# Particles for SQA Higher Physics 1. Antiparticles
2. The Photon Model
3. Weak interactions

Every particle has a corresponding antiparticle. An antiparticle is a subatomic particle which has the same mass and rest energy, and the opposite charge as the corresponding particle. Some common antiparticles are given with their corresponding particle in the diagram. The rest energy in MeV of neutrons and antineutrons is 938.3, of protons and antiprotons is 939.6, of electrons and positrons is 0.51 and of neutrinos and antineutrinos is 0. The Photon model states that electromagnetic radiation is released in energy packets at discrete energy values called photos. The energy of a photon can be calculated using the following equation: Energy = hf = hc/λ, where h is the Planck constant 6.63 × 10⁻³⁴ Js, f is the frequency, c is the speed of light in a vacuum and λ is the wavelength. Atoms will radioactively decay through the weak interaction, which occurs through the exchange particle of W bosons, which come in the forms of W⁻ and W⁺ bosons. β⁻ and β⁺ decay occurs through the weak interaction. In β⁻ decay, a neutron will decay into a proton, an electron and an anti-electron neutrino through the W⁻ boson as the exchange particle. In β⁺ decay, a proton will decay into a neutron, a positron and an electron neutrino through the W⁺ boson as the exchange particle. Two more examples of the weak interaction are electron capture and electron-proton collisions. Electron capture occurs when a proton absorbs an electron and releases a neutron and an electron neutrino through the W⁺ boson exchange particle. Electron-proton collisions also result in the same products of a neutron and an electron neutrino through the W⁻ boson exchange particle. # 1

Draw a table to show the rest energies, in MeV, of an antineutrino, an antiproton, a positron and an antineutrino.

image # 2

Give two examples of weak interactions which occur through the W⁺ boson.

Two examples of weak interactions are β⁺ decay and electron capture. # 3

Use the information for the given photon to calculate its wavelength.

Using E = hc/λ, 1 × 10⁻¹⁹ J = (6.63 × 10⁻³⁴ Js × 3 × 10⁸ m/s)/λ, therefore λ = 1.989 × 10⁻⁶ m. # 4

Explain what an electron-proton collision results in.

When an electron and a proton collide, they will produce a neutron and an electron neutrino through the W⁻ boson exchange particle. # 5

What is the frequency of a photon with energy 9 × 10⁻¹⁷ J?

Using E = hf, 9 × 10⁻¹⁷ J = 6.63 × 10⁻³⁴ Js × f, therefore f = 1.36 × 10¹⁷ Hz (to 3 significant figures). End of page