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132–138Nuclear and Particle PhysicsStretch

Particle Accelerators and Detectors

Particle physics runs on two pieces of equipment: an accelerator\textbf{accelerator} to give charged particles enormous energies, and a detector\textbf{detector} to work out what emerges when they collide. Both rely entirely on the electric and magnetic field physics you already know.

40 min Topic 8 · Nuclear and Particle Physics

What you'll be able to do

  • Describe the operation of a linear accelerator and a cyclotron
  • Use r=p/BQr = p/BQ for a charged particle in a magnetic field
  • Interpret particle tracks in a detector to infer charge and momentum
  • Use ΔE=c2Δm\Delta E = c^{2}\Delta m and the units MeV and GeV/c2c^{2}
1

Linear accelerators

A linac\textbf{linac} is a long line of hollow metal drift tubes connected alternately to a high-frequency alternating supply. Inside a tube there is no field, so a particle drifts at constant speed. In the gaps between tubes it is accelerated by the p.d.

The supply reverses each time the particle is inside a tube, so every gap it reaches has the right polarity to push it forward. A particle of charge QQ gains energy QVQV at each gap.

Because the particle speeds up but the alternating supply has a fixed frequency, it must spend the same time in each tube. Successive tubes therefore get longer\textbf{longer}.

Example 1A proton passes through 40 gaps in a linac, each with an accelerating p.d. of 2.5 MV. Find its final kinetic energy in MeV and joules.
1Energy per gap =eV=2.5= eV = 2.5 MeV.
2Total =40×2.5=100= 40 \times 2.5 = 100 MeV.
3100×1.60×10−13100 \times 1.60\times10^{-13} J.
Answer100 MeV=1.6×10−11 J100\ \mathrm{MeV} = 1.6\times10^{-11}\ \mathrm{J}

Tip — An electron-volt is the energy gained by one electron charge through 1 V. A charge of ee through VV volts gains exactly VV eV, which saves conversions.

2

The cyclotron

A cyclotron has two hollow D-shaped electrodes, the dees\textbf{dees}, between the poles of a large magnet. The magnetic field bends particles into semicircles inside the dees, and an alternating p.d. across the gap accelerates them each time they cross.

With each crossing the particle speeds up and its radius increases, since r=p/BQr = p/BQ. It spirals outwards until it leaves at the edge.

The key insight is that the time for each semicircle does not depend on speed. A faster particle travels a bigger circle in the same time, so a fixed-frequency supply stays in step. This holds only while the particle is non-relativistic; at very high speeds its effective mass increases and it falls out of step.

r=pBQr = \dfrac{p}{BQ}
Radius of path in a magnetic field.
f=BQ2πmf = \dfrac{BQ}{2\pi m}
Cyclotron frequency, independent of speed.
Example 1Protons in a cyclotron move in a 1.2 T field. Find the frequency of the accelerating supply, and the momentum of protons leaving at radius 0.50 m.
1f=1.2×1.60×10−192π×1.67×10−27=1.83×107f = \dfrac{1.2 \times 1.60\times10^{-19}}{2\pi \times 1.67\times10^{-27}} = 1.83\times10^{7} Hz.
2p=BQr=1.2×1.60×10−19×0.50p = BQr = 1.2 \times 1.60\times10^{-19} \times 0.50.
Answerf=1.8×107 Hzf = 1.8\times10^{7}\ \mathrm{Hz}; p=9.6×10−20 kg m s−1p = 9.6\times10^{-20}\ \mathrm{kg\,m\,s^{-1}}.
3

Reading particle tracks

Detectors reveal charged particles through the ionisation they cause along their path. Neutral particles leave no track, but they can be inferred from gaps — a track that starts from nowhere, or a "V" of two opposite tracks appearing where a neutral particle decayed.

A magnetic field is applied across the detector. From the tracks you can deduce: the sign of charge\textbf{sign of charge} from the direction of curvature, using Fleming’s left-hand rule; the momentum\textbf{momentum} from the radius, since p=BQrp = BQr; and a loss of energy from a track that spirals inwards as the particle slows.

Tighter curvature means lower momentum. Two particles with equal and opposite charges created at the same point curve in opposite directions — the signature of pair production.

Example 1A track of an electron in a 0.25 T field has a radius of 4.0 cm. Find the electron’s momentum.
1p=BQr=0.25×1.60×10−19×0.040p = BQr = 0.25 \times 1.60\times10^{-19} \times 0.040.
Answer1.6×10−21 kg m s−11.6\times10^{-21}\ \mathrm{kg\,m\,s^{-1}}

Tip — For track diagrams, state the rule you used for direction and the relationship between radius and momentum explicitly.

4

Mass, energy and new particles

In a high-energy collision, kinetic energy can become mass: new particles appear that were not there before. The conversion follows ΔE=c2Δm\Delta E = c^{2}\Delta m.

Particle masses are often quoted in MeV/c2\mathrm{MeV}/c^{2} or GeV/c2\mathrm{GeV}/c^{2} — the energy equivalent divided by c2c^{2}. A proton has mass 938 MeV/c2938\ \mathrm{MeV}/c^{2}, so creating a proton and an antiproton needs at least 1876 MeV.

In annihilation\textbf{annihilation} the reverse happens: a particle meets its antiparticle and both masses become energy, usually as two photons travelling in opposite directions so momentum is conserved.

In every interaction, charge and momentum are conserved. Momentum conservation is why collisions between two beams moving in opposite directions release far more usable energy than firing a beam at a fixed target.

ΔE=c2Δm\Delta E = c^{2}\Delta m
Mass–energy equivalence.
Example 1An electron and a positron, each essentially at rest, annihilate. Find the energy of each photon produced. The electron mass is 9.11×10−319.11\times10^{-31} kg.
1Total mass =2×9.11×10−31= 2 \times 9.11\times10^{-31} kg.
2Total energy =1.822×10−30×(3.00×108)2=1.64×10−13= 1.822\times10^{-30} \times (3.00\times10^{8})^{2} = 1.64\times10^{-13} J.
3Two photons share it equally.
Answer8.2×10−14 J8.2\times10^{-14}\ \mathrm{J} each, about 0.51 MeV.

Equation recap

r=p/BQr = p/BQ
Track radius.
f=BQ/2πmf = BQ/2\pi m
Cyclotron frequency.
Ek=QVE_{k} = QV
Energy gained per gap.
ΔE=c2Δm\Delta E = c^{2}\Delta m
Mass–energy.

Common mistakes to avoid

Saying the magnetic field accelerates particles in a cyclotron.
The electric field across the gap does the work; the magnetic field only bends the path.
Thinking linac tubes are equal length.
They lengthen so transit time stays matched to the supply.
Saying a tighter track means higher momentum.
Smaller rr means smaller pp for the same field and charge.
Having annihilation produce one photon.
Two photons are needed to conserve momentum.

Key takeaways

  • Linacs accelerate in gaps between drift tubes that lengthen along the line.
  • Cyclotrons use a magnetic field to bend and a gap p.d. to accelerate; f=BQ/2πmf = BQ/2\pi m.
  • Track radius gives momentum, p=BQrp = BQr; curvature direction gives charge sign.
  • Kinetic energy can become new mass: ΔE=c2Δm\Delta E = c^{2}\Delta m.
  • Charge and momentum are conserved in all particle interactions.

Test yourself

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