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PHYS1008 Nuclear Physics Introduction

Objectives: by the end of this you should

understand how nuclei are made up of protons and neutrons
understand what isotopes are
use conservation laws to explain what can happen in nuclear processes
understand fusion and fission
understand radioactive decays
understand how the sun works
understand why neutrinos matter

Proton

discovered as nucleus of H by Rutherford and Blackett (1921) in

α + N ⇒ O + p

M_proton = 1.67x10^-27 kg

Neutron

discovered by Chadwick (1932) as penetrating massive neutral particle (i.e., not a γ)
M_neutron = 1.68x10^-27 kg. since neutrons and protons are so similar in mass, convenient to think of them as the same particle (a nucleon) which comes in two flavours!

Atomic Number Z 
= charge on nucleus
 = N_protons

This defines chemistry

Mass number A = N_protons + N_neutrons
Isotopes: nuclei with different A but same Z.

Usual notation is ^A_ZX_N but often just write a nucleus as (e.g) ³⁵S, since the name implies Z. In this notation:

Conservation laws:

Much of nuclear physics is governed by conserved by conservation laws.

Mass-energy conservation

```
E = mc²
```
Usually easier to quote elementary particle masses in terms of energy, measured in eV

e.g. "mass" of electron

= mc² = 9.1x10^-31x(3x10⁸)²/1.6x10^-19 
  = 511 keV = .511 MeV

Conservation of charge
n ⇒ p + γ
is forbidden: algebraic sum of charges at end of reaction = sum at start

Conservation of baryon number:
The process
p ⇒ e⁺ + γ
is not forbidden by anything: however experimentally it does not occur.
No of (protons + neutrons) is always the same in a reaction: Easiest to say that p & n carry baryon number B = 1, rest carry 0

The sizes of things:

a proton is about 1 fm (=10^-15m) in diameter.

A nucleus is a close-packed array of protons and neutrons, so radius
```
R ~ R₀ A^1/3 where R₀~ 1.4 fm
```
In turn, this allows us to estimate the energies involved: if a proton is trapped in a nucleus 5 fm in radius
```
δxdp ~ h 
```
and δx ~10^-14 m
\color{red}{ E = \frac{{p^2 }}{{2m}} \approx \frac{{h^2 }}{{2m\left( {\delta x} \right)^2 }} \sim 10MeV}
Typical energy levels in nuclei tend to be separated by 10 MeV, compared to 10 eV in atoms.

Nuclei build up in much the same way as atoms in the periodic table: force are much more complicated, so cannot really solve for the energy. Light nuclei:
Note that there are no stable elements of A = 5 or A = 8:

if you make ⁸Be it will decay instantly into 2 ⁴He.

Roughly speaking, need equal numbers of p and n: if a nucleus is badly out of balance, it will undergo a decay.
Many nuclei can be created but only a few are stable: this shows nuclei up to oxygen.

Whole pattern shows N ~ Z for light, N > Z for heavy.

Radio-Activity and Decays

Becquerel:

Radio-activity and nuclear decays: have already seen 3 varieties.

Simplest conceptually is γ decay: just as atoms have energy levels, so do nuclei. . However, they are more complicated:

two kinds of particles
protons and neutrons each have their own levels
Forces are more complicated (often use SHO)

One of the protons (or neutrons) can make a transition if there is a gap.Energy is much higher than in atoms: ~ 10 MeV.

β-decays

β-decays: an isolated neutron will decay.
```
n ⇒ p + e^-
```
This led to a huge problem:the electron came out with varying energy.

Fermi (1933) proposed that the missing energy is carried off by an invisible particle, the neutrino, so the real reaction is:
```
n ⇒ p + e^- + ν
```
This can also happen in a nucleus if the energies are favorable: e.g. could have
¹²B ⇒ ¹²C + e^- + ν

Antiparticles:

For every particle with given quantum numbers, there is a corresponding anti-particle with the properties flipped:

e.g. electron has charge e = -1.6x10^-19 C
positron has same mass, charge = 1.6x10^-19 C
e.g. proton \color{red}{p} has charge = 1.6x10^-19 C, m = 1.67x10^-27 kg, B=1
e.g. anti-proton \color{red}{{\bar p}} has charge = -1.6x10^-19 C, m = 1.67x10^-27 kg, B=-1

More Conservation Laws:

Conservation of lepton number:
Number of (electrons + neutrinos) is conserved
L is lepton number
L = 1 for e^-, ν L = -1 for e⁺
Note that since electron has L = 1, we must have an anti-neutrino along with it
\color{red}{n = p + e^ - + \bar \nu }

Picture shows many pairs created in a hydrogen bubble chamber. \color{red}{\gamma + nucleus \Rightarrow e^ + + e^ - + nucleus}
Picture from Fermilab
Minimum energy of γ?
Also positron will annihilate:
```
 e⁺ e^- ⇒ γ + γ
```

Conservation of angular momentum ("spin"):

Mainly important because some particles carry spin ½: e.g. n ⇒ p + e^- is allowed by all previous arguments, but ½ ⇒ ½ + ½ is not: instead
```
n  ⇒ p  +  e^- + ν , ½ ⇒ ½ + ½ + (-½) 
```

which is allowed.

α-decay

Occurs in heavy nuclei. He nucleus is very tightly bound, since it has 2 protons, 2 nucleons which can be in the lowest level since they each have spin.

Heavy nuclei are unstable, since they have repulsive forces between the protons. Hence

Radioactive Decays

All radioactive decays have a similar behaviour. The decay occurs totally at random, hence the probability of observing a decay is proportional to the number of nuclei: P ∝N.
This reduces the number of nuclei available to decay: i.e.
```
δN = -λNδt or dN = -λN
              dt
```
This is easy to solve:
```
N = N₀ e^-λt
```
Half-life: time taken for half the nuclei to decay. τ_1/2 = ln(2)/λ. After τ_1/2, 1/2 of the nuclei will have decayed .

e.g ¹³N has a half-life of 10 mins: if we start with 1000 nuclei, how much would be left after 30 minutes?

250 atoms
333 atoms
125 atoms
62 atoms

This allows us to do radioactive dating: e.g. carbon dating.

In the atmosphere, some of the ¹⁴N ⇒ ¹⁴C by cosmic rays.
This gets incorporated in living things, substituting for ¹²C .
When the object dies, no more ¹⁴C is absorbed, and what is already there decays back to ¹⁴N, with a half-life of 5700 years.

e.g. the Turin Shroud, (supposedly used to wrap Christ in when he was lowered from the Cross)
Proportion of ¹⁴C which is 89.5% of that of current materials.
How old is the shroud?

Forces

To understand nuclei in any more detail, we need to look at nuclear forces

Strong force is the force that binds together nucleons to make nuclei, weak force is force that causes β-decay. Believe there are only 4 forces in nature

Note: "feels" means that this is what the force couples to: e.g. gravity does not care whether a particle is charged, only whether it has mass.
Range: if it is ∞ then F ∝ 1/r², else it cuts off at distance shown
Strength: roughly the relative strength of the forces at a distance of 1 fm.
Although (e.g.) strong force>>>E.M at 1 fm (=10^-15 m), it vanishes totally beyond 10^-14 m.
E.M >>> Gravity, but it tends to cancel out since most matter is electrically neutral, whereas mass accumulates.
The weaker the force, the more particles feel it!

Particles: of the 450 (or so) elementary particles, only 5 are important to nuclei

Strength also gives us (very roughly) the depth that a particle will penetrate matter without interacting:
e.g. a proton will penetrate a few mm, a X-ray photon a few cm, a neutrino several parsecs!
These conservation laws let us make up an extended particle table. The numbers are all conserved:

e.g. why doesn't

n ⇒ p e γ

happen?"

It doesn't conserve charge Q
It doesn't conserve Lepton number L
It doesn't conserve Baryon number B
It doesn't conserve energy

How do we really know that there are neutrinos?

\color{red}{n = p + e^ - + \bar \nu }
but we can't see the ν. We can reverse the process \color{red}{\nu + n = p + e^ - }
so we get electrons created out of "nothing"
Of the other particles, note the muon μ: appears to be identical to the electron except that it is 210 x heavier and seems to come with its own number:
it even has its own neutrino ν_μ.
The μ^- can even make atoms
WHY does the muon exist? We don't know
And there is an even heavier lepton: the τ (tau) which is m_τ~ 3500 m_e and its neutrino ν_τ
WHY does the tau exist?? We don't know
How many more are there?
None!

Fusion

Energy involved: E = m₀c² (Heard of this before?). e.g. consider
```
D + D ⇒ ⁴He
```
e.g. mass of deuterium (a proton + a neutron) = 2.014103 u
mass of helium 4 (⁴He) (2 protons + 2 neutrons) = 4.002603
so can make ⁴He from 2 D, with some mass left over.
in units of u = 1.660531x10^-27 kg
In terms of energy,
```
 E =  δm u c²
```
E =δm 1.660531x10^-27 x (3x10⁸)² so what is energy released?
This doesn't sound like much but 1 kg of D₂ contains how many atoms?
So how much energy do we get by making 1 kg of He⁴

What reactions can happen?

p + ²H ⇒ ³He
is exothermic, gives ~ 1 MeV
⁴He + p ⇒ ⁵Li
doesn't happen (⁵Li is unstable)
⁴He + d ⇒ ⁶Li
is endothermic (i.e. won't normally happen)

Would T + T ⇒ ⁶Li happen?

Yes, since T is unstable
No, since it doesn't conserve charge
No, since it doesn't conserve baryon number

As a rule of thumb, most reactions up to Fe are exothermic, any past that are endothermic

Since stars get their energy via fusion reactions the most important curve is binding energy vs A
(Put differently): fusion reactions can occur for light elements, since medium-heavy elements are more stable, fission reactions can occur for heavy elements.

How do stars work?

Why do stars need to be hot (or why is there no cold fusion?)
There is a large repulsive E.M. force at large distances between two protons: only if r < 1 fm is the force attractive (fm=femtometre =fermi =10^-15m)
Height of this barrier is \color{red}{E = \frac{{kZ_1 eZ_2 e}}{{R^2 }}}
Hence to "ignite" thermonuclear fusion, it would appear to need temp ~ 1MeV,
which is E = ³/₂ kT,
so T ~ 8x10⁹ K.
This is temp reached in H-bomb: however the sun can work more slowly! (Central temp is 13x10⁶ K)

Quantum Mechanics tells us that particles can tunnel through barrier:

For a particle of given energy. Prob of tunnelling

P(E)~ exp(-1/√(E)) which gets larger for large E
However there are very few nuclei with this energy
Ignition Temp T₀~ 10⁷K
Rates depend very strongly on temp and charge of nuclei: e.g. for pp, R ~ T⁴

Star starts off as H + ⁴He: what reactions can occur?

p + p ⇒ ²He
can't occur, ²He doesn't exist
p + p ⇒ d + e⁺ +ν
is the initial process
and is very slow (the ν has only weak interactions, and so the whole process must go by that)

Once past this, the reactions are simple:

```
p + d ⇒ ³He + γ
d + ³He ⇒ ⁴He + p
```
or various variations:
Most important is the "executive summary"
```
4 p ⇒ ⁴He + 2 e⁺  + 2ν + 27 MeV
```
And this is what keeps us warm!

Detectors:

Actually the sun produces 4x10²⁶ watts,
which is 10³⁹ ν's/s ⇒ 9x10¹⁴ ν's m^-2s^-1
(You didn't notice?)
How do we know if it is true?
Look for the ν's!
Require a target which will give a suitable end-product when hit by a ν : e.g.
```
³⁷Cl +ν ⇒ ³⁷Ar + e^-
```
³⁷Ar is radioactive and decays after 35 days (back to Cl + e )
Rates: 1 SNU (solar neutrino unit) = 10^-36 captures/atom/second
Expectation in 1961 20 SNU's, now 5 SNU's

Davis expt (133 tonnes) of Cl (in the form of C₂Cl₄: perchlorethylene).
Find Ar atoms by flushing system with He, picks up Ar, then collect it and watch for decays
How many? 5 SNU implies 1 atom of ³⁷Cl converted/day (a needle in a haystack?). Run for 3 months, search for Ar³⁷
Now results show 7.9 SNU's expected 2.1 observed.
Davis gets Nobel Prize 2003 (Bahcall, who did the calculations, should have done as well!)
Where have all the ν's gone?
- Bad experiment?
- We don't understand the sun?
- Maybe the sun has gone out!
- Nuclear physics wrong?

Need totally new experiment (and M$60)

Sudbury Neutrino Observatory (SNO) How does this work? . Need to go deep to get away form cosmic rays

SNO collaboration

in the Creighton mine (note the other bit of astronomy: the Sudbury basin is an old meteorite crater!)

SNO collaboration

SNO uses 1000 tons of D₂
ν + d ⇒ e^- + p + p
allows ν's to be seen immediately.
Must be pure (radioactivity gives false positives) < 10^-15 parts Radium
SNO collaboration

Light (photons) are picked up by photo-tubes
SNO collaboration

15000 look like this
SNO collaboration

Produces "rings" of light which allow incident particle to be reconstructed

Answers in 2001
- ~~Bad experiment?~~
- ~~We don't understand the sun?~~
- ~~Maybe the sun has gone out!~~
- Nuclear physics wrong?
The ν's produced in the sun change to another kind of ν on the way over.
Why? We don't know, but it means that ν's have mass.
One consequence: the amount of mass in the universe due to ν's is more than all the mass of the "ordinary" matter combined.
To finish up with: some applications of nuclear physics