Nuclear Physics

1
Nuclear Physics
2
Discovery of the Nucleus

• Prior to 1911, the atom was thought to be like a
  plum pudding, with its positive charge spread
  out through the entire volume of the atom (the
  pudding), and the electrons (the plums)
  vibrating at fixed points within this sphere of
  charge.
• Ernest Rutherford showed via simple experiment
  that the atom could not be structured like this
  at all, but rather had all its positive charge
  concentrated in a tiny nucleus, approximately 104
  times smaller than the atom itself.

• Rutherford fired energetic ?-particles (Helium
  nuclei) at a thin piece of gold foil, and
  measured the extent to which they were deflected
  as they passed through.
• He found that most of the ?-particles were
  deflected through small angles, but a small
  fraction of them were scattered through very
  large angles approaching 180! In Rutherfords
  words it was as if a 15-inch shell was fired at
  a piece of tissue paper and it came back and hit
  you!.

Alpha source
Gold foil
?
Detector
3
Discovery of the Nucleus

• Rutherford deduced that the maximum deflecting
  force that could act on an ?-particle as it
  passed through an atom with a plum pudding type
  charge distribution, would be far too small to
  deflect it by even as much as 1 degree. The only
  way such a large (180) deflection could take
  place was to have all the positive charge
  concentrated tightly at the atoms centre. In
  that way the incoming ?-particle could get very
  close to it and in doing so experience the full
  force of the charge.

How close?? Consider a 5.30MeV ?-particle (that
has a charge of 2e) that heads directly towards
the nucleus of an atom of gold (which has a
charge of 79e). The repulsive force the
?-particle feels from the gold nucleus will cause
it to come to rest, or in other words, its
kinetic energy will be converted into
electrostatic potential energy, with 5.30 MeV
(1/4??0)(Q?QAu/d) putting in the appropriate
numbers gives d 4.29 x 10-14m, which is about
10-4 x smaller than size of atom ? it is mostly
empty space!!
4
Nuclear Properties

• Some nuclear terminology
• Nuclei are made up of protons and neutrons.
• The number of protons is called the atomic number
  or proton number and is represented by the symbol
  Z.
• The number of neutrons is called the neutron
  number and is represented by the symbol N.
• The total number of neutrons and protons in a
  nucleus is called its mass number A, where A Z
  N.
• Nuclides with the same atomic number Z but
  different neutron numbers N are called isotopes.
• As an example, the element gold has 30 isotopes
  ranging from 175Au to 204Au. Only one of them,
  197Au, is stable the remaining 29 are
  radioactive, decaying by the spontaneous emission
  of a particle, with average lives ranging from
  seconds to a few months.

5
Nuclear Properties

• Distribution of nuclides in the Z-N plane
• It is interesting to examine where naturally
  occurring nuclides both stable and radioactive
  lie in a plot of atomic number (Z) versus
  neutron number (N).

• The stable nuclides are represented by the green
  data points they are seen to lie in a
  well-defined central band.
• The radionuclides (radioactive nuclides) tend to
  lie on either side of and at the upper end of the
  band defined by the stable nuclides (light-orange
  shading).

6
Nuclear Properties

• Nuclear Radii
• A convenient unit for measuring the sizes of
  atomic nuclei is the femtometer (10-15m) or
  fermi (shortened to fm).
• The size and structure of nuclei is best
  determined by bombarding them with high-energy
  electrons and observing the way the nuclei
  deflect them. The energy of the electrons must be
  200MeV in order for them to be
  structure-sensitive probes.
• Such experiments show that the nucleus has a
  characteristic mean radius R given by
• R RoA?,
• where A is the mass number and Ro 1.2fm.
• Nuclear masses
• These are measured in atomic mass units, u,
  chosen so that the atomic mass of 12C is exactly
  12u. 1u 1.661 x 10-27 kg.

7
Nuclear Properties

• Nuclear Binding Energies
• When the masses of nuclei are measured with great
  precision (via mass spectrometry), it is found
  that they are slightly smaller than the value
  derived by summing the masses of the individual
  protons and neutrons in the nuclei. This is
  called the mass defect (?m).
• The energy associated with this mass defect, E
  ?mc2, is what binds the nucleus together, and is
  called the nuclear binding energy. It can also be
  thought of as the total energy required to tear a
  nucleus apart into its constituent protons and
  neutrons.
• If we divide the binding energy of a nucleus by
  its mass number, we get the binding energy per
  nucleon.
• The figure on the next slide shows a plot of the
  binding energy per nucleon versus mass number (A)
  for the various nuclides.

8
Binding Energy

• We see that the binding energy per nucleon curve
  rises sharply with increasing A, in the range
  2ltAlt12, and then flattens off to reach a peak at
  56ltAlt81. This range in A is the region of
  greatest stability, where nuclei have the highest
  possible binding energy per nucleon.
• The binding energy curve then shows a gradual
  decline in going to higher A values beyond the
  peak region.

9
Binding Energy

• This behavior means that it is energetically
  favorable for nuclides at the low A end to fuse
  into higher mass nuclei ? nuclear fusion.
• It also means that it is energetically favourable
  for nuclides at the high A end to split into
  smaller nuclei ? nuclear fission.

10
Nuclear Force

• Since the nucleus contains positively charged
  protons (as well as neutral neutrons), it must be
  bound together by a very different force to the
  electromagnetic force that binds electrons to the
  nucleus.
• It must be a strongly attractive force that
  overcomes the electromagnetic repulsion between
  the protons, and contains all the protons and
  neutrons in such a small volume. It must also be
  short range, since its influence is known not to
  extend far beyond the nuclear surface.
• This force is called the nuclear strong force
  which binds the constituents of neutrons and
  protons quarks together.

11
Radioactive Decay

• As we have seen from the figure shown in slide 5,
  a significant fraction of nuclides are
  radioactive (as indicated by the light-orange
  shading) that is, a nucleus will spontaneously
  emit a particle, thereby transforming itself into
  a different nuclide. This is a completely
  statistical process in that there is no way to
  predict whether and when a nucleus will decay
  rather all we can do is talk about the chance of
  it happening.
• The statistical nature of the decay can, however,
  be expressed mathematically, in that if a sample
  contains N radioactive nuclei, then the rate at
  which nuclei decay is proportional to N
• -dN/dt ?N
• where ? is the disintegration constant which has
  a characteristic value for every radionuclide.
• Integrating the above decay equation gives
• N N0e-?t
• where N0 is the number of radioactive nuclei in
  the sample at t0 and N is the number remaining
  at any subsequent time t.

12
Radioactive Decay

• It is often the case that the actual decay rate R
  (-dN/dt) is of most interest, and we can derive
  this by differentiating the previous equation
• R -dN/dt ? N0e-?t R0e-?t
• where R0 ? N0 is the initial decay rate at t0.
  The decay rate R is also referred to as the
  activity of the radionuclide.
• R is measured in disintegrations per second, or
  counts per second (as measured by a Geiger
  counter), or in curies (Ci) where 1 Ci 3.70 x
  1010 disintegrations per second.

The half-life ? of a radioactive nucleus is
defined as the time after which both N and R are
reduced to one-half their initial values.

• Putting R ½ R0 in the above equation and
  substituting ? for t, we have
• ½ R0 R0e-??
• and solving for ? yields
• ? (ln 2)/?

13
Radioactive Dating

• If the half-life of a given radionuclide is
  known, then in principle it can be used as a
  clock to measure a time interval ? radioactive
  dating.
• This is best understood by working through the
  following example

Q. Analysis of potassium and argon atoms in a
moon rock shows that the ratio of the number of
stable 40Ar atoms present to the number of
radioactive 40K atoms is 10.3. Assume that all
the argon atoms were produced by the decay of
potassium atoms, with a half-life of 1.25x109
yrs. How old is the rock? A. If N0 potassium
atoms were present at the time the rock was
formed, the number of potassium atoms remaining
at the time of analysis is NK N0e-?t where t
is the age of the rock. For every potassium atom
that decays, an argon atom is produced. Hence the
number of argon atoms present is NAr N0 - NK
14
Radioactive Dating
We cannot measure N0, so we need to eliminate it
from the equations ?t ln (1 NAr/NK). Solving
for t and replacing ? with (ln 2)/?, we get t
? ln (1 NAr/NK)/ln 2 1.25 x
109ln (110.3)/ln 2 4.37 x 109 yrs
15
Alpha Decay

• Nuclear decay occurs, sooner or later, whenever a
  nucleus containing a certain number of nucleons
  resides in an energy state which is not the
  lowest possible one for a system with that number
  of nucleons.
• Nuclear decay divides itself into 3 categories
  ?-decay, ?-decay and ?-decay. The first of these,
  ?-decay, involves the spontaneous emission of an
  ? particle (that is, a Helium nucleus 2He) from a
  nucleus of large atomic number (Zgt82).
• This decay takes place spontaneously because it
  is energetically favoured, the mass of the parent
  nucleus being greater than the mass of the
  daughter nucleus plus the mass of the ? particle.
  The reduction in nuclear mass in the decay is
  primarily due to a reduction in the Coulomb
  energy of the nucleus when its charge Ze is
  reduced by the charge 2e carried away by the ?
  particle.
• The energy made available in the decay is the
  energy equivalent of the mass difference, and it
  is carried away by the ? particle. As confirmed
  by experiment, all the ? particles emitted in the
  decay of a particular species have the same
  unique kinetic energy (mass difference)c2.

4
16
Alpha Decay

• This ?-decay energy, E, can be written in terms
  of the atomic masses of the parent, daughter, and
  ? particle nuclei
• E MZ,A (MZ-2,A-4 M2,4)c2
• Note that the ?-decay energies range from 8.9MeV
  to 4.1MeV.

Illustrative example an ? particle is emitted by
the parent nucleus 84Po. Estimate the Coulomb
potential it feels at the nuclear surface and
plot the sum of the Coulomb and nuclear
potentials acting on the ? particle in various
locations. Solution if we approximate the
daughter nucleus and the ? particle as uniformly
charged spheres, the Coulomb repulsion potential
energy when they are just touching will be Vo
2Ze2/4??or Where 2e is the ? particle charge,
Ze is the daughter nucleus charge, and r is the
sum of the radii of the ? particle and daughter
nucleus (which in this case 8.0F).
212
17
? decay example continued So we have Vo 2 x
82 x (1.6 x 10-19)2/1.1 x 10-10 x 8.0 x 10-15
4.8 x 10-12 J 30 MeV

• The plot shows the total (Coulomb nuclear)
  potential acting on the ? particle. As it
  approaches the nucleus, it feels the repulsive
  Coulomb potential increasing in inverse
  proportion to the separation between the centres
  of the ? particle and nucleus, and reaches a
  value of Vo when this distance equals r.
• Inside the surface it feels a rapid onset of the
  strong attractive nuclear potential.
• Also indicated is the ?-decay energy, E8.9MeV,
  which is the energy of the emitted ? particle.

• The E value is considerably smaller than the
  height of the Coulomb barrier, which is 30MeV
  classically, the ? particle would appear trapped
  inside the barrier, but is able to escape through
  the process of quantum tunneling.

18
Alpha Decay and the Z-N diagram
?-decay involves a loss of 2 protons and 2
neutrons, and hence causes nuclei to move
diagonally downwards in the Z-N diagram,
allowing radioactive nuclei to migrate down to
the stable zone (green band)
19
Beta Decay

• There are three different ?-decay processes
• Electron emission where a negatively charged
  electron is emitted by the nucleus, so Z
  increases by one, N decreases by one, and A
  remains fixed (effectively n ? e- p within
  nucleus)
• Electron capture where a nucleus captures a
  negatively charged electron, and Z decreases by
  one, N increases by one, and A remains fixed
  (effectively p e- ? n)
• Positron emission where a nucleus emits a
  positively charged electron, and Z decreases by
  one, N increases by one, and A remains fixed
  (effectively p ? e n)
• Conditions for each of these forms of ?-decay to
  occur
• E mZ,A (mZ1,A me)c2 gt 0
• E (mZ,A me) mZ-1,Ac2 gt 0
• E mZ,A (mZ-1,A me)c2 gt 0

me electron mass
Mass defect
20
Beta Decay

• Experimental studies of ?-decay show that the
  electrons (either positive or negative) that are
  emitted, have a broad spectrum of kinetic
  energies, Ke, as shown in the figure. Initially,
  this was a very mysterious and disturbing result,
  since you would expect the electron to carry away
  all the decay energy associated with this process
  (i.e. given by the E value equations on the
  previous page). Where did the missing energy go?

• Problem solved by Pauli in 1931, who postulated
  that a particle, now called the anti-neutrino, is
  also emitted in the ?-decay process, but is not
  normally detected because its interaction with
  matter is extremely weak.

21
Beta Decay and the Z-N diagram

• ?-decay causes nuclei to move diagonally upwards
  or downwards as shown, and hence allows
  radioactive nuclei at intermediate values of N
  and Z to migrate towards the stable zone.

positron emission electron capture
electron emission
22
?- and ?-decay in the Z-N diagram

• In reality, it is the combination of successive
  ?- and ?-decays that allow nuclei to migrate to
  the stable zone.

23

• Often nuclear decay is accompanied by the
  emission of a ?-ray, which is a photon that
  carries away the energy released when a nucleus
  makes a transition from a higher excited state to
  a lower excited state.
• An example is shown in the diagram where the
  parent nucleus undergoes ?-decay to one of three
  different energy levels of the daughter nucleus
  (the lowest being the ground state). Where the
  decay is to one of the two excited states, it is
  followed by ?-ray emission, as the daughter
  nucleus drops down to the lower energy states.
• IMPORTANT ?-rays typically have energies of
  several MeV (highly energetic!), since this is
  the separation between the different energy
  levels of the nucleus.

Gamma Decay
E (MeV)
parent
5
?
4
?
?
3
?
2
?
1
0
24
Nuclear Fission

• In 1932, James Chadwick discovered the neutron.
  Subsequently, physicists realised that this
  particle was a very useful nuclear projectile,
  since being neutral, it had no Coulomb barrier to
  overcome as it approached the nucleus. They had
  particular success in producing new, lower mass
  nuclei by bombarding uranium salts with neutrons.
• A typical fission event is seen in the case of
  235U, which can absorb a neutron, producing 236U
  in a highly excited state. This nucleus undergoes
  fission, splitting into two fragments 140Xe and
  94Sr, which each rapidly emit a neutron
• 235U n ? 236U ? 140Xe 94Sr 2n
• The fragments 140Xe and 94Sr are both highly
  unstable and undergo ?-decay (with the emission
  of an electron) until each reaches a stable end
  product
• 140Xe ? 140Cs ? 140Ba ? 140La ? 140Ce
• 94Sr ? 94Y ? 94Zr

25
Nuclear Fission and its applications

• We see from the previous example of 235U fission,
  for every single neutron that is required to
  trigger the process, two neutrons are produced by
  it.
• Hence the process is self-sustaining since these
  neutrons are capable of triggering further
  fission events ? creates a CHAIN REACTION.
• This chain reaction, plus the release of energy
  associated with each fission event
• disintegration energy, Q (for fission of 235U)
  209MeV (LARGE!!)
• ? it can be used as the basis for a nuclear bomb
  (where process happens rapidly) or a nuclear
  reactor (where process is kept under control and
  made to occur slowly).

26
Nuclear Fusion

• We have already seen from examination of the
  binding energy curve for nuclides that it is
  energetically favourable for two light nuclei to
  combine to form a single larger nucleus ? a
  process called nuclear fusion.
• However, the process is hindered by the Coulomb
  barrier that the nuclei have to surmount in
  order to overcome the repulsion between them due
  to each having a positive charge.

Reminder the Coulombic potential energy
associated with this repulsion is given by
V(1/4??o)(e2/r)Z1Z2, where Z1 and Z2 are the
proton numbers of each nuclei, and r is their
separation. When just touching, r is typically
several Fermis ( 10-15m), and V200(Z1Z2) keV.
The nuclei need kinetic energies of this order to
overcome the Coulomb barrier.

• When the temperature is high enough that the
  thermal motions of the nuclei alone are
  sufficient to penetrate the Coulomb barrier, then
  thermonuclear fusion takes place.

27
Nuclear Fusion
How high does the temperature have to be? Energy
associated with thermal motion, KkT (k
Boltzmanns constant 1.38 x 10-23 J/K). So if K
200keV ? T (200 x 103 x 1.6 x
10-19)/1.38x10-23 2.32 x 109 K

• A place which might be hot enough for
  thermonuclear fusion to take place is at the
  centre of a star. However, taking the Sun as an
  example, it would at first glance appear not to
  be hot enough with T?(centre)107 K. But we do
  know that the Sun is in fact powered by the
  thermonuclear conversion of hydrogen into helium
  at its centre why is this so?
• The above calculation refers to a mean or most
  probably energy, whereas the gas at the centre of
  the Sun has a Maxwellian distribution of
  energies, with a long tail which extends to much
  higher values.
• The barrier heights that have been calculated
  represent peak values and barrier tunneling can
  occur at energies significantly below these
  values.

28
Thermonuclear fusion in the Sun

• The Sun radiates energy at the rate of 3.9 x 1026
  W and has been doing so for 4-5 billion years. It
  is impossible to explain this rate of energy
  output over such a long period by non-nuclear
  processes
• Chemical burning e.g., combustion of coal and
  oxygen, would only last for 1000 years
• Conversion of gravitational potential energy to
  thermal energy solar lifetime too short by a
  factor of 500.
• Instead, the Sun is powered by thermonuclear
  fusion in which hydrogen is burned into helium
  via a multi-step process called the proton-proton
  (p-p) cycle. The various steps are as follows
• 1H 1H ? 2H e ? (Q0.42MeV)
  x2
• e e- ? ? ? (positron-electron
  annihilation, Q1.02MeV) x2
• 2H 1H ? 3He ? (Q 5.49MeV)
  x2
• 3He 3He ? 4He 1H 1H (Q 12.86MeV)
• Q(total) 2(0.421.025.49) 12.86 26.7 MeV

29
Thermonuclear fusion in the Sun

• The Sun radiates energy at the rate of 3.9 x 1026
  W and has been doing so for 4-5 billion years. It
  is impossible to explain this rate of energy
  output over such a long period by non-nuclear
  processes
• Chemical burning e.g., combustion of coal and
  oxygen, would only last for 1000 years
• Conversion of gravitational potential energy to
  thermal energy solar lifetime too short by a
  factor of 500.
• Instead, the Sun is powered by thermonuclear
  fusion in which hydrogen is burned into helium
  via a multi-step process called the proton-proton
  (p-p) cycle. The various steps are as follows
• 1H 1H ? 2H e ? (Q0.42MeV)
  x2
• e e- ? ? ? (positron-electron
  annihilation, Q1.02MeV) x2
• 2H 1H ? 3He ? (Q 5.49MeV)
  x2
• 3He 3He ? 4He 1H 1H (Q 12.86MeV)
• Q(total) 2(0.421.025.49) 12.86 26.7 MeV

solar neutrino
30
Thermonuclear fusion in the Sun

• Important notes
• The first step in the process where two protons
  collide to successfully form a deuteron is
  extremely rare (a 1 in 1026 chance!!) hence this
  step provides a bottle-neck which regulates the
  thermonuclear process in the Sun (and stars in
  general).
• In spite of this bottle-neck, there are so many
  protons undergoing collisions, that 1012 kg of
  deuterium is produced in this way each second!
• Overall, the p-p cycle amounts to the combination
  of 4 protons and 2 electrons to form an
  ?-particle, 2 neutrinos, and 6 ?-rays.

31
The Solar Neutrino Problem

• We see that step 1. of the p-p cycle produces a
  neutrino, and hence the thermonuclear fusion
  taking place at the centre of the Sun produces a
  steady stream of neutrinos which travel
  unhindered from the Suns core to the Earth (and
  through it!).
• However, the number of neutrinos detected on
  Earth is found to be between ? and ½ the number
  predicted from theoretical calculations.
• Unlikely that the observations were wrong
  instead either the models of the Sun were wrong
  or the models of neutrino behaviour (and hence
  the standard model of physics) were wrong ?
  SOLAR NEUTRINO PROBLEM!
• This problem remain unsolved for about 40 years,
  until 2002 when it was shown by the
  Super-Kamiokande and Sudbury Neutrino
  Observatories that the neutrino actually had a
  mass.
• This was in contradiction to the Standard Model
  of Particle Physics where neutrinos have zero
  mass, and the relative fractions of the different
  types electron-neutrino, muon-neutrino, and
  tau-neutrino were invariant.

32
The Solar Neutrino Problem

• It was known from theory that if neutrinos did
  have mass, then they could change from one type
  to another ? NEUTRINO OSCILLATIONS. In particular
  the electron-neutrinos, which are the type
  produced by the p-p cycle in the Sun and the only
  type that could be detected by the neutrino
  observatories like Super-Kamiokande and Sudbury,
  can change into the other two types (muon and
  tau) during their journey from the Sun to the
  Earth.
• Hence the solar neutrino problem was solved, with
  theoretical calculations showing that with
  neutrino oscillations, the number of
  electron-neutrinos expected to be observed is
  35 of the number that leave the Sun, consistent
  with experiment.