index.Rmd

--- 
title: "Inside the Atom"
author: "Dr Maire Gorman"
date: "`r Sys.Date()`"
site: bookdown::bookdown_site
output: bookdown::gitbook
documentclass: book
bibliography: [book.bib, packages.bib]
biblio-style: apalike
link-citations: yes
github-repo: rstudio/bookdown-demo
description: "Notes for Inside the Atom"
---

# Prerequisites {-}

```{r setup, include=FALSE}
knitr::opts_chunk$set(echo = TRUE)
```
Dr Maire Gorman   
Room 1.21  
maire.gorman@bristol.ac.uk  
Course notes courtsey of Prof. Helen Heath  

This course is a brief introduction to Nuclear and Particle Physics which will be built on, for most of you, in the Nuclear and Particle Physics course in the second year. 

## Books
The material in these lectures is almost all covered in chapters 40 and 41 of Tipler and Mosca. I’ve indicated the sections in Tipler which correspond to each section in the section titles. The semi-empirical mass formula in Lecture 2 is not covered in Tipler and so you might like to look at Williams "Nuclear & Particle Physics" which is the recommended text for the 2nd year Nuclear and Particle Physics Course. Alternatively, there are many 
good internet sites which cover this. 

## Learning Materials
- *These notes*
- Pre-recorded videos and associated PowerPoint slides 
- Live lectures (Monday 5pm, Friday 9am)
- Workshop material 
- Core Physics II problem book 
- Exam past papers 

Please note that you should expect to augment these resources with your own reading. 

## Course Breakdown 
### Section A – Inside the Atom

#### Lecture 1 
- Basics; Notation, Units
- Binding Energy

#### Lecture 2 
- The Liquid Drop Model
- The Semi-Empirical Mass Formula

#### Lecture 3
- Radioactive Decay

### Section B - Inside the Nucleus

#### Lecture 4
- Properties of Fundamental Particles
- Quark Model of Hadrons

#### Lecture 5
- The Fundamental Forces

#### Lecture 6
- Beta-decay in the quark model


## Learning Outcomes
By the end of this course you should be able to:

**1. Quote**\

- the approximate size of atoms, nuclei and other subatomic particles and their charges.
- the approximate range of fundamental and internucleon forces the masses of their carriers.
- the approximate mass, the charges and the flavour quantum numbers of the quarks. 
- the approximate interaction times associated with different forces.
- the approximate value of nuclear binding energies. 

**2. Explain the meaning of**\

- the superscripts and subscripts used with atomic symbols to denote the composition of nuclei.
- the terms hadrons, baryons and mesons and determine the quark content of particles given the 
quantum numbers.
- the terms isotope, isotone and isobar.
- the use of the letters A, Z and N when describing nuclei.

**3. Use**\

- the mass, momentum and energy units of MeV/c2, u, MeV/c and MeV and convert to SI units.
- Avogadro’s number to perform simple calculations involving numbers of atoms.

**4. Describe**\

- the principles behind the liquid drop model of the nucleus
- what is meant by saturation of a force. 
- how the atomic number, Z, varies with the neutron number, N, for stable and long lived nuclei.
- how the binding energy per nucleon varies with increasing mass number, A 
- what is meant by magic numbers.
- what is meant by a radioactive series and the transformations that happen in such a series.
- what is meant by an isospin multiplet
- the changes that take place in alpha, beta and gamma decay
- why quarks are confined within hadrons
- how the electron kinetic energy spectrum in beta decay indicates the existence of the neutrino.
- beta decay in terms of the quark transformations.
- the origins of the terms of the Semi Empirical Mass formula (SEMF).

**5. Calculate**\

- mass defects and binding energies in nuclei using atomic masses or the semi empirical mass formula
- the activity of radioactive sources from decay constants/half lives/mean lives and vice versa.
- the energy released in different types radioactive decay using the SEMF or measured atomic masses.
- estimated quark masses from the masses of members of hadron multiplets.
- the estimated of the range of forces from the mass of the exchange particle. 

**6. Apply**\

- the laws of conservation of energy, charge, lepton number, flavour and baryon number to determine i) if interactions are allowed or forbidden, ii) the force by which interactions would normally proceed. 

# Basics

## Preparation exercises

What are the charges, masses and approximate sizes of the following? 

- Protons
- Neutrons
- Electrons
- Hydrogen Atom
- Gold Atom
- Gold Nucleus 

## Notation (40-1)

Z = atomic number = number of protons   
N = number of neutrons   
A = mass number = N+Z   
Isotopes - Have the same no of protons  
Isotones - Have the same no of neutrons  
Isobars - Have the same mass number (same number of nucleons/baryons)  

## Units (40-1)
In nuclear physics the energies we deal with are usually of the order of $\mathrm{MeV}$.e. this is the typical magnitude of the kinetic energies of particles produced in radioactive decay or involved in nuclear interactions. We usually give masses in units of $\mathrm{MeV/c^2}$ or atomic mass unit $\mathrm{u}$ where $1\mathrm{u} = 931.494013 \pm 0.000037$ $\mathrm{MeV/c^2}$ which is 1/12  of the mass of a $^{12}_{6}$C atom. 

## Avogadro's Number (17-3)
Avogadro’s number, $N_{A} = 6.022\;x\;10^{23}$, is the number of Carbon-12 atoms in 1 mole of Carbon-12. One mole of Carbon-12 has a mass of 12 $\mathrm{g}$. One mole of any atomic element contains $N_{A}$ atoms of that element and has a total mass in grams numerically equal to the relative atomic mass.

## Mass Defect (40-1)
In order for a nucleus to form the total energy of the nucleus must be less than the total energy of the constituent parts; remembering that mass is a form of energy. The mass of any nucleus is therefore less than the mass of its constituent protons and neutrons. We define the mass defect ($\Delta m$) as the “loss” of mass which occurs when the nucleus is formed. 

>For example:
>The mass of an alpha particle (the helium nucleus, $^{4}_{2}He)$ is $3727.379$ $\mathrm{MeV/c^{2}}$. 
>The total mass of the constituents (two protons and two neutrons) is
>$2m_{p}+ 2m_{n}= 2(938.272) + 2(939.565) = 3755.764$ $\mathrm{MeV/c^{2}}$. 
>The mass defect is therefore $3755.764 − 3727.379 = 28.295$ $\mathrm{MeV/c^{2}}$. 

## Binding Energy (40-1)
The binding energy of a nucleus is the energy released when the constituent parts of a nucleus are put together. The binding energy (BE) and the mass defect $\Delta m$ are related by

\begin{equation*}
BE = (\Delta m) c^{2}
\end{equation*}

The binding energy for the Helium nucleus from the example above is 28.295 $\mathrm{MeV}$. 

The mass of a nucleus composed of Z protons and (A-Z) neutrons is given by
\begin{equation*}
m(A, Z) = Zm_{p} + (A-Z)m_{n} - B(A, Z)/c^{2}
\end{equation*}
where $Zm_{p}$ is the mass of the protons, $(A-Z)m_{n}$ is the mass of the neutrons and $B(A, Z)$ is the binding energy.
\begin{equation*}
\mathrm{Nuclear\;mass} = \mathrm{Constituent\;masses} – \mathrm{Nuclear\; Binding\;Energy}/c^{2}
\end{equation*}

Often atomic masses are used in the calculation because it is easier to measure the masses of atoms than the masses of nuclei. The atomic mass is 
\begin{equation*}
M(A, Z) = Zm_{p} + (A-Z)m_{n} + Zm_{e} - B(A, Z)/c^{2} - b(Z)/c^{2}
\end{equation*}
where $Zm_{p}$ is the mass of the electrons and b(Z) is the **atomic** binding energy. Atomic binding energies are small compared to nuclear binding energies and can usually be neglected. 

## Summary 
You should 

- Be able to use the notation for nuclei. 
- Be able to use and convert between atomic mass units and $\mathrm{MeV/c^{2}}$. 
- Understand what is meant by binding energy and be able to calculate BE from nuclear or atomic masses.
- Use Avogadro’s number to calculate the number of atoms of an element given the mass. 

## Example
In stars “hydrogen burning” combines four protons to make a Helium ($^{4}_{2}$He) nucleus. What is the energy released when 1 g of Hydrogen is “burnt”?

## Problem book questions
4.1 - 4.8

# SEMF

## Preparation exercises
If a nucleus, with atomic number $Z$ and mass number $A$ is a spherical “string bag” full of spherical neutrons and protons which are the same size. How you would expect the following quantities to depend on $A$ and $Z$ ?

- The total mass
- The volume of the bag
- The surface area of the bag
- The repulsive force between the protons

## The Semi-Empirical Mass Formula
The semi empirical mass formula is a formula which attempts to describe how the mass of a nucleus depends on $A$ and $Z$. The SEMF has a number of terms each describing one physical property. The constants multiplying 
each term are determined by fits to data on nuclear masses. Most of the principles are covered in T&M 40-1 but the formula is not quoted. See *Williams* or any other nuclear physics text book. Note that since the constants are based on fits to data the exact values will depend on the data used and will therefore vary slightly.

## The liquid drop model (40-1)
We can scatter particles from each other and learn about their size and shape. E.g. Rutherford scattering initially indicated that the nucleus was very small. 

Experiments show that most nuclei are approximately spherical with radius 
$R = R_{0}A^{1/3}$, where $R_{0}$ is a constant of order 1 $\mathrm{fm}$. The volume of a nucleus is proportional to $R^{3} =R_{0}^{3}A$. The mass of a nucleus is also roughly proportional to $A$ so the densities of nuclear matter in different atoms (mass/volume) are approximately the same. 

It is possible to visualise the nucleus as a small drop of liquid containing particles which are rather strongly 
bound to each other.

> Nuclei have the same interior density so mass depends on $r^{3}$ and the BE/nucleon is approximately constant. 

>For liquid drips, the interior density doesn't depend on drop size and the heat of vaporization is proportional to mass. 

## BE per nucleon (40-3)
<figure>
    <img src="Figures/L2-BE-nucleon.PNG" alt="Binding energy per nucleon plot">
    <figcaption>Binding energy per nucleon<figcaption>
<figure>  

## Z versus N (40-1)
<figure>
    <img src="Figures/LE-ZN-plot.PNG" alt="ZN plot">
    <figcaption>Atomic number versus Number of neutrons.<figcaption>
<figure>  

You can make these plots for these sections yourself using the spreadsheet on the Blackboard site, these are exercises 4.5 and 4.6 in the problems book. The information is displayed in similar form in plots 40-1 and 40-3 in Tipler.

## Semi-Empirical Mass Formula (atomic mass units)
This is the form of the SEMF in atomic mass units. Identify the following terms: Surface term, proton masses, Coulomb term, Volume term, Pairing term, Neutron masses, Asymmetry term. 

\begin{equation*}
M(Z, A) = + 1.008142Z + 1.008982(A-Z) \\ 
- 0.01692A  + 0.01912A^{2/3} + 0.000763\frac{Z^{2}}{A^{1/3}} + 0.10178\frac{\left(Z-\frac{A}{2}\right)^2}{A} - 0.012\frac{\delta}{A^{1/2}} 
\end{equation*}

## SEMF (MeV)
\begin{equation*}
M(A, Z) = Zm_{p} + (A-Z)m_{n} - B(A, Z)/c^{2}
\end{equation*}

\begin{equation*}
B(A, Z) = a_{V}A - a_{S}A^{2/3} - a_{C}\frac{Z^{2}}{A^{1/3}} - a_{A}\frac{\left(A-2Z\right)^{2}}{A} +  a_{p}\frac{\delta}{A^{1/2}}
\end{equation*}

where
\begin{equation*}
\delta = 
\begin{cases}
  +1 & \text{if Z and N are even} \\
  0 & \text{if Z or N are odd} \\
  -1 & \text{if Z and N are odd}
\end{cases}
\end{equation*}

\begin{equation*}
a_{V} = 15.56\;\mathrm{MeV} \\
a_{S} = 17.23\;\mathrm{MeV} \\
a_{C} = 0.70\;\mathrm{MeV} \\
a_{A} = 23.29\;\mathrm{MeV} \\
a_{p} = 12.00\;\mathrm{MeV} 
\end{equation*}

## Volume Term 
The *Volume Term* is also known as the *Bulk Term* and arises from the observation that the BE/nucleon is approximately constant. This suggests that each nucleon is only bound to near neighbours; adding a new nucleon to the outside of a nucleus doesn’t affect the energy of those in the centre. The internucleon force is described as *saturated*. 

The fact that that the BE/nucleon doesn’t change indicates that the nuclear force that binds the nucleons together is short range and does not distinguish between protons and neutrons. Both the total binding energy and the volume are proportional to the number of nucleons (A). 

## Surface Term 
Nucleons at the surface of a nucleus are not bound on all sides and therefore the binding energy for a surface nucleon is lower than for the other nucleons. There is a reduction in binding energy which is proportional to the number of nucleons on the surface of the nucleus which is, in turn, proportional to the square of the radius of the nucleus. We noted that $R \propto A^{1/3}$, therefore $R^{2} \propto A^{2/3}$. 

## Coulomb Term 
The electrostatic repulsion between protons reduces the binding energy. The Coulomb energy of a sphere with charge $Ze$ uniformly distributed throughout and radius $R$ is

\begin{equation*}
\frac{3(Ze)^{2}}{5(4\pi \epsilon_{0})R}
\end{equation*}

(You can try and prove this for yourself – it is good practice in integration and relevant to the Fields lectures– think of the work done adding a shell of charge of thickness $dr$)

The Coulomb term is proportional to $Z^{2}$. Increasing $A$ without increasing $Z$, i.e. adding more neutrons to a nucleus reduces the Coulomb term because the radius of the nucleus increases and therefore the average distance between the protons increases. The Coulomb term becomes more important for higher mass nuclei and these tend to have $N$ > $Z$ to reduce this term. 



## Quantum effects
The terms considered so far would imply that the largest binding energy was for a nucleus with $Z$ = 0 i.e the most stable configuration would be a collection of neutrons because the attractive strong force acts on all nucleons but the repulsive electrostatic force only acts on protons. 

If a particle is confined it can only have discrete values for its energy (c.f. potential well). Protons and neutrons are confined in the nucleus and therefore their energies can only take certain fixed values. Protons and neutrons are governed by the Pauli exclusion principle and therefore each proton energy level can only contain two protons and each neutron energy level can only contain two neutrons. In the ground state the lowest energy levels are occupied. 


## Asymmetry Term
The most important contribution to the binding energy of a nucleus is the strong force which affects protons and neutrons equally. This means that the energy levels for protons and neutrons are very similar. If the energy levels for protons and neutrons were identical the lowest energy state would always have the same number of neutrons and protons. The asymmetry term represents the tendency to have $N$ = $Z$ = $A/2$ and is zero if this is the case. 

## Pairing Term
The energy levels for protons and neutrons are not identical because the protons experience the Coulomb force which reduces their binding energy. If the number of protons and neutrons is the same then the highest proton energy level will be higher than the highest neutron energy level. As the levels are not quite the same for protons and neutrons and each level has two particles the binding energy is usually largest when both proton and neutron levels are filled, i.e. there are an even number of protons and neutrons. 

Stable nuclei are more likely to occur when both $Z$ and $N$ are even (see exercise 4.6 in the problems book) i.e. protons and neutrons are most strongly bound when they are in spin up/spin down pairs. Nuclei where both the number of protons and the number of neutrons are odd tend to be unstable to beta decay. 

If a nucleus has odd $A$ then one of $N$ and $Z$ will be odd and one even. Odd-even nuclei have a pairing term of zero. For even $a$ nuclei, $N$ and $Z$ are either both odd or both even. The odd-odd version has the smallest binding energy and the largest mass and the sign in front of the pairing term should be chosen to make this the case. 

*Note: The asymmetry term reflects the fact that the energy levels of neutrons and protons are similar. The pairing term reflects the fact that because the neutron and proton levels are slightly different it is normally energetically favourable to “finish”filling the neutron or proton level*

>Common Student Error. The term odd-odd means odd N and odd Z NOT odd A and odd Z. 


## Magic Numbers
The SEMF gives a good description of the general trends for the mass/BE in nuclei. However some nuclei show significant deviation from the average behaviour. Nuclei where $Z$ and $N$ are equal to magic numbers are found to be particularly stable. The magic numbers are 2, 8, 20, 28, (40), 50, 82 and 126. For example tin has a magic atomic number of 50 and has the largest number of stable isotopes of any element.


<figure>
    <img src="Figures/L2-Data-SEMF.PNG" alt="Data for SEMF as functioin of number of nucleons">
    <figcaption>Discrepancies in the SEMF<figcaption>
<figure>  


## Summary
You should be able to identify and describe the origin of each of the terms in the SEMF

You should be able to use the formula to

- Determine the binding energies of nuclei.
- Determine which isotopes are likely to be stable

## Example
Consider the nuclides with $A$ = 27. 

a. Use the semi-empirical mass formula to find the value of $Z$ for which the mass is a minimum.
b. What does the SEMF predict for the binding energy of this nuclide?
\begin{equation*}
m_{p} = 938.272\;\mathrm{MeV/c^{2}} \\
m_{n} = 939.572\;\mathrm{MeV/c^{2}} \\
m_{He} = 3729.402\;\mathrm{MeV/c^{2}} \\
m_{e} = 0.511\;\mathrm{MeV/c^{2}}.
\end{equation*}


## Problem book questions
- 4.5 and 4.6 (these are the spreadsheet exercises to produce the plots for sections 2.2 and 2.3) 
- 4.9, 4.10,4.11, 4.12. 


# Radioactive decay

## Preparation exercises
- How many different types of radioactive decay can you think of?
- What happens in each?

## Introduction 
We’ll consider three main types of radioactive decay:  

- Beta-decay: emission of an electron or positron
- Alpha decay: emission of a doubly charged Helium nucleus
- Gamma decay:emission of a photon from the excited state of a nucleus


## Decay Rates (40-2)
Radioactive decays are random processes governed by a fixed probability per unit time $\lambda$ which is inversely proportional to the mean life $\tau$ i.e. $\lambda = \frac{1}{\tau}$. 

The half-life $\tau_{1/2}$ is the time taken for half of a given sample to decay. The activity is the number of decays per second: 
\begin{equation*}
\mathrm{Activity} = \lambda N. 
\end{equation*}

We can derive an expression for the number of particles remaining after time $t$ as follows:
\begin{equation*}
\mathrm{d}N = - \lambda N \mathrm{d}t
\end{equation*}

\begin{equation*}
\int_{N_{0}}^{N}\frac{\mathrm{d}N}{N} = -\int_{t'=0}^{t'=t}\lambda \mathrm{d}t'
\end{equation*}

\begin{equation*}
ln\left(\frac{N}{N_{0}} \right) = -\lambda t
\end{equation*}

\begin{equation*}
N = N_{0}\exp(-\lambda t) =  N_{0}\exp(-t/\tau)
\end{equation*}

Here $N_{0}$ is the original number of undecayed particles. After one half-life has elapsed the number of particles remaining is $N = \frac{N_{0}}{2}$. Hence we can use the above expression to obtain expressions relating half-life ($\tau_{1/2}$), mean lifetime ($\tau$) and $\lambda$ as follows: 

\begin{equation*}
\tau_{1/2} = \frac{ln(2)}{\lambda} =  \tau ln(2) 
\end{equation*}

## Beta Decay
There are three forms of beta decay which we’ll consider, electron emission $\beta^{-}$, positron emission $\beta^{+}$ and k-capture ("electron capture"). 

> Common Student Error - Confusion often arises with calculating the energies involved in beta decay. This is 
because we use atomic masses and we are lazy when writing the equations. Ensure you understand the 
discussion below.

Consider
\begin{equation*}
^{A}_{Z}X \rightarrow ^{A}_{Z+1}Y + e^{-} + \bar{\nu}_{e}
\end{equation*}

Naively we might think that, assuming the neutrino mass is zero, the kinetic energy of the decay products is
given by

\begin{equation*}
KE =\left[m\left(^{A}_{{Z}}X\right) - m\left(^{A}_{{Z+1}}Y\right) - m_{e}\right]c^{2}
\end{equation*}

However we need to take into account that the atom of $Y$ will be left positively charged. The mass we need in the energy calculation is not the atomic mass of $Y$, $m\left(^{A}_{Z+1}Y\right)$ but the mass of $Y^{+}$ which is $m\left(^{A}_{Z+1}Y\right)\;-\;m_{e}$. Note that there is a very small additional correction due to the change in atomic binding energy but we will ignore this. The equation for the kinetic energy of the emitted particles is therefore 

\begin{equation*}
KE = \left[m\left(^{A}_{Z}X\right) - \left(m\left(^{A}_{Z+1}Y\right) - m_{e}\right) -m_{e}\right]c^{2} \\
\end{equation*}

\begin{equation*}
KE = \left[m\left(^{A}_{Z}X\right) -m\left(^{A}_{Z+1}Y\right)\right]c^{2}
\end{equation*}

<figure>
    <img src="Figures/L2-Beta-decay.PNG" alt="Summary table for beta decay and electron capture.">
    <figcaption>Summary table<figcaption>
<figure>  

## Beta decay Parabolas
In beta decay $A$ does not change. Rearranging the SEMF to group terms in powers of $Z$ gives

\begin{equation*}
M(A, Z) = Zm_{p} + (A-Z)m_{n} - \left(a_{V}A - a_{S}A^{2/3} - a_{C}\frac{Z^{2}}{A^{1/3}} 
-a_{A}\frac{(A-2Z)^{2}}{A} + a_{p}\frac{\delta}{A^{1/2}}\right)/c^{2}
\end{equation*}

\begin{equation*}
M(A,Z) = \left(Am_{n} - a_{V}\frac{A}{c^{2}} + a_{S}\frac{A^{2/3}}{c^{2}} + a_{A}\frac{A}{c^{2}} + a_{p}\frac{\delta}{c^{2}}\right)  \\
- Z\left( (m_{p} - m_{n}) + 4\frac{a_{A}}{c^{2}}\right) \\
+ Z^{2}\left(\frac{a_{C}}{A^{1/3}c^{2}} + \frac{4a_{A}}{Ac^{2}}\right)
\end{equation*}

For constant $A$ the coefficients of all three terms are constant except for the pairing term. As the expression is quadratic in $Z$, plotting $M(A, Z)$ versus $Z$ for fixed $A$, will yield points on parabolas.

For odd values of $A$, $M(Z)$ v $Z$ is one parabola as all the nuclei will be odd-even so the pairing term is zero in all cases. Even values of $A$ give two parabolas, offset from each other in $M(Z)$, by a value which corresponds to the difference between the odd-odd and even-even values of the pairing term, $2 × a_{p}/(c^{2}A^{1/2})$.

<figure>
    <img src="Figures/L2-Beta-decay-nodes.PNG" alt="Beta Decay Parabolas">
    <figcaption>Beta Decay Parabolas<figcaption>
<figure>  
  
## Alpha Decay
\begin{equation*}
^{A}_{Z}X \rightarrow ^{A-4}_{Z-2}Y^{--} + ^{4}_{2}He^{++}
\end{equation*}


For alpha decay to occur $M(A, Z)$ > $M(A-4, Z-2) + M(4,2)$. In this case if the atomic masses are used then the two additional electrons on $Y$ are included in the helium mass. If the mass of the alpha particle is used instead then the two electrons need to be added on the right hand side.

Following alpha decay the remaining nucleus may be unstable and decay again. Each alpha decay removes the same number of protons and neutrons but the heavy elements that alpha decay tend to have more neutrons than lighter elements. Chains of alpha decays therefore produce daughter nuclei which are rather rich in neutrons and so eventually the produced daughter nuclei will be unstable to beta decay.

The mass number changes by four in alpha decay and by zero in beta decay and so there are four possible radioactive series with mass numbers, $4n$, $4n+1$, $4n+2$ and $4n+3$ (where $n$ is an integer). It is not possible to transform between the four series by alpha and beta decay

## Radioactive Series (40-2)
<figure>
    <img src="Figures/L2-Radioactive-series.PNG" alt="Radioactive series, Z vs N plot">
    <figcaption>Radioactive Series<figcaption>
<figure>  

## Gamma Decay (40-2)
When a nucleus decays by alpha or beta decay the resulting nucleus can be left in an excited state. Nuclei are bound systems governed by Quantum Mechanics and there are defined energy levels. Transitions between energy levels produce photons with defined energies.
    
## Summary 
- You should be able to perform calculations using the expressions for rates of decay of radioactive nuclei. 
- You should understand the changes that happen in alpha, beta and gamma decay.
- You should understand which decays are possible for different nuclei based on their masses. 

## Example
The atomic masses of three nuclei with mass number 50 are given in the table below. 
<figure>
    <img src="Figures/L2-E-Table.PNG" alt="Summary table for Ti, V and Cr">
    <figcaption>Table of values for Ti, V and Cr<figcaption>
<figure>  

a. State the possible decay mode(s) of the Vanadium isotope and calculate the energy released for each mode. 
b. The Titanium isotope is stable. The Vanadium and Chromium isotopes have lifetimes ~1017 years. What is the approximate rate of radioactive decays in a 500g block of Vanadium 50? 
c. What is the mechanism by which Chromium 50 can decay?

## Problem book questions 
4.16-4.25 – this is the selection of questions labelled “Radioactive Decay”. 



# Fundamental Particles

## Preparation exercises
What are the properties of the following?
- Quarks 
- Leptons
- Hadrons
- Baryons
- Mesons
Which are fundamental? What do we mean by fundamental?

## Spin (10-4)
The fundamental matter and force particles we discuss in this lecture and the next have intrinsic properties which cannot be changed. Two such properties are the mass and the electric charge. In addition fundamental particles have an intrinsic angular momentum. For the fundamental matter particles the measured value of the component of the angular momentum along an axis is quantised and can take one of two possible values $+\frac{1}{2}\hbar$ or $-\frac{1}{2}\hbar$. We refer to these particles as spin \frac{1}{2} particles.

## The Matter Particles (41-1)
Matter particles are the cosmic bricks. They are fermions (particles with spin $\frac{1}{2}$) and therefore they obey the Pauli exclusion principle. The exclusion principle means that matter has volume and fills space. For example electrons in atoms occupy orbitals which have physical dimensions. When atoms are in close contact the exclusion principle keeps the atoms apart. 

## The Standard Model (41-7)
There are twelve matter particles which we currently regard as fundamental. These particles interact via four fundamental forces. The mathematical description of these interactions is known as **the standard model of particle physics** and this model successfully describes and predicts the interactions of particles over a wide range of energies.

## Quarks and Leptons (41-3, 41-4)
The 6 types ("flavours") of leptons are: electron, electron neutrino, muon, muon neutrino, tau, tau neutrino. 

The 6 types ("flavours") of quarks are: up, down, charm, strange, top and bottom. 
In lecture 5 we will discuss which types of interactions can change quark flavour from one to another.

## Antiparticles (41-2)
For each of the fundamental matter particles there is an antiparticle which has the opposite quantum numbers, e.g. (and perhaps most obviously) electric charge. The antiparticle has the same mass as the particle. Particle and antiparticle can annihilate to produce energy in the form of gamma rays or pairs of other particles.


## Confinement (41-4)
Quarks and gluons (the carrier of the strong force – see section 5) have a property called colour. This is the charge associated with the strong force. Quarks come in three colours, red green and blue and antiquarks in three anticolours. It is an experimental fact that no coloured particles (quarks or gluons) have been observed directly. It is therefore assumed that coloured particles are confined inside colourless combinations. Particles made up of quarks are known as hadrons. Until recently only a few combinations have been observed. These are

- Baryons: three quarks bound together
- Antibaryons: three antiquarks bound together
- Mesons: a quark and an antiquark

There is growing evidence for more exotic states
- Tetraquarks: two quarks and two antiquarks
- Pentaquarks: four quarks and an antiquark
- Glueballs: states made of gluons only

## Hadron Structure (41-4)
The existence of quarks was first proposed to explain the observed patterns of hadron masses. A large number of hadrons had been observed in experiments and these were grouped by various properties including their spin and strangeness. In the 1960s it was observed that hadrons occurred in sets with the same spin, strangeness and almost the same mass but with different charges. These sets were called multiplets. The particles within a multiplet were given the same symbol. For example there are three $\Sigma$ baryons ($\Sigma^{-}$, $\Sigma^{0}$, $\Sigma^{+}$) which have masses that are the same to a few parts in a thousand and which all have the same strangeness.


An initial idea to explain multiplets was that the particles in each multiplet had the same value of a property called isospin, with the different particles corresponding to different measured values of the projection of isospin. You can compare isospin with the concept of electron spin. The electron has spin $\frac{1}{2}$ and the measured values of the spin can be $+\frac{1}{2}\hbar$ or $-\frac{1}{2}\hbar$. The proton and the neutron were considered to be two aspects of an isospin ½ object called the nucleon. With isospin $\frac{1}{2}$ there are two “measured’’ values. The “measured” value of the proton isospin ($I_{3}$) is $+\frac{1}{2}$ and for the neutron it is $-\frac{1}{2}$. 


We can categorize the multiplets by considering strangeness of the particle and isospin as two separate dimensions. For the moment we’ll look at the spin $\frac{1}{2}$  baryons and the spin $\frac{3}{2}$ baryons. We can plot these on two dimensional plots with strangeness on the y–axis and the projection of isospin ($I_{3}$) on the x-axis. Plots like the ones below predicted the existence of undiscovered particles. In particular the properties of the $\Omega^{-}$ were predicted before it was found. 

<figure>
    <img src="Figures/L4-Baryons.PNG" alt="Summary table for Ti, V and Cr">
    <figcaption>Schematic of different types of Baryons.<figcaption>
<figure>  

>Note: Tables 41-1 and 41-3 in Tipler that list particle properties have several mistakes. 

It is important to understand that isospin and strangeness were used before quarks were known to exist. Now we know of the quark substructure of hadron $I_{3}$ is half the number of up quarks minus the number of down quarks. The similarity in the mass of particles in the same isospin multiplet is due to the similarity in the mass of the up and down quarks (or, if we consider the bare quark masses, that the up and down quarks are very light and their contribution to the mass of hadrons is very small). The strangeness of a particle is simply due to the number of strange quarks. The strange quark is defined to have strangeness -1 thus a particle with two strange quarks has strangeness -2 and a particle with an antistrange quark has strangeness +1

As with nuclei we can consider the mass of a hadron, $m(H)$ to be a combination of the component masses and the mass defect due to the released binding energy. 
\begin{equation*}
m(H)c^{2} = m_{q1}c^{2} + m_{q2}c^{2} + m_{q3}c^{2} - B(H) - b(H)   
\end{equation*}

Where $m_{qn}$ is the mass of the quark n, B(H) is the binding energy due to the nuclear force and b(H) is the electromagnetic contribution to the binding energy. With this sort of model and measured hadron masses it is possible to make estimates of the quark masses and the binding energies in hadrons.
    
## Summary 
You should know the properties of quarks, leptons and hadrons.
Be able to use simple arguments to make estimates of quark masses.
    
## Example
Using the information in the baryon spin 3/2 decuplet plot (above) and the masses given below (in $\mathrm{MeV/c^{2}}$)
a. State the quark content of the three $\Sigma^{*}$ baryons. 
b. Estimate the masses of the strange and light quarks.
c. Hence predict the mass of the $\Omega^{-}$ baryon.  
   
\begin{equation*}
M(\Delta) = 1232.0 \\
M(\Sigma^{+}) = 1382.8 \\
M(\Sigma^{0}) = 1383.7 \\
M(\Sigma^{-}) = 1387.2 \\
M(\Xi^{0}) = 1531.8 \\
M(\Xi^{-}) = 1535.0
\end{equation*}

## Problem book questions 
4.26-4.30 – the section labelled hadron structure


# The fundamental forces (41-5)

## Preparation exercises
The uncertainty principle can be expressed as $\Delta E \Delta t < \hbar$. This means that a particle with mass m can be created 
from the vacuum for a time $\Delta t$ provided that $mc^{2}\Delta t < \hbar$. 

If the created particle travels at the speed of light what is the mass of a particle that can travel a typical inter-nucleon distance of $10^{-15}$ m?

## Fundamental Forces (41-5)
Classically, particles with charge are surrounded by fields and interact via these fields. Classical fields are stored energy. We know that energy is quantized so quantum fields are made up of quanta of energy. The quantum explanation of fields is that forces arise from the exchange of field quanta. These quanta can be produced from the vacuum surrounding a particle with charge, provided they exist for a limited time. We can consider these quanta to be particles but they are bosons rather than the fermions that are the matter particles. Bosons do not obey the Pauli exclusion principle.

## Range of forces (41-4)
The gravitational and electromagnetic force are long range because the carriers are massless this means that a carrier with very low energy can be produced. The carriers of the weak force have masses $>\;80\;\mathrm{GeV/c^{2}}$ which means that the weak force has a very short range. 

## Confinement (41-4)
The carriers of the strong force are massless but the force is short range due to gluon-gluon interactions which occur because the gluons carry colour charge. This is the reason that coloured objects, quarks and gluons, are confined within hadrons. There are no equivalent interactions in the electromagnetic force as photons are not electrically charged and cannot interact with each other. 

## The Higgs Boson
The final particle in the standard model of particle physics is the Higgs boson which couples to the fundamental particles with a strength which depends on their mass. Without the Higgs field all fundamental particles would be massless. The discovery of a Higgs boson with a mass of $125\; \mathrm{GeV/c^{2}}$ was announced by the CMS and ATLAS collaborations on July 4th 2012.

## A non-fundamental force (41-5)
5.2 A non-fundamental force (41-5)
The forces in the table above are thought to be fundamental. Fundamental matter particles are those where the particle is (as far as we know) indivisible. Fundamental force particles have carriers which are indivisible. It is possible to have force carriers which are more complex objects and an example of such a force is the strong force which holds protons and neutrons together inside hadrons and is carried by gluons. 

A rough calculation based on the range of the inter-nucleon force (see the pre-lecture exercise) implies a carrier of mass $\sim 200\;\mathrm{MeV/c^{2}}$. The carriers of the strong force between protons and neutrons are the pions which have mass $\sim 135\;\mathrm{MeV/c^{2}}$. Remember that the pions are combinations of a quark and anti-quark and this means that the carrier of the inter-nucleon force is not fundamental.

>Note: For some time the pion and muon were confused because they have very similar masses although their interactions are different. Cecil Powell won the Nobel Prize for his work, in Bristol, on studying cosmic ray interactions with photographic emulsions. One of the important piece of work that Powell did was to conclusively demonstrate that there were two particles, the pion and the muon, with similar masses. He did this by observing the decay of pions into muons and then into electrons.

## Interactions (41-3)
An interaction is where one or more particles change through the exchange of a force particle. The initial state, the particle(s) that are there to start with, transforms into the final state, the particle(s) present at the end. There are conservation laws which govern which interactions are allowed and what is possible will depend on the force which mediates the interaction.

In all interactions energy, momentum, angular moment and electric charge must be conserved in changing between the initial and final states. In addition since quarks can only be produced in a pair with an antiquark the total number of quarks does not change. This is sometimes called the law of baryon conservation since it is only possible to make the three quarks needed for a baryon if three antiquarks are also produced as an antibaryon i.e. the total number of baryons is unchanged. Mesons can be produced provided there is sufficient energy. Leptons are also only produced in a pair with an antilepton but here the rule is more specific in that you can only make an lepton of electron type (i.e. electron or electron neutrino) with an antiparticle of electron type (i.e. positron or electron antineutrino). This law is conservation of lepton number. 

**The strong interaction** only affects particles which carry colour charge; quarks and gluons. Quarks interacting via the strong interaction can scatter, annihilate or be produced in pairs. In all these interactions the total number of each flavour of quark remains constant. A quark can “cancel” with an antiquark of the same flavour so an up quark can annihilate with an antiup but not with an antidown. 

**The EM interaction** only acts between charged particles but charged particles can also produce photons in the final state or absorb photons that were present in the initial state. In any EM interaction the total number of each type of particle remains constant but a particle can “cancel” with its antiparticle.

**The weak interaction** can

- change the flavour of quarks 
- change a charged lepton into its neutrino
Note: the charge of the particle must also change when it changes type. 


## Gluon/photon/$Z^{0}$
All pair production or annihilations that can proceed via a photon can also proceed via a $Z^{0}$. All quark antiquark pair production or annihilation that can proceed via a gluon can also proceed via a photon or $Z^{0}$. When different interactions compete (i.e when an interaction can happen in more than one way) the strong is usually fastest, then EM, then weak.

Typical times, for decays, caused by the different interactions are
\begin{equation*}
\tau_{\mathrm{strong}} \sim 10^{-24} s,  \\
\tau_{\mathrm{EM}} \sim 10^{-20} s,  \\
\tau_{\mathrm{weak}} \sim 10^{-15} s. 
\end{equation*}


## Conservation Laws
1. Electric Charge is always conserved
2. Energy is always conserved
3. Momentum is always conserved
4. Angular momentum is always conserved
5. Lepton number for each lepton flavour is always conserved
6. Quark number or baryon number is always conserved
7. Quark flavour is conserved except in weak interactions. 

**Note**: : The total number of particles of each type is the number of particles of that type minus the number of antiparticles.

<figure>
    <img src="Figures/L5-cons-laws.PNG" alt="Conservation-laws">
    <figcaption>Applying conservation laws.<figcaption>
<figure>  

## Summary
You should be able recall the properties of the four fundamental forces and their carriers. You should be able to state whether given transitions are allowed or forbidden and if allowed by which force they would proceed. 
    
## Example
The decay of the $\Xi^{-}$ initiates the sequence of decays below. Classify the decays as strong, EM or weak. 

\begin{align*}
\begin{split}
a.\;& \Xi^{-} \rightarrow K^{-} + \Sigma^{0} \\
b.\; & \Sigma^{0} \rightarrow \lambda^{0}  + \gamma \\
c.\; & \Lambda^{0} \rightarrow p + e^{-} + \bar{\nu}_{e} \\
d.\; & K^{-} \rightarrow \pi^{-} + \pi^{0}  \\
e.\; & \pi^{-} \rightarrow \mu^{-} + \bar{\nu}_{\mu} \\
f.\; & \mu^{-} \rightarrow e^{-} + \nu_{\mu} + \bar{\nu}_{e}
\end{split}
\end{align*}

Quark compositions:
\begin{equation*}
\Xi^{-} = ssd \\
\Sigma^{0}, \Lambda^{0} = sud  \\
K^{-} = \bar{u}s
\end{equation*}

## Problem book questions
4.31-4.34.

#  Beta decay revisited (40-2)

## Beta decay at the quark level
At the nuclear level we look at beta-decay as the transformation of a proton to a neutron or vice-versa. Since the proton = uud and the neutron = udd at the quark level beta decay involves the transition between an up and a down quark. 

## Nucleon Stability
The combination of a proton plus an electron is lighter than the neutron so free neutrons can beta decay to protons with a half-life of about 15 minutes. The proton is the lightest baryon and, as yet, we don’t know of any process by which free protons can decay. Proton decay would require violation of baryon number conservation. Experiments looking for proton decay have put limits of $\sim 10^{33}$ years on its lifetime – the exact value depends on the particular decay channel

## beta-energy spectrum 
The beta-decay spectrum provided evidence for the existence of the neutrino. The neutrino was first postulated by Pauli in 1933 but the first reactions were only observed directly in 1959.

## Kurie Plots
To measure the mass of the neutrino detailed studies have been made of the endpoint of the beta spectrum. Traditionally this was done using a Kurie plot which should be a straight line for zero neutrino mass. This is not a spelling mistake as the Kurie is Franz Kurie not the Nobel Prize Winner whose name is a homonym of Kurie. Current experiments do direct fits to the energy spectra.

### Electron and Positron energy spectra from $\beta$-decay of $^{64}Cu$
<figure>
    <img src="Figures/L6-Beta-decay.PNG" alt="Beta decay">
    <figcaption>from Cook and Langer Phys Rev 73, 601 (1948)<figcaption>
<figure>  

## Neutrino Mass
Neutrino masses are very small and for the purposes of calculations in this course may be assumed to be zero. The current limit on the electron neutrino mass is $m_{\nu_{e}} < 2 \mathrm{eV/c^{2}}$. This limit been determined by looking at the decay of tritium.
    
Experiments have recently confirmed that neutrinos oscillate i.e. that as they propagate through the universe they change from one type to another. In order for neutrino oscillation to occur they must have mass (this is not on this syllabus).
   
<figure>
    <img src="Figures/L6-Kurie-plot.PNG" alt="Kurie plot">
    <figcaption>A Kurie plot from tritium decay from Hamilton, Alford and Gross. Phys Rev 9 1521-1525. <figcaption>
<figure>  

## Theory of beta decay
The initial attempt to explain beta decay considered it as a contact interaction. i.e. the interaction happens at a point. If the interaction happens at a point the only thing that can affect the rate is the energy available. This assumption of a point interaction works for slow decays but predicts too high rates for modest energies. 
    
<figure>
    <img src="Figures/L6-schematic.PNG" alt="Schematic">
    <figcaption>Schematic of $\beta^{-}$ decay <figcaption>
<figure>  
    
We now know that the interaction actually proceeds via W exchange and understanding the interaction in these terms gives the observed energy dependence. The probability for weak interactions is usually small because a $W$ or $Z$ boson has to be produced. We can do a similar calculation as for the nuclear force (in lecture 5) and show that the weak force is limited to about $10^{-18}$ m. 
        
<figure>
    <img src="Figures/L6-examples.PNG" alt="Examples">
    <figcaption><figcaption>
<figure>  
    
    
    
    
## Summary
You should know that 

- Beta decay at the quark level arises when a down quark changes to an up quark or vice versa
- The spectrum of beta decay was evidence for the existence of the neutrino.
- Detailed studies of the spectrum can put limits on the neutrino mass.
- The energy dependence of beta-decay indicates that it arises from particle exchange.   
    
## Example 
This question concerns the isobars $^{3}He$ and $^{3}H$. Throughout this question you should ignore atomic binding energies. 
\begin{equation*}
\mathrm{^{3}He\;mass:}\;3.016029\;u \\
\mathrm{^{3}H\;mass:}\;3.016049\;u \\
\mathrm{Neutron\;mass:}\;1.008665\;u \\
\mathrm{Proton\;mass:}\; 1.007276\;u \\
\mathrm{Electron\;mass:}\;5.49\;×\;10^{-4}\;u
\end{equation*}
    
a Show that the difference in nuclear binding energy between the two isobars is $\approx\;0.7 6\;\mathrm{MeV}$.  

b. What is the cause of the difference in binding energy calculated in part a.?

c. One of these isobars is unstable.
    i. Which isobar is unstable?
    ii. Write down the products of the decay of the unstable isotope.
    iii. Show that the decay in ii. is energetically allowed.