Why does nuclear density remain approximately constant across the periodic table? Use the concepts of saturation and the short range of the strong force in your answer.
Apply strong force principles to predict stability and decay
- Explain why a stable uranium nucleus ($Z=92$) requires 146 neutrons ($N/Z \approx 1.59$), while oxygen-16 ($Z=8$) requires only 8 neutrons ($N/Z = 1$). Use the properties of the strong force and Coulomb repulsion in your answer.
- Describe two differences between the strong nuclear force and the electromagnetic force. Use these to explain why large nuclei tend to be unstable.
- A student claims: "Iron-56 cannot be used in nuclear reactors because neither fission nor fusion releases energy from it." Assess this claim, explaining what happens to binding energy per nucleon when iron undergoes fission or fusion.
- Predict whether carbon-14 ($^{14}_{6}\text{C}$, which has too many neutrons for its proton number) would most likely undergo alpha decay, $\beta^-$ decay, or $\beta^+$ decay. Justify your prediction.
A fresh five-question set drawn from this lesson's bank — feedback shown immediately. +5 XP per correct · +25 XP all correct
Pick your answer, then rate your confidence — that tells the system what to drill next.
ApplyBand 4(3 marks) 1. A nucleus of lithium-7 ($^7_3\text{Li}$) has mass 7.016003 u. Using $m_p = 1.007276$ u and $m_n = 1.008665$ u: (a) Calculate the mass defect $\Delta m$ in u. (b) Calculate the binding energy in MeV (1 u = 931.5 MeV/$c^2$). (c) Calculate the binding energy per nucleon in MeV.
1 mark: correct $\Delta m$ · 1 mark: correct $E_b$ · 1 mark: correct $E_b/A$
AnalyseBand 6(5 marks) 2. (a) Outline four key properties of the strong nuclear force. (b) Explain why the strong force does not affect electrons orbiting the nucleus. (c) Define mass defect and binding energy and explain their relationship via $E = mc^2$. (d) Using the binding energy per nucleon curve, explain why both fusion of light nuclei and fission of heavy nuclei can release energy. (e) Explain why very heavy nuclei ($A > 200$) tend to be radioactively unstable, referring to the competing forces.
1 mark: four strong force properties · 1 mark: electrons unaffected · 1 mark: mass defect/BE relationship · 1 mark: fusion and fission explanation · 1 mark: heavy nuclei instability
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Multiple choice
MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set drawn from the lesson bank.
Short Answer — Model Answers
Q1 (3 marks): (a) $\Delta m = 3(1.007276) + 4(1.008665) - 7.016003 = 3.021828 + 4.034660 - 7.016003 = 0.040485$ u (1 mark). (b) $E_b = 0.040485 \times 931.5 = 37.71$ MeV (1 mark). (c) $E_b/A = 37.71/7 = 5.39$ MeV/nucleon (1 mark).
Q2 (5 marks): (a) Short range (~1–3 fm); charge-independent (p-p, n-n, p-n equally); saturated (each nucleon interacts only with immediate neighbours); repulsive below ~0.5 fm (1 mark). (b) Electrons orbit at ~$10^{-10}$ m — approximately $10^5$ times larger than the nuclear radius (~$10^{-15}$ m). The strong force only operates over ~1–3 fm, so it is completely negligible at electron orbital distances; electrons only experience the electromagnetic force (1 mark). (c) Mass defect $\Delta m = Zm_p + Nm_n - m_{\text{nucleus}}$: the nucleus is lighter than the sum of its free nucleons because energy is released when nucleons bind. By $E = mc^2$, this mass difference represents the binding energy $E_b = \Delta m \cdot c^2$ — the energy needed to completely disassemble the nucleus (1 mark). (d) The binding energy per nucleon curve peaks at Fe-56. Light nuclei fusing move to higher $E_b/A$ → mass decreases → energy released. Heavy nuclei fissioning produce fragments with higher $E_b/A$ than the original nucleus → energy released. Both processes move products toward the Fe-56 peak (1 mark). (e) The strong force is short-range and saturates: adding protons increases strong attraction only locally. The electromagnetic force is long-range: every proton repels every other proton ($F \propto 1/r^2$), so Coulomb repulsion grows as $Z(Z-1)/2$. For $A > 200$, the cumulative Coulomb repulsion from many protons exceeds the short-range strong force binding, making all such nuclei unstable (1 mark).
At the start you were asked about the force that holds 79 protons inside a gold nucleus 14 fm across — the problem Hideki Yukawa solved in 1935 by predicting a force carrier of mass ~200 MeV (the π meson, discovered by Cecil Powell in 1947 using cosmic ray plates from 5,220 m altitude in Bolivia). Review your predictions:
- Did you predict the strong nuclear force must be strongly attractive and short-range, as Yukawa's meson exchange model requires? Correct — the force overcomes Coulomb repulsion at ~1–3 fm, but it saturates and cannot stabilise arbitrarily large nuclei.
- Did you predict the strong force doesn't affect electrons because of its short range? Correct — electrons orbit at ~10⁻¹⁰ m, roughly 10⁵ times the nuclear radius, far beyond the ~3 fm range that Yukawa's pion exchange produces.
- Did you predict heavy nuclei are unstable because Coulomb repulsion accumulates while the strong force saturates? Correct — for large Z, the long-range electromagnetic repulsion between many protons exceeds the short-range strong attraction, destabilising the nucleus.
Extend: (a) Calculate the total binding energy of uranium-235 in MeV, given $E_b/A \approx 7.59$ MeV/nucleon. (b) A fission event splits U-235 into two roughly equal fragments, each with $E_b/A \approx 8.5$ MeV/nucleon. Estimate the energy released per fission event in MeV. (c) Explain why nuclear power stations use fission rather than fusion, despite fusion releasing more energy per kilogram of fuel.
Five timed questions on the strong nuclear force and nuclear stability. Beat the boss to bank a tier — gold (perfect + fast), silver (80%+), or bronze (cleared).
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