Nuclear physics §

Where the force comes from §

Quantum chromodynamics confines quarks into protons and neutrons (see Elementary particles). Each nucleon is color-neutral, so — like two neutral atoms — two nucleons feel almost no force at large distance. But their quark clouds polarize each other, leaving a residual attraction at short range, analogous to the van der Waals force between molecules (see molecular bonds). This residual force is what binds nuclei.

Properties §

• Short range: effective range ≈ 1–2 fm ($1 \text{ fm} = 10^{-15}$ m); beyond ~3 fm it is negligible. Yukawa explained the exponential falloff as exchange of a massive particle: $V(r) \propto -e^{-r/\lambda}/r$ with $\lambda = \hbar/(m_\pi c) \approx 1.4$ fm — the pion (mass ≈ 140 MeV) is the force carrier.
• Repulsive core: below ~0.5 fm the force turns strongly repulsive, which is why nuclei have a nearly constant density (saturation) rather than collapsing.
• Charge independence: approximately the same for p–p, n–n, and p–n pairs in the same spin-isospin channel.
• Spin dependence: the deuteron (p + n) is bound only in the spin-triplet channel ($B = 2.22$ MeV); the spin-singlet state is unbound. A tensor component mixes S and D waves.
• Saturation: each nucleon interacts mainly with its nearest neighbors, so total binding energy grows roughly as $A$, not $A^2$.

The potential between two nucleons §

For two protons, the effective potential is the sum of the short-range nuclear well and the long-range Coulomb repulsion $V_C(r) = 1.44 \text{ MeV·fm}/r$:

The Coulomb tail looks tiny on this scale — but it is the whole story of alpha decay and fission: for an $\alpha$ particle facing a heavy daughter ($Z \approx 90$), the same $1/r$ term reaches ≈ 30 MeV at the nuclear surface, forming a barrier that the $\alpha$ can only escape by quantum tunneling.

Example: why the deuteron is fragile §

The deuteron’s binding energy (2.22 MeV) is tiny compared with typical nuclear wells (~40 MeV) because it is bound only by the tensor part of the force in the spin-triplet channel — there is no bound p–p or n–n nucleus. This is direct evidence that the nuclear force depends on spin and isospin, not just position.

Authoritative sources

• Krane, Introductory Nuclear Physics, Ch. 4 “The force between nucleons”
• Yukawa, On the Interaction of Elementary Particles (1935)

Step 1: the liquid drop model §

Because the nuclear force is short-ranged and saturating, a nucleus behaves like a droplet of incompressible fluid: each nucleon contributes a fixed volume energy, only surface nucleons lose neighbors, protons repel each other via Coulomb, and the Pauli exclusion principle penalizes proton–neutron imbalance. The Bethe–Weizsäcker semi-empirical mass formula:

with $a_V \approx 15.8$ MeV (volume), $a_S \approx 18.3$ MeV (surface), $a_C \approx 0.714$ MeV (Coulomb), $a_A \approx 23.2$ MeV (asymmetry), and a pairing term $\delta = \pm 12/\sqrt{A}$ (positive for even–even nuclei, negative for odd–odd, zero for odd $A$).

Example. For iron-56 ($A = 56$, $Z = 26$): volume term $15.8 \times 56 = 884.8$; surface $18.3 \times 56^{2/3} = 267.9$; Coulomb $0.714 \times 26 \times 25 / 56^{1/3} = 121.3$; asymmetry $23.2 \times 16/56 = 6.6$; pairing $+1.6$. Total $B \approx 490.6$ MeV, i.e. $B/A \approx 8.76$ MeV — within 0.4% of the measured 8.79 MeV.

Step 2: where the liquid drop fails — magic numbers §

The liquid drop formula predicts binding energy varying smoothly with $Z$ and $N$. But experiment shows certain nuclei are exceptionally stable and abundant: those with

the magic numbers. Nuclei with both $N$ and $Z$ magic (doubly magic: ${}^4$He, ${}^{16}$O, ${}^{40}$Ca, ${}^{48}$Ca, ${}^{208}$Pb) are spherical, rigid, and tightly bound. A smooth droplet model cannot explain this — just as classical physics could not explain the discrete spectral lines that revealed electron shells.

Step 3: the shell model §

The nuclear shell model (Mayer and Jensen, 1949) treats each nucleon as moving independently in the average potential well created by all the others (a Woods–Saxon well, similar in spirit to the electron shells of atomic physics). Because nucleons are fermions, the Pauli exclusion principle forces them to fill discrete levels: closed shells correspond to the magic numbers.

The crucial difference from atoms is the spin–orbit coupling: the nuclear spin–orbit splitting is strong and has the opposite sign to the atomic case (where it is a small relativistic effect). A level with orbital angular momentum $\ell$ splits into $j = \ell + 1/2$ (lower) and $j = \ell - 1/2$ (higher), and the large gaps after the lowered $j = \ell + 1/2$ levels reproduce the magic numbers 28, 50, 82, and 126.

The shell model explains why ${}^{4}$He (closed 1s shell), ${}^{16}$O (closed 1p shell), and the calcium isotopes ($Z = 20$) are so stable, and why neutron shell closures at $N = 126$ shape the abundance peaks of the heavy elements formed in nucleosynthesis.

Authoritative sources

• Krane, Introductory Nuclear Physics, Ch. 5 “Nuclear models”
• Mayer, On Closed Shells in Nuclei (Phys. Rev. 74, 1948)
• Weizsäcker, Zur Theorie der Kernmassen (1935)

Step 1: the curve §

Plotting $B/A$ against mass number $A$ gives the most important single picture in nuclear energy physics:

Three features to read off the curve:

1. Rapid rise up to $A \approx 20$: each added nucleon is grabbed by the short-range force (this is why nucleosynthesis in stars releases energy).
2. Broad peak at iron ($B/A \approx 8.8$ MeV): iron-group nuclei are the most tightly bound — the ash of nuclear burning, and the endpoint of exothermic fusion.
3. Slow decline beyond $A \approx 60$: the Coulomb repulsion of the growing proton population wins out (the $Z^2/A^{1/3}$ term in the mass formula), so very heavy nuclei are less bound per nucleon and can release energy by splitting.

Step 2: reading energy release off the curve §

The energy released by a rearrangement is the total binding energy gained:

• Fission: uranium at $B/A \approx 7.6$ MeV splitting into fragments near $A \approx 120$ at $B/A \approx 8.5$ MeV gains ≈ 0.9 MeV per nucleon → ≈ 200 MeV per fission.
• Fusion: four protons → one ${}^4$He climbs from 0 (free protons) to 7.07 MeV per nucleon; after paying the proton→neutron conversions and the positron rest masses, ≈ 26.7 MeV per helium made (this is the Sun’s energy source).

Step 3: energy per kilogram §

Because nuclear binding energies (~MeV per nucleon) are ~10⁶ times larger than chemical bond energies (~eV per molecule), nuclear fuels are enormously more energy-dense than chemical ones:

Applications. Fission reactors exploit the neutron multiplication of the fission reaction (see Nuclear reactions) to sustain a controlled chain reaction. Fusion powers the stars (see Cosmology) and drives terrestrial fusion-energy research, where the D–T reaction is the leading candidate because it has the largest cross-section at reachable temperatures.

Authoritative sources

• Krane, Introductory Nuclear Physics, Ch. 13 “Nuclear fission” and Ch. 14 “Nuclear fusion”

Step 1: the conservation laws §

Whatever happens in a nuclear reaction, the books must balance: total charge $Z$, total baryon number $A$ (for nucleons), lepton number, energy, and momentum are conserved. Writing a reaction is therefore bookkeeping: match the superscripts (mass numbers) and subscripts (atomic numbers) on both sides.

Step 2: the main reaction types §

• Elastic scattering: the projectile and target exchange kinetic energy but keep their identity and internal state: $p + p \to p + p$. This is how Rutherford discovered the nucleus, and how neutron moderators slow neutrons in reactors.
• Inelastic scattering: the target is left in an excited state: $n + {}^{12}\text{C} \to n' + {}^{12}\text{C}^*$, followed by ${}^{12}\text{C}^* \to {}^{12}\text{C} + \gamma$ (4.44 MeV). The outgoing neutron carries less energy.
• Charge-exchange (transfer) reactions: a nucleon is exchanged, changing the product’s species: $p + {}^{14}\text{N} \to n + {}^{14}\text{O}$ is a (p,n) reaction that converts a target proton into a neutron, moving the residual nucleus one step up in $Z$ — not mere scattering, since the projectile changes identity too.
• Fusion: light nuclei combine: ${}^{2}\text{H} + {}^{3}\text{H} \to {}^{4}\text{He} + n + 17.6$ MeV. Coulomb repulsion makes the rate extremely temperature-sensitive (see the binding energy curve for why it releases energy).
• Fission: a heavy nucleus splits: $n + {}^{235}\text{U} \to {}^{141}\text{Ba} + {}^{92}\text{Kr} + 3n$, releasing ≈ 200 MeV and 2–3 neutrons — the neutrons can fission further nuclei, enabling the chain reaction at the heart of reactors.

Step 3: the Q-value §

The Q-value is the rest-mass difference converted to kinetic energy:

Example. For the D–T reaction, using atomic masses $m({}^{2}\text{H}) = 2.014102$ u, $m({}^{3}\text{H}) = 3.016049$ u, $m({}^{4}\text{He}) = 4.002603$ u, $m(n) = 1.008665$ u: $m_i = 5.017522$ u, $m_f = 5.011268$ u, so $Q = 0.018883 \times 931.5 \approx 17.6$ MeV — exothermic, as the binding-energy curve predicts.

Authoritative sources

• Krane, Introductory Nuclear Physics, Ch. 11 “Nuclear reactions”
• Particle Data Group, Review of Particle Physics

Step 1: alpha decay — tunneling through the Coulomb barrier §

An $\alpha$ particle (a helium-4 nucleus) is preformed inside a heavy nucleus, trapped by the Coulomb barrier described in the nuclear force section. Classically it could never escape; quantum mechanically it tunnels through the barrier with a tiny but nonzero probability per collision — which is why alpha half-lives range from microseconds to billions of years for nearly the same decay energy (the Geiger–Nuttall law).

Example. ${}^{238}$U $\to {}^{234}$Th $+ \alpha + 4.27$ MeV, with $t_{1/2} = 4.5 \times 10^9$ yr — long because the 4.2 MeV $\alpha$ sits far below the ≈ 29 MeV barrier top.

Step 2: beta decay — the weak interaction at work §

In beta-minus decay a neutron converts to a proton via the weak interaction, emitting an electron and an antineutrino through a $W^-$ boson (see Elementary particles):

Beta decay moves a nucleus along the valley of stability (changing $N$ and $Z$ at fixed $A$) and is the dominant decay mode of the neutron-rich fission fragments. The neutrino carries away variable energy — the observation of its continuous electron spectrum was the first evidence for the neutrino.

Step 3: gamma decay — photons from nuclear levels §

After alpha or beta decay the daughter nucleus is often left in an excited state. It sheds the excitation by emitting a high-energy photon — nuclear electromagnetic radiation at the MeV scale. Gamma decay changes neither $Z$ nor $A$.

Example. ${}^{99m}$Tc $\to {}^{99}$Tc $+ \gamma$ (140 keV, $t_{1/2} = 6.0$ h): the workhorse of nuclear medicine, injected into patients for imaging because the 140 keV photon escapes the body and is easily detected.

Step 4: a complete decay scheme §

The classic decay scheme of cobalt-60 ties all three ideas together: the beta decay creates the excited daughter, and the gamma cascade de-excites it.

Note the division of labor: the weak interaction (β⁻) changes the species of nucleus; the electromagnetic interaction (the two γ’s) then drains the excitation energy of the daughter ${}^{60}$Ni. The two sharp gamma lines (1173 and 1332 keV) are used to calibrate gamma detectors.

Step 5: the decay law and radiometric dating §

Each unstable nucleus decays with a fixed probability per unit time $\lambda$, independent of age and chemistry:

Example. Carbon-14 ($t_{1/2} = 5730$ yr) is produced in the atmosphere and taken up by living organisms. When they die, the intake stops and the ${}^{14}$C/${}^{12}$C ratio decays exponentially — measuring it dates organic remains up to ~50 000 years (see the example problem below).

Authoritative sources

• Krane, Introductory Nuclear Physics, Ch. 6 “Radioactive decay”, Ch. 8 “Alpha decay”, Ch. 9 “Beta decay”, Ch. 10 “Gamma decay”
• Geiger & Nuttall, The ranges of the α particles from radioactive substances (1911)

A uranium-235 nucleus absorbs a thermal neutron and fissions into barium-141 and krypton-92, releasing 3 neutrons. Given the atomic masses: $m({}^{235}\text{U}) = 235.044$ u, $m(n) = 1.009$ u, $m({}^{141}\text{Ba}) = 140.914$ u, $m({}^{92}\text{Kr}) = 91.926$ u, compute the Q-value of the reaction in MeV. (1 u = 931.5 MeV/$c^2$.)