# Antiparticles and Photons

## Antiparticles and Photons Revision

**Particles and Antiparticles**

As far as we know, every **particle** in existence has a corresponding **antiparticle**. They share similar properties, aside from charge, and are created and annihilated together. In fact, a **photon** can create a **particle** and its corresponding **antiparticle**.

**Photon**

**Photons** are packets of electromagnetic **energy**. **Photons** can exist at a wide range of frequencies, represented by the electromagnetic spectrum, which is a continuous spectrum showing all possible frequencies of electromagnetic **energy**.

The **frequency** and **wavelength **of a **photon** are linked through the following equation:

\textcolor{aa57ff}{f = \dfrac{c}{\lambda}}

- \textcolor{aa57ff}{f} is the
**frequency**in Hertz \left(\text{Hz}\right). - \textcolor{aa57ff}{c} is the
**speed of light in a vacuum**, \textcolor{aa57ff}{3 \times 10^8 \: \text{ms}^{-1}}. - \textcolor{aa57ff}{\lambda} is the
**wavelength**in metres \left(\text{m}\right).

Electromagnetic **energy** is represented as **photons**, and the **energy** of a **photon **is dependent on the frequency or wavelength of the radiation through the following equation:

\textcolor{aa57ff}{E = hf = \dfrac{hc}{\lambda}}

- \textcolor{aa57ff}{E} is the
**energy of the photon**, often in joules or electronvolts \left(\text{J or eV}\right). - \textcolor{aa57ff}{h} is
**Planck’s constant**, \textcolor{aa57ff}{6.63 \times 10^{-34} \: \text{Js}} - \textcolor{aa57ff}{f} is the
**frequency**in Hertz \left(\text{Hz}\right).

Remember that 1 \: \text{eV} = 1.6 \times 10^{-19} \: \text{J}.

**Antiparticles**

Every **particle** has a corresponding **antiparticle**.

An **antiparticle** has the same **mass** and **rest energy** as the particle but has the **opposite charge**.

The **rest energy** is the energy equivalent of a particle when that particle is at rest. It is commonly measured in \text{MeV} \left(1 \: \text{MeV} = 1 \times 10^6 \: \text{eV} = 1.6 \times 10^{-13} \: \text{J}\right).

The datasheet will provide the **mass** (in \text{kg}) and** rest energy** in (in \text{MeV}) for each particle.

Particle |
Symbol |
Relative Charge |
Mass \left(\textbf{kg}\right) |
Rest Energy \boldsymbol{\left(\textbf{MeV}\right)} |

proton | p | +1 | 1.673 \times 10^{-27} | 938.3 |

antiproton | \bar{p} | -1 | 1.673 \times 10^{-27} | 938.3 |

neutron | n | 0 | 1.675 \times 10^{-27} | 939.6 |

antineutron | \bar{n} | 0 | 1.675 \times 10^{-27} | 939.6 |

electron | e^- | -1 | 9.11 \times 10^{-31} | 0.511 |

positron | e^+ | +1 | 9.11 \times 10^{-31} | 0.511 |

neutrino | v_e / v_{\mu} | 0 | 0 | 0 |

antineutrino | \bar{v}_e / \bar{v}_{\mu} | 0 | 0 | 0 |

There are two types of **neutrinos** that you need to know, the **electron neutrin****o** and **muon neutrino**. They both have **antiparticles**. As **neutrinos** are so small, it is assumed that their **mass** and **rest energy** is zero.

**Pair Production**

New particles are always formed together in **pairs**; the particle and its antiparticle equivalent. For example, if a proton is formed then an antiproton must also be formed.

Matter and antimatter can be created from **energy**. **Mass **can be turned into** energy** and **energy** can turn into **mass**. This comes from Einstein’s Special Theory of Relativity.

The **rest energy** of a particle (measured in \text{MeV}) is the **energy** equivalent of a particle’s** mass**. When energy is converted into **mass**, equal amounts of matter and antimatter are produced.

One way this can occur is through **collisions** of high-speed particles. If you fire two protons at high speed at each other, there is a significant amount of **energy** at the collision point. If there is enough **energy**, then that **energy** may be converted to **mass** and form a **particle** and its corresponding **antiparticle**. For example, an electron and positron can be formed. This is an example of **pair production**.

**Photons** play a key role in **pair production** as **energy** is in the form of a **photon** before it is converted to matter and antimatter.

A single **photon** (usually a gamma ray) must have enough energy to produce both the **particle** and **antiparticle**. To conserve momentum, this usually happens near the **nucleus** of an atom. After the formation of the particles, they move in opposite directions, again for momentum conservation. Observations in particle physics experiments may show the particles having curved paths. This is due to there being a magnetic field present.

The most commonly produced **pair** is the **electron** and **positron** as they have a lower combined **rest energy** than other pairs. This is shown below in the diagram.

To calculate the minimum** energy** a photon needs to undergo **pair production**, you use the following equation:

\textcolor{00bfa8}{E_{\text{min}} = hf = 2E_{\text{rest}}}

- \textcolor{00bfa8}{E_{\text{min}}} is the
**minimum energy of the photon**in joules or electronvolts \left(\text{J or eV}\right). - \textcolor{00bfa8}{h} is
**Planck’s constant**, 6.63 \times 10^{-34} \: \text{Js} . - \textcolor{00bfa8}{E_{\text{rest}}} is the
**rest energy of the particles**in joules or electronvolts \left(\text{J or eV}\right). - \textcolor{00bfa8}{f} is the
**frequency**in Hertz \left(\text{Hz}\right).

**Example: Calculations involving Pair Production**

Calculate the minimum frequency a photon needs to produce a proton through pair production.

**[2 marks]**

**First, let’s get the correct equation needed to solve the question and rearrange:**

E_{\text{min}} = hf = 2E_{\text{rest}}

f = \dfrac{2E_{\text{rest}}}{h}

**We need to calculate the rest energy of a proton, you will be given this in the data sheet:**

\text{Proton Rest Energy} = 938.3 \: \text{MeV}

**In pair production, an antiparticle is produced, which will have the same rest energy:**

\text{Antiproton Rest Energy} = 938.3 \: \text{MeV}

**Convert the rest mass energy of one particle to joules:**

E_{\text{rest}} = 938.3 \: \text{MeV} = 938.3 \times 10^6 \times 1.6 \times 10^{-19} \: \text{J} = 1.50128 \times 10^{-10} \: \text{J}

**Substitute into the equation to find frequency:**

f = \dfrac{2 \left(1.50128 \times 10^{-10}\right)}{6.63 \times 10^{-34}} = 4.53 \times 10^{23} \: \text{Hz}

**Annihilation**

When a **particle** meets its **antiparticle**, both particles are turned back into energy through the process of annihilation.

Antiparticles only exist for a very short period of time as they quickly meet a particle.

Two **photons** are produced, whose minimum combined** energy** will equal the rest **masses **of the particles. The minimum energy of a** energy** released due to **annihilation** is given by the following equation:

\textcolor{10a6f3}{E_{\text{min}} = hf_{\text{min}} = E_{\text{rest}}}

- \textcolor{10a6f3}{E_{\text{min}}} is the
**minimum energy of the photon**in joules or electronvolts \left(\text{J or eV}\right). - \textcolor{10a6f3}{f_{\text{min}}} is the
**minimum frequenc****y**in Hertz \left(\text{Hz}\right). - \textcolor{10a6f3}{E_{\text{rest}}} is the
**rest energy of the particles**in joules or electronvolts \left(\text{J or eV}\right). Note that the**rest energy**of the**particle**and**antiparticle**is the**same**.

The photons will travel in opposite directions in order to conserve momentum.

**Example: Calculations involving Annihilation**

Calculate the wavelength of the photons released when a proton and antiproton annihilate each other.

**[2 marks]**

**First, let’s get the correct equation needed to solve the question:**

E_{\text{min}} = hf = E_{\text{rest}}

**We need to find the wavelength, therefore we use the wave equation to find the correct equation:**

f = \dfrac{c}{\lambda}

\dfrac{hc}{\lambda} = E_{\text{rest}}

\dfrac{hc}{E_{\text{rest}}} = \lambda

**We need to use the rest energy of a proton, you will be given this in the data sheet. Remember that both a particle and its antiparticle will have the same rest mass:**

\text{Proton Rest Energy} = 938.3 \: \text{MeV}

**The rest energy needs to be converted to joules:**

E_{\text{rest}} = 938.3 \: \text{MeV} = 938.3 \times 10^6 \times 1.6 \times 10^{-19} \: \text{J} = 1.5 \times 10^{-10} \: \text{J}

**Substitute into the equation to find the wavelength:**

\lambda = \dfrac{6.63 \times 10^{-34} \times 3.0 \times 10^8}{1.5 \times 10^{-10}} = 1.3 \times 10^{-15} \: \text{m}

## Antiparticles and Photons Example Questions

**Question 1: **Calculate the wavelength required for a photon to have an energy of 3.5 \times 10^4 \: \text{MeV}.

**[3 marks]**

\begin{aligned} E &= \dfrac{hc}{\lambda} \\ \\ \boldsymbol{\lambda} &\boldsymbol{= \dfrac{hc}{E}} \\ \\ \boldsymbol{\lambda} &= \dfrac{6.63 \times 10^{-34} \times 3.0 \times 10^8}{5.6 \times 10^{-9}} = \boldsymbol{3.6 \times 10^{-17}} \: \textbf{m} \end{aligned}

**Question 2:** Why might a collision between an electron and positron not produce any particles?

**[1 mark]**

**The energy produced from the collision is below the combined rest mass of a particle and antiparticle. **

**Question 3:** Calculate the minimum frequency a photon needs in order to produce a neutron.

**[2 mark]**

E = hf and E_{\text{min}} = 2E_{\text{rest}}. We also know \text{Rest energy of a neutron} = 939.6 \: \text{MeV} .

\begin{aligned} hf_{\text{min}} &= 2E_{\text{rest}} \\ \\ \boldsymbol{f_{\textbf{min}} }&\boldsymbol{= \dfrac{2E_{\textbf{rest}}}{h}} \\ \\ f_{\text{min}} &\boldsymbol{=} \dfrac{2 \left(939.6 \times 10^6 \times 1.6 \times 10_{-19}\right)}{6.63 \times 10^{-34}} = \boldsymbol{4.5 \times 10^{23}} \: \textbf{Hz} \end{aligned}