# Nuclear Physics

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```					Nuclear and Particle Physics

3 lectures:
Nuclear Physics
Particle Physics 1
Particle Physics 2

1
Nuclear Physics Topics

 Composition of Nucleus
 features of nuclei
 Nuclear Models
 nuclear energy
 Fission
 Fusion
 Summary

2
 Energy - electron-volt
 1 electron-volt = kinetic energy of an electron when
moving through potential difference of 1 Volt;
o 1 eV = 1.6 × 10-19 Joules
o 1 kW•hr = 3.6 × 106 Joules = 2.25 × 1025 eV
o 1 MeV = 106 eV, 1 GeV= 109 eV, 1 TeV = 1012 eV

 mass - eV/c2
o   1 eV/c2 = 1.78 × 10-36 kg
o   electron mass = 0.511 MeV/c2
o   proton mass = 938 MeV/c2 = 0.938 GeV/ c2
o   neutron mass = 939.6 MeV/c2

 momentum - eV/c:
o 1 eV/c = 5.3 × 10-28 kg m/s
o momentum of baseball at 80 mi/hr
 5.29 kgm/s  9.9 × 1027 eV/c
 Distance
o 1 femtometer (“Fermi”) = 10-15 m
3
 Antoine Becquerel (1896): serendipitous discovery of radioactivity:
penetrating radiation emitted by substances containing uranium
 Antoine Becquerel, Marie Curie, Pierre Curie (1896 – 1898):
o three kinds of radiation, with different penetrating power
(i.e. amount of material necessary to attenuate beam):
 “Alpha (a) rays” (least penetrating – stopped by paper)
 “Beta (b) rays” (need 2mm lead to absorb)
 “Gamma (g) rays” (need several cm of lead to be attenuated)
o three kinds of rays have different electrical charge:
a: +, b: -, g: 0
 Ernest Rutherford (1899)
o Beta (b) rays have same q/m ratio as electrons
o Alpha (a) rays have same q/m ratio as He nucleus
o Alpha (a) rays captured in container show He-like emission spectrum
4
Geiger, Marsden, Rutherford expt.
 (Geiger, Marsden, 1906 - 1911) (interpreted by Rutherford, 1911)
 get a particles from radioactive source
 make “beam” of particles using “collimators”
(lead plates with holes in them, holes aligned in straight line)
 bombard foils of gold, silver, copper with beam
 measure scattering angles of particles with scintillating screen (ZnS)

5
6
Geiger Marsden experiment: result
 most particles only slightly deflected (i.e. by small
angles), but some by large angles - even backward
 measured angular distribution of scattered particles did
not agree with expectations from Thomson model (only
small angles expected),
 but did agree with that expected from scattering on
small, dense positively charged nucleus with diameter
< 10-14 m, surrounded by electrons at 10-10 m
Ernest
Rutherford
1871-1937

7
Proton
 “Canal rays”
 1898: Wilhelm Wien:
opposite of “cathode rays”
 Positive charge in
nucleus (1900 – 1920)
 Atoms are neutral
o positive charge needed to cancel electron’s negative charge
o Rutherford atom: positive charge in nucleus
 periodic table  realized that the positive charge of
any nucleus could be accounted for by an integer
number of hydrogen nuclei -- protons

8
Neutron
 Walther Bothe 1930:
 bombard light elements (e.g. 49Be) with alpha particles 
 Irène and Frédéric Joliot-Curie (1931)
 pass radiation released from Be target through paraffin wax 
protons with energies up to 5.7 MeV released
 if neutral radiation = photons, their energy would have to be 50
MeV -- puzzle
 puzzle solved by James Chadwick (1932):
particle with mass approximately equal to that of the proton”
 identified with the “neutron” suggested by Rutherford in 1920
 observed reaction was:
a (24He++) + 49Be  613C*
13
6 C*     612C + n
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Beta decay -- neutrino
 Beta decay puzzle :
o decay changes a neutron into a proton
o apparent “non-conservation” of energy
o apparent non-conservation of angular momentum
 Wolfgang Pauli predicted a light, neutral,
feebly interacting particle (called it neutron,
later called neutrino by Fermi)

 Although accepted since it “fit” so well, not
actually observed initiating interactions until
1956-1958 (Cowan and Reines)                       10
Puzzle with Beta Spectrum
a, b, g
 Both a, g discrete spectrum          F. A. Scott, Phys. Rev. 48, 391 (1935)
because
Ea, g = Ei – Ef
 But b spectrum continuous

 Energy conservation violated??
 Bohr:: “At the present stage
of atomic theory, however,
we may say that we have no
argument, either empirical or
theoretical, for upholding the
energy principle in the case
of β-ray disintegrations”
11
Desperate Idea of Pauli

12
Positron
 Positron (anti-electron)
 Predicted by Dirac (1928) -- needed for relativistic
quantum mechanics
 existence of antiparticles doubled the number of
known particles!!!

Positron track going
plate

 P.A.M. Dirac
 Nobel Prize (1933)
 member of FSU faculty
(1972-1984)
 one of the greatest physicists of the 20th century     13
Structure of nucleus
 size (Rutherford 1910, Hofstadter 1950s):
 R = r0 A1/3, r0 = 1.2 x 10-15 m = 1.2 fm;
 i.e. ≈ 0.15 nucleons / fm3
 generally spherical shape, almost uniform density;
 made up of protons and neutrons
 protons and neutron -- “nucleons”;
are fermions (spin ½), have magnetic moment
 nucleons confined to small region (“potential well”)
  occupy discrete energy levels
 two distinct (but similar) sets of energy levels,
one for protons, one for neutrons
 proton energy levels slightly higher than those of
neutrons (electrostatic repulsion)
 spin ½  Pauli principle                               14
 only two identical nucleons per energy level
Nuclear Sizes - examples
1

r  ro (A )   3
ro = 1.2 x 10-15 m

Find the ratio of the radii for the following nuclei:

1H, 12C, 56Fe, 208Pb, 238U

1     1    1        1       1
3     3    3        3       3
1 : 12 : 56 : 208 : 238

1 : 2.89 : 3.83 : 5.92 : 6.20
15
A, N, Z
 for natural nuclei:
 Z range 1 (hydrogen) to
92 (Uranium)
 A range from 1 ((hydrogen)
to 238 (Uranium)
 N = neutron number = A-Z
 N – Z = “neutron excess”;
increases with Z
 nomenclature:
 ZAXN or AXN or
A
X or X-A
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Atomic mass unit

 “atomic number” Z
Number of protons in nucleus

 Mass Number A
Number of protons and neutrons in nucleus
Atomic mass unit is defined in terms of the
mass of 126C, with A = 12, Z = 6:
1 amu = (mass of 126C atom)/12
1 amu = 1.66 x 10-27kg
1 amu = 931.494 MeV/c2
17
Properties of Nucleons
 Proton
 Charge = 1 elementary charge e = 1.602 x 10-19 C
 Mass = 1.673 x 10-27 kg = 938.27 MeV/c2
=1.007825 u = 1836 me
 spin ½, magnetic moment 2.79 eħ/2mp

 Neutron
 Charge = 0
 Mass = 1.675 x 10-27 kg = 939.57 MeV/c2
= 1.008665 u = 1839 me
 spin ½, magnetic moment -1.9 eħ/2mn

18
Nuclear masses, isotopes

 Nuclear masses measured, e.g. by mass
spectrography
 masses expressed in atomic mass units (amu),
energy units MeV/c2
 all nuclei of certain element contain same number
of protons, but may contain different number of
neutrons
 examples:
 deuterium, heavy hydrogen 2D or 2H;
heavy water = D2O (0.015% of natural water)
U),
 U- 235 (0.7% of natural U), U-238 (99.3% of natural 19
Nuclear energy levels: example

Problem: Estimate the lowest possible energy of a neutron contained
in a typical nucleus of radius 1.33×10-15 m.

E = p2/2m = (cp)2/2mc2

x p = h/2                    x (cp) = hc/2
(cp) = hc/(2 x) = hc/(2 r)
(cp) = 6.63x10-34 Js * 3x108 m/s / (2 * 1.33x10-15 m)
(cp) = 2.38x10-11 J = 148.6 MeV

E = p2/2m = (cp)2/2mc2 = (148.6 MeV)2/(2*940 MeV) = 11.7 MeV
20
Nuclear Masses, binding energy
 Mass of Nucleus  Z(mp) + N(mn)
 “mass defect” m = difference
between mass of nucleus and mass of
constituents
 energy defect = binding energy EB
EB = m c2
 binding energy = amount of energy that
must be invested to break up nucleus
into its constituents
 binding energy per nucleon = EB /A
21
Nuclear Binding Energy
 Nuclear binding energy =
difference between the      1 amu =         931.49 MeV
energy (or mass) of the
nucleus and the energy
m(proton)             1.00782
(or mass) of its            m(neutron)            1.00867
constituent neutrons and
protons.                                A=             56
 = (-) the energy needed                  Z=            26
to break the nucleus
apart                                   N=             30
 Average binding energy
per nucleon = total
Mass (amu)          55.92066
binding energy divided by   Ebinding (MeV)    -505.58094
the number of nucleons
(A).                        EB/A(MeV)            -9.02823
 Example: Fe-56

22
Problem – set 4
 Compute binding energy per nucleon for
 42He          4.00153 amu
 168O          15.991 amu
 5626Fe        55.922 amu
 23592U        234.995 amu
 Is there a trend?
 If there is, what might be its significance?
 note:
 1 amu =       931.5 MeV/c2
 m(proton) = 1.00782 amu
 m(neutron)= 1.00867 amu
23
Binding energy per nucleon



24

 Alpha Decay
 AZ  A-4(Z-2) + 4He
o Number of protons is conserved.
o Number of neutrons is conserved.
 Gamma Decay
 AZ*  AZ + g
o An excited nucleus loses energy
by emitting a photon.

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Beta Decay

 Beta Decay
 AZ  A(Z+1) + e- + an anti-neutrino
o A neutron has converted into a proton,
electron and an anti-neutrino.
 Positron Decay
 AZ  A(Z-1) + e+ + a neutrino
o A proton has converted into a neutron,
positron and a neutrino.
 Electron Capture
 AZ + e-  A(Z-1) + a neutrino
o A proton and an electron have converted
into a neutron and a neutrino.
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Electron capture:
 Several decay processes:
a decay: A    A- 4
A
X + e - Z -AY +
Z    X  Z -2Y + 2 He
4                                  Z               1

e.g.,210Po206Pb+ 2 He
84    82
4
e.g.,12N + e - 12C +
7          6

b- decay:
~
g decay:
A
X  Z +1Y + e - + 
A
Z
~
A
Z   X * ZAX + g
e.g ., Tc  Rb + e + 
99
43
99
44
-

e.g.,99Tc* 99Tc + g (140keV )
b+ decay:                                         43     43

A
Z   X Z -AY + e + +
1

e.g.,12N 12C + e + +
7    6                                                       27
 Activity A =                   dN
A    .
number of                  dt
decays per unit time
 decay constant  =
probability of decay       dN
 -N .
per unit time              dt
 Rate of decay 
- t
number N of nuclei       N (t )  N 0 e .
 Solution of diff.
equation (N0 = nb. of             

nuclei at t=0)                     t e -t dt
  t dN               
1
 Mean life  = 1/ 
0

 dN  e  dt

- t           
0                     28
Nuclear decay rates
Nuclear Decay

1000.0
Nuclei Remaining

800.0
- t
600.0                                         N (t )  N 0e .
400.0
At t = 1/,
200.0
N is 1/e (0.368)
0.0                                         of the original
0.0   1.0     2.0   3.0   4.0   5.0   amount
Time(s)

29
Nuclear (“strong”) force
 atomic nuclei small -- about 1 to 8fm
 at small distance, electrostatic repulsive forces
are of macroscopic size (10 – 100 N)
 there must be short-range attractive force
between nucleons -- the “strong force”
 strong force essentially charge-independent
 “mirror nuclei” have almost identical binding
energies
 mirror nuclei = nuclei for which n  p or p  n
(e.g. 3He and 3H, 7Be and 7Li, 35Cl and 35Ar);
slight differences due to electrostatic
repulsion
 strong force must have very short range – <<
atomic size, otherwise isotopes would not have
same chemical properties
30
Strong force -- 2
 range: fades away at distance ≈ 3fm
 force between 2 nucleons at 2fm distance ≈
2000N
 nucleons on one side of U nucleus hardly
affected by nucleons on other side
 experimental evidence for nuclear force from
scattering experiments;
 low energy p or a scattering: scattered
particles unaffected by nuclear force
 high energy p or a scattering:
particles can overcome electrostatic
repulsion and can penetrate deep enough to
enter range of nuclear force                 31
N-Z and binding energy vs A
 small nuclei (A<10):
 All nucleons are within range of strong force
exerted by all other nucleons;
 add another nucleon  enhance overall cohesive force
 EB rises sharply with increase in A
 medium size nuclei (10 < A < 60)
 nucleons on one side are at edge of nucl. force range from
nucleons on other side  each add’l nucleon gives diminishing
return in terms of binding energy  slow rise of EB /A
 heavy nuclei (A>60)
 adding more nucleons does not increase overall cohesion due
to nuclear attraction
 Repulsive electrostatic forces (infinite range!) begin to have
stronger effect
 N-Z must be bigger for heavy nuclei (neutrons provide
attraction without electrostatic repulsion
 heaviest stable nucleus: 209Bi
– all nuclei heavier than 209Bi are unstable
32
EB/A vs A

33
Nuclear Models – liquid drop model
 liquid drop model (Bohr, Bethe, Weizsäcker):
 nucleus = drop of incompressible nuclear fluid.
 fluid made of nucleons, nucleons interact
strongly (by nuclear force) with each other,
just like molecules in a drop of liquid.
 introduced to explain binding energy
and mass of nuclei
 predicts generally spherical shape of nuclei
 good qualitative description of fission
of large nuclei
 provides good empirical description
of binding energy vs A
34
Bethe – Weizsäcker formula for binding energy
 Bethe - Weizsäcker formula:
 an empirically refined form of the liquid drop model for the
binding energy of a nucleus of mass number A with Z protons
and N neutrons
 binding energy has five terms describing different aspects
of the binding of all the nucleons:
o   volume energy
o   surface energy
o    Coulomb energy (electrostatic repulsion of the protons,)
o    an asymmetry term (N vs Z)
o    an exchange (pairing) term (even-even vs odd-even vs odd-odd
number of nucleons)

B(A, Z)  a V A - a S A   2/3         Z 2
- a C 1/3 - a Sy m
Z - N  - λ a A -3/4
2

P
A                A
35
“liquid drop” terms in B-W formula

36
Independent Particle Models
 assume nucleons move inside nucleus
without interacting with each other
 Fermi- gas model:
 Protons and neutrons move freely within nuclear volume,
considered a rectangular box
 Protons and neutrons are distinguishable and so move
in separate potential wells
 Shell Model
 formulated (independently)
by Hans Jensen and Maria Goeppert-Mayer
 each nucleon (proton or neutron) moves in the average
potential of remaining nucleons, assumed to be spherically
symmetric.
 also takes account of the interaction between a nucleon’s
spin and its angular momentum (“spin-orbit coupling”)
 derives “magic numbers” (of protons and/or neutrons) for
which nuclei are particularly stable: 2, 8, 20, 28, 50, 82, 126, ..
37
Fermi-Gas Model of Nucleus
 Ground State
Potential well
 In each potential well,
the lowest energy states
are occupied.

 Because of the Coulomb
repulsion the proton well
is shallower than that of
the neutron.
 Therefore, as Z increases
 But the nuclear energy        we would expect nuclei to
is minimized when the         contain progressively
maximum energy level is       more neutrons than
protons and neutrons
 U has A = 238, Z = 92
38
Collective model
 collective model is “eclectic”, combining aspects
of other models
 consider nucleus as composed of “stable core”
of closed shells, plus additional nucleons outside
of core
 additional nucleons move in potential well due to
interaction with the core
 interaction of external nucleons with the core
 agitate core – set up rotational and
vibrational motions in core, similar to those that
occur in droplets
 gives best quantitative description of nuclei
39
Nuclear energy
   very heavy nuclei:
 energy released if break up into two medium sized nuclei
 “fission”
 light nuclei:
 energy released if two light nuclei combine -- “fuse” into a
heavier nucleus – “fusion”

40
Nuclear Energy - Fission

+ about 200 MeV energy   41
Fission

42
Nuclear Fusion

43
Sun’s Power Output

Unit of Power
 1 Watt = 1 Joule/second
 100 Watt light bulb = 100 Joules/second

 Sun’s power output
 3.826 x 1026 Watts
 exercise: calculate sun’s power output
using Stefan-Boltzmann law (assume sun
is a black body)

44
The Proton-Proton Cycle            1H + 1H → 2H + e+ + 
e+ + e- → g + g
2H + 1H → 3He + g
1 pp collision in 1022 → fusion!
3He   + 3He → 4He + 1H + 1H

4H → 4He

Deuterium creation 3He creation           4He   creation

45
Super Kamiokande: Solar Neutrinos

Solar neutrino

Electron

46
A Nearby Super-Giant

47
Life of a 20 Solar Mass Super-Giant
 Hydrogen fusion
 ~ 10 million years
 Helium fusion
 ~ 1 million years
 Carbon fusion
 ~ 300 years
 Oxygen fusion
 ~ 9 months
 Silicon fusion
 ~ 2 days

48
http://cassfos02.ucsd.edu/public/tutorial/SN.html
Carbon fusion

7.65 MeV above 12C ground state

49
Oxygen fusion

7.19 MeV
7.12 MeV

50
Supernova 1987A   Before

After

51
Summary
 nuclei made of protons and neutrons,
held together by short-range strong nuclear force
 models describe most observed features,
still being tweaked and modified
 no full-fledged theory of nuclei yet
 development of nuclear theory based on
QCD has begun
 nuclear fusion is the process of energy
production of Sun and other stars
 we (solar system with all that’s in it)
are made of debris from dying stars        52

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