Free Pearson Edexcel International A Level Physics lesson

Atoms, accelerators and detectors

Use energetic probes to infer structures too small to see directly.

What you’ll learn

  • Interpret scattering evidence.
  • Explain linac and cyclotron operation.
  • Connect probe energy, particle tracks and lifetime.

Rare large deflections reveal a compact nucleus

An atom’s proton number Z counts nuclear protons and determines its element. Its nucleon number A counts protons plus neutrons; neutron number is A − Z. Isotopes share Z but differ in neutron number. A neutral atom has Z electrons, whereas ionisation changes electron count without changing the nucleus.

In alpha-scattering experiments most alpha particles passed through thin metal foil with little deflection, while a very small fraction scattered through large angles. A diffuse positive-charge model could not easily produce those rare strong reversals. The observations supported a tiny, concentrated, positively charged nucleus containing most of the mass, with much of the atom’s volume relatively empty.

The evidence constrains a model rather than providing a direct photograph. The alpha particle is repelled electrically by the nucleus, and a close approach can cause a large deflection. Thin foil limits multiple scattering; repeated small deflections through thick material would complicate interpretation. Atomic models changed as new scattering and spectral evidence exposed limitations in earlier descriptions.

Neutron number = A − Z; neutral electron count = Z; nuclear charge = +Ze

Symbols and units

  • A is nucleon number and Z is proton number, both integer counts
  • e is positive elementary charge in C
  • nuclear charge is +Ze.

How the equation explains the idea

  1. Every nucleon is a proton or neutron, so neutron count is A − Z. A neutral atom has Z electrons to balance nuclear charge.
  2. For A = 14 and Z = 6, there are 8 neutrons and 6 protons. Ionisation changes electron count, not A or Z; a 2+ ion has two fewer electrons than its neutral atom.

Worked example

Identify an isotope

  1. A nucleus has A = 27 and Z = 13.
  2. It contains 13 protons and 27 − 13 = 14 neutrons.
  3. A neutral atom has 13 electrons; a 3+ ion has 10 electrons.

Ionisation changes electrons, not proton or nucleon number.

Keep this idea. Use both the common weak deflections and rare strong deflections when explaining nuclear evidence.

Electric gaps add energy and magnets return the beam

Heating a cathode gives some electrons enough energy to escape the surface: thermionic emission. An electric field between the cathode and a positive anode accelerates the released electrons. Their kinetic-energy gain has magnitude eV when they cross an accelerating potential difference V. A magnetic field can then steer the beam, but cannot supply this energy gain by itself.

A linear accelerator uses alternating electric fields across gaps between drift tubes. Particles are shielded from the field while inside the conducting tubes, and the field reverses so the next gap accelerates them again. At non-relativistic increasing speeds, drift-tube lengths increase to match the required time between gaps; the timing must stay synchronised with the radio-frequency voltage.

A cyclotron uses a perpendicular magnetic field to bend particles into semicircles inside dees and an alternating electric field across their gap to accelerate them. In the classical model the orbital period is 2πm/(B|q|), independent of speed, while the orbit radius increases. At high relativistic energies that fixed-frequency synchronism fails. Detectors instead use ionisation to register passage and magnetic deflection to analyse charged tracks.

ΔEk = |q|Vaccelerating; classical cyclotron T = 2πm/(B|q|); r = p/(B|q|)

Symbols and units

  • q is particle charge in C
  • Vaccelerating is accelerating voltage magnitude in V
  • ΔEk is energy gain in J
  • m is mass in kg
  • B is magnetic field in T
  • T is orbital period in s.

How the equation explains the idea

  1. An accelerating potential transfers energy |q|V when its polarity accelerates the particle. In a perpendicular magnetic field, r = mv/(B|q|) in the classical model.
  2. The circular period is 2πr/v. Substitute r to obtain T = 2πm/(B|q|), independent of speed only while relativistic changes and other model failures are negligible.

Worked example

Electron gun energy

  1. An electron starts with negligible kinetic energy and crosses an accelerating 2500 V.
  2. ΔEk = 1.60 × 10⁻¹⁹ × 2500 = 4.00 × 10⁻¹⁶ J.
  3. This is 2500 eV = 2.50 keV.

The electric potential difference supplies the kinetic energy.

Keep this idea. Acceleration, steering and detection are different jobs performed by different physical interactions.

Small structures need short-wavelength probes

To resolve a structure, a probe wavelength must be comparable with or smaller than the feature. Matter-wave wavelength follows λ = h/p, so higher momentum gives shorter wavelength. High-energy beams can probe nucleon substructure and can create new particles in collisions; neither an optical microscope nor a low-momentum beam resolves arbitrarily small scales.

Charged particles can ionise a detector medium, leaving a measurable signal, while magnetic curvature helps infer charge sign and momentum. At an interaction vertex, momentum vectors and electric charge must balance when all products are included. A gap or missing transverse momentum can suggest an unobserved neutral particle, but detector acceptance and resolution must also be considered.

Unstable particles have a mean lifetime measured in their rest frame. A rapidly moving particle can be observed to survive longer in the laboratory frame, allowing it to travel farther than a simple rest-lifetime calculation would predict. Atmospheric muons and accelerator beams make this effect significant. This is relativistic time dilation, not evidence that the particle’s speed exceeds c; no relativistic equation is required here.

λ = h/p; interaction: Σp before = Σp after and Σq before = Σq after

Symbols and units

  • λ is de Broglie wavelength in m
  • h is Planck’s constant in J s
  • p is momentum magnitude in kg m s⁻¹
  • q is signed charge in C.

How the equation explains the idea

  1. From λ = h/p, a tenfold momentum increase gives a tenfold shorter wavelength, helping probe smaller length scales.
  2. At a collision, conserve total momentum as a vector and total charge with signs. An unobserved neutral particle can carry momentum despite leaving no charged track; track count alone is not momentum balance.

Worked example

A wavelength target

  1. Use h = 6.63 × 10⁻³⁴ J s and target λ = 1.0 × 10⁻¹⁵ m.
  2. p = h/λ = 6.63 × 10⁻¹⁹ kg m s⁻¹.
  3. This is a high momentum for a microscopic particle; classical speed formulas may be unsuitable.

Shorter structures require larger probe momentum.

Keep this idea. Use wavelength for resolution, conservation for tracks and frame dependence for fast-particle lifetime.

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