Particle accelerators, detectors and creating particles

Core ideas

Linear accelerator (linac)

A line of hollow cylindrical drift tubes sits in an evacuated pipe. The tubes are connected alternately to the two terminals of a high-frequency alternating pd.

• Inside a tube there is no electric field, so the particle drifts at constant speed.
• In each gap there is an electric field. The pd reverses while the particle is inside a tube, so every gap it reaches accelerates it.
• The time spent in each tube must be half a period of the supply. The particle gets faster along the line, so each tube must be longer than the one before: $L = v \times \dfrac{T}{2} = \dfrac{v}{2f}$.
• The pipe is evacuated so the particles do not collide with air molecules.

Cyclotron

Two hollow D-shaped electrodes (“dees”) sit in a uniform magnetic field perpendicular to their faces, with an alternating pd across the gap between them.

• The electric field in the gap accelerates the particle. The magnetic field provides the centripetal force, so the particle moves in semicircles inside the dees.
• $BQv = \dfrac{mv^2}{r}$ gives $r={\displaystyle }$, so the radius grows as the particle speeds up and it spirals outwards.

Synchrotron (CCEA, OCR B)

Particles travel round a ring of fixed radius. The magnetic field is increased as their momentum rises, keeping $r = p/BQ$ constant, and radio-frequency cavities accelerate them once each lap. Two beams can travel in opposite directions and be made to collide head-on, as at the Large Hadron Collider.

Detectors and tracks

Charged particles ionise the medium they pass through, leaving a visible track in a cloud chamber or bubble chamber, or an electrical signal in a modern detector. Neutral particles leave no track. A neutral particle only shows up as a gap between tracks, often a V where it decays into two charged particles.

With a magnetic field applied across the detector:

• The direction of curvature gives the sign of the charge (Fleming’s left-hand rule, with the current in the direction of motion of a positive charge).
• The radius gives the momentum: $p = BQr$.
• A spiral means the particle is losing kinetic energy by ionising, so $p$ and $r$ both decrease.
• Momentum is a vector: in a decay or collision, the components of momentum along and perpendicular to the original direction are each conserved.

Creating particles and why high energies are needed

In a collision, kinetic energy can be converted into the rest energy of new particles: $\Delta E = c^2 \Delta m$. Masses are quoted in MeV/c² or GeV/c². For example, a rest energy of 938 MeV means a mass of 938 MeV/c². To get kilograms, convert the energy to joules and divide by $c^2$.

• Fixed target: momentum must still be conserved after the collision, so the products must keep moving and some of the beam energy stays as kinetic energy. It cannot all become mass.
• Colliding beams with equal and opposite momenta: the total momentum is zero, so in principle all the energy is available for new mass.
• Resolving structure: to probe a size $d$, the de Broglie wavelength must be comparable to it, $\lambda = \dfrac{h}{p} \lesssim d$. For m this needs . For a particle moving close to , GeV.

Worked examples

Exam technique

• “Explain why the tubes get longer”: give both parts. The particle speeds up, and the time in each tube must stay fixed at half the period of the supply.
• Derivations: start from $BQv = mv^2/r$ and state that the magnetic force is perpendicular to the velocity, so it provides the centripetal force.
• Track questions: say what the curvature shows (the charge), what the radius shows (the momentum) and what a changing radius shows (energy loss). Use the field direction and Fleming’s left-hand rule explicitly.
• Unit ladders: eV → J (× $1.60 \times 10^{-19}$). MeV/c² → kg: multiply by , then divide by .

Common mistakes

Quick recap

• A linac accelerates particles in the gaps. They drift inside the tubes, and the tubes lengthen because $v$ increases while the time in each tube stays at $T/2$.
• In a cyclotron the E-field accelerates and the B-field bends. $r = p/BQ$, and $f = BQ/2\pi m$ is independent of speed.