PHYS-420 · Lesson 1 of 12 · 21 min
Why nuclear and particle physics: the structure of matter at the smallest scales
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Every atom you have studied so far has a tiny, heavy core. The nucleus holds almost all of an atom's mass in a region about \(10^{-14}\) m across, a hundred thousand times smaller than the atom itself. Inside it, protons and neutrons are held together by a force strong enough to overcome the electric repulsion of up to 92 protons crowded into that space. And the protons and neutrons are themselves made of smaller things: quarks, bound by gluons. Nuclear and particle physics is the study of these inner layers of matter: what they are made of, what holds them together, how they change, and what laws govern the smallest pieces we know.
This course follows that path inward. The first half is nuclear physics: the size, mass and spin of nuclei; models that explain their binding and their structure; radioactive decay; nuclear reactions; and the uses of nuclear physics for energy, medicine, dating and astrophysics. The second half is particle physics: detectors and accelerators, the zoo of particles and the conservation laws that organize it, quarks and the strong force, and finally the Standard Model, the Higgs boson and the open questions beyond it. Throughout, we build on PHYS-271, which introduced the nucleus, binding energy and radioactivity, and on PHYS-373, whose WKB treatment of alpha decay and Born approximation for scattering we will use directly.
Length and energy scales
The natural length unit of nuclear physics is the femtometre, \(1\ \text{fm} = 10^{-15}\) m, which physicists also call the fermi. A proton has a charge radius of about 0.84 fm; a medium nucleus is a few fermis across; the heaviest nuclei reach about 7 fm. The natural energy unit is the mega-electronvolt, MeV. Nuclear binding energies are about 8 MeV per nucleon, a million times the few electronvolts that bind outer electrons to atoms. That ratio is why nuclear fuels hold a million times more energy per kilogram than chemical fuels, a fact we will quantify in lesson 7.
Two constants make nuclear arithmetic quick. The combination \(\hbar c = 197.3\) MeV fm converts between energy and inverse length, and \(e^2/4\pi\varepsilon_0 = \alpha\hbar c = 1.44\) MeV fm gives Coulomb energies directly: two protons 1.44 fm apart have 1 MeV of electrostatic energy. The atomic mass unit, \(1\ \text{u} = 931.494\ \text{MeV}/c^2\), converts measured masses into energies.
To see structure of size \(d\), a probe must have a wavelength no longer than about \(d\). For a photon or an ultrarelativistic particle, \(E = hc/\lambda = 2\pi\hbar c/\lambda\), so resolving a nucleus a few fermis across needs particles of a few hundred MeV, and resolving the inside of a proton needs GeV energies. This simple estimate explains why the history of the field is a history of ever larger accelerators: each new layer of structure demanded a shorter wavelength and therefore a higher energy.
Notation for nuclei
A nucleus is specified by its proton number \(Z\), which fixes the chemical element, and its neutron number \(N\). Their sum \(A = Z + N\) is the mass number. We write \({}^{A}_{Z}\mathrm{X}_N\), or more often just \({}^{A}\mathrm{X}\) or X-\(A\): \({}^{238}\mathrm U\) has \(Z = 92\), \(N = 146\). Nuclei with the same \(Z\) are isotopes, with the same \(N\) isotones and with the same \(A\) isobars. About 250 nuclides are stable, and more than 3,000 radioactive ones have been made and studied. Plotting them on a chart of \(N\) against \(Z\) shows a narrow valley of stability that starts along \(N = Z\) and bends toward neutron excess for heavy nuclei, a shape we will explain in lesson 3.
Mass and binding energy
The mass of a nucleus is less than the total mass of its separate protons and neutrons. The difference, times \(c^2\), is the binding energy \(B\): the energy needed to pull the nucleus completely apart. Using atomic masses, which include the electrons, the electron masses cancel if we use the hydrogen-atom mass for each proton:
Atomic masses are measured to better than one part in \(10^8\) with Penning traps and mass spectrometers, so binding energies are known to a few keV. The program below converts tabulated masses into binding energies and binding energy per nucleon, and estimates radii with the empirical rule \(R = r_0A^{1/3}\), \(r_0 \approx 1.2\) fm, which we derive from scattering data in lesson 2.
# Scales of nuclear and particle physics, and binding energies from atomic masses
import math
hbar_c = 197.327 # MeV fm
u = 931.494 # MeV per atomic mass unit
m_H, m_n = 1.007825, 1.008665 # hydrogen atom and neutron masses (u)
print("Probe energy needed to resolve a length d (photon or ultrarelativistic particle)")
for name, d in (("atom", 1.0e5), ("nucleus (A = 100)", 5.6), ("proton", 0.84), ("quark scale", 0.01)):
E = 2 * math.pi * hbar_c / d # E = h c / lambda with lambda = d
print(f" {name:20s} d = {d:9.2f} fm E ~ {E:12.1f} MeV")
print()
# measured atomic masses (u)
nuclides = [("H-2", 1, 2, 2.014102), ("He-4", 2, 4, 4.002603), ("C-12", 6, 12, 12.000000),
("O-16", 8, 16, 15.994915), ("Fe-56", 26, 56, 55.934936), ("Ni-62", 28, 62, 61.928345),
("Sn-120", 50, 120, 119.902202), ("Pb-208", 82, 208, 207.976652), ("U-238", 92, 238, 238.050788)]
print("nuclide B (MeV) B/A (MeV) R = 1.2 A^(1/3) fm")
for name, Z, A, M in nuclides:
B = (Z * m_H + (A - Z) * m_n - M) * u
print(f" {name:7s} {B:9.2f} {B / A:8.3f} {1.2 * A ** (1 / 3):5.2f}")
rho = 3 / (4 * math.pi * 1.2 ** 3)
print()
print(f"nucleon number density ~ {rho:.3f} per fm^3; mass density ~ {rho * 1.6726e-27 * 1e45:.2e} kg/m^3")Probe energy needed to resolve a length d (photon or ultrarelativistic particle) atom d = 100000.00 fm E ~ 0.0 MeV nucleus (A = 100) d = 5.60 fm E ~ 221.4 MeV proton d = 0.84 fm E ~ 1476.0 MeV quark scale d = 0.01 fm E ~ 123984.2 MeV nuclide B (MeV) B/A (MeV) R = 1.2 A^(1/3) fm H-2 2.22 1.112 1.51 He-4 28.30 7.074 1.90 C-12 92.16 7.680 2.75 O-16 127.62 7.976 3.02 Fe-56 492.26 8.790 4.59 Ni-62 545.26 8.795 4.75 Sn-120 1020.54 8.505 5.92 Pb-208 1636.44 7.867 7.11 U-238 1801.70 7.570 7.44 nucleon number density ~ 0.138 per fm^3; mass density ~ 2.31e+17 kg/m^3
Three features of the output set the agenda for the nuclear half of the course. First, \(B/A\) is roughly constant, near 8 MeV, from oxygen to uranium: each nucleon is bound to only a few neighbours, not to all the others, so the nuclear force must be short ranged and saturating. Second, \(B/A\) rises steeply among light nuclei, peaks near iron and nickel at about 8.8 MeV, and falls slowly toward uranium. Energy is therefore released by fusing light nuclei and by splitting heavy ones, which is the physics of both stars and reactors. Third, the radius grows as \(A^{1/3}\), so the volume is proportional to \(A\) and the density of nuclear matter is the same in every nucleus: about 0.14 nucleons per cubic fermi, or \(2 \times 10^{17}\) kg per cubic metre. A teaspoon of nuclear matter would weigh as much as a mountain, and a neutron star is essentially one enormous nucleus held together by gravity.
The four interactions
Four fundamental interactions are known. Gravity is negligible for single particles. Electromagnetism binds atoms and also pushes protons apart inside the nucleus. The strong interaction binds quarks into protons and neutrons and, as a residual effect, binds nucleons into nuclei; its range is about 1 fm. The weak interaction changes one kind of quark or lepton into another and drives beta decay and the fusion reactions in the Sun; its range is even shorter, about \(10^{-3}\) fm. Their relative strengths at nuclear distances, roughly 1 for the strong force, \(1/137\) for electromagnetism and \(10^{-6}\) or less for the weak force, explain the time scales of decays: strong decays take \(10^{-23}\) s, electromagnetic transitions \(10^{-16}\) s or longer, and weak decays anything from milliseconds to billions of years.
A short history
Nuclear physics began by accident. In 1896 Henri Becquerel found that uranium salts fogged photographic plates in the dark, and in 1898 Marie and Pierre Curie isolated polonium and radium, far more active than uranium. Ernest Rutherford, working at McGill University in Montreal from 1898 to 1907, sorted the radiations into alpha, beta and gamma rays and, with Frederick Soddy, showed that radioactivity transforms one element into another. In Manchester in 1909, Hans Geiger and Ernest Marsden fired alpha particles at gold foil, and in 1911 Rutherford explained the rare large-angle scatterings by a tiny, massive, positively charged nucleus. In 1919 he observed the first artificial nuclear reaction, turning nitrogen into oxygen with alpha particles. James Chadwick discovered the neutron in 1932, completing the picture of the nucleus as protons and neutrons.
Particle physics grew from cosmic rays and then from accelerators. The positron (1932), the muon (1936), the pion (1947) and strange particles (late 1940s) were found in cosmic-ray tracks; from the 1950s, accelerators produced hundreds of new hadrons. Murray Gell-Mann and George Zweig proposed quarks in 1964 to organize them, and deep inelastic electron scattering at SLAC from 1968 saw point-like constituents inside the proton. The electroweak theory of Glashow, Weinberg and Salam, quantum chromodynamics for the strong force, and the discoveries of the charm quark (1974), the \(W\) and \(Z\) bosons (1983), the top quark (1995) and the Higgs boson (2012) built the Standard Model. Canada has played a large part throughout, from Rutherford's Montreal work and the ZEEP reactor at Chalk River (1945) to TRIUMF in Vancouver and the Sudbury Neutrino Observatory, whose director Arthur McDonald shared the 2015 Nobel Prize in Physics.
Why it matters
Nuclear physics generates about a tenth of the world's electricity and roughly 15 parts in 100 of Canada's, mostly from CANDU reactors in Ontario; it supplies the isotopes used in tens of millions of medical procedures a year, from technetium-99m imaging to cobalt-60 sterilization and lutetium-177 cancer therapy; it dates archaeological finds and rocks; and it explains why the Sun shines and where the elements came from. Particle physics answers what the world is made of at the deepest level we can reach, and its tools, from accelerators to detectors to the World Wide Web (invented at CERN in 1989), have spread far beyond the field. This course gives you the physical ideas behind all of these, with enough mathematics to calculate.
Worked problem
Binding energy of helium-4
Using \(M({}^4\mathrm{He}) = 4.002603\) u, \(M({}^1\mathrm H) = 1.007825\) u and \(m_n = 1.008665\) u, find the binding energy of He-4 and compare its binding energy per nucleon with that of the deuteron, 1.11 MeV.
Show solution
- Total mass of the parts: \(2(1.007825) + 2(1.008665) = 4.032980\) u.
- Mass defect: \(4.032980 - 4.002603 = 0.030377\) u.
- Binding energy: \(0.030377 \times 931.494 = 28.30\) MeV.
- Per nucleon: \(28.30/4 = 7.07\) MeV, more than six times the deuteron's 1.11 MeV.
Answer: \(B = 28.3\) MeV and \(B/A = 7.07\) MeV. The alpha particle is exceptionally tightly bound, which is why it is emitted as a unit in alpha decay and why it is the main product of stellar hydrogen burning.
Common mistakes
MistakeUsing nuclear masses for the nucleus but atomic masses for hydrogen in the binding energy formula.
InsteadUse atomic masses throughout with \(M({}^1\mathrm H)\) for each proton, so the electron masses cancel.
MistakeThinking the nucleus is a scaled-down atom with a central attracting body.
InsteadThere is no central body: nucleons bind each other through a short-range force, which is why \(B/A\) is nearly constant.
MistakeAssuming a probe of energy \(E\) resolves sizes of \(\hbar c/E\) in every case.
InsteadFor a massive non-relativistic probe the wavelength is \(h/p\), not \(hc/E\); use the momentum.
MistakeTreating the weak interaction as just a weaker version of electromagnetism.
InsteadIt changes particle identity (flavour) and has a very short range because its carriers are massive.
MistakeExpecting nuclear density to grow with \(A\) as atomic density does.
InsteadNuclear volume grows in proportion to \(A\), so the density is roughly the same in every nucleus.
Summary and key terms
- Nuclei are a few fermis across; binding energies are about 8 MeV per nucleon, a million times atomic binding.
- Resolving a size \(d\) needs \(E \gtrsim 2\pi\hbar c/d\): hundreds of MeV for nuclei, GeV for the inside of nucleons.
- \(B = [ZM({}^1\mathrm H) + Nm_n - M]c^2\); \(B/A\) peaks near iron, so fusion of light and fission of heavy nuclei release energy.
- \(R \approx 1.2A^{1/3}\) fm, so nuclear density is about 0.14 nucleons per fm³ in every nucleus.
- Four interactions; the strong and weak forces are short ranged and dominate nuclear and particle processes.
- Mass number
- \(A = Z + N\), the total number of nucleons.
- Binding energy
- Energy needed to separate a nucleus into free protons and neutrons.
- Fermi (fm)
- \(10^{-15}\) m, the natural nuclear length unit, also called a femtometre.
- Isotope
- Nuclei with the same \(Z\) and different \(N\).
- Valley of stability
- The band of stable nuclides on a chart of \(N\) against \(Z\).
Practice problems
Work each problem on paper first, then reveal the answer to check.
What photon energy has a wavelength of 1 fm?
Show answer
\(E = 1240\ \text{MeV fm}/1\ \text{fm} = 1240\) MeV, about 1.2 GeV.
Estimate the radius of \({}^{125}\mathrm{Te}\).
Show answer
\(R = 1.2 \times 125^{1/3} = 1.2 \times 5 = 6.0\) fm.
Compute the Coulomb energy of two protons 2 fm apart.
Show answer
\(1.44/2 = 0.72\) MeV.
Using the program's table, how much energy would be released by fusing two O-16 nuclei into an S-32 nucleus (\(B/A = 8.493\) MeV)?
Show answer
\(32 \times 8.493 - 2 \times 127.62 = 271.8 - 255.2 = 16.5\) MeV.
Why can't a 1 MeV electron resolve the structure inside a proton?
Show answer
Its wavelength \(h/p\) is about 870 fm, a thousand times the proton's size.
A neutron star has mass \(1.4M_\odot = 2.8 \times 10^{30}\) kg. Estimate its radius from nuclear density.
Show answer
Volume \(= 2.8 \times 10^{30}/2.3 \times 10^{17} = 1.2 \times 10^{13}\) m³, so \(R = (3V/4\pi)^{1/3} \approx 14\) km.
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