Nuclear Physics_1.pdfmjjii7uuyy6uuuuuu778888

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1. Introduction
Muhammad Nur
Zaenul Muhlisin
Physics Departement
Diponegoro University

2. Nuclear Physics
•Why study nuclear physics
• Why nuclear physics is difficult
•Notation & Units
•The Nucleus
•Binding Energy
•Radioactivity
•Half-life
•Biological Effects of Radiation
•Induced Nuclear Reactions
•Fission and Fusion

3. Why Study Nuclear Physics?
•Understand origin of different nuclei
–Big bang: H, He and Li
–Stars: elements up to Fe
–Supernova: heavy elements
•We are all made of stardust
•Need to know nuclear cross sections  experimental nuclear
astrophysics is a hot topic.

4. Practical Applications
•Nuclear fission for energy generation.
–No greenhouse gasses
–Safety and storage of radioactive material.
•Nuclear fusion
–No safety issue (not a bomb)
–Less radioactive material but still some.
•Nuclear transmutation of radioactive waste with neutrons.
–Turn long lived isotopes  stable or short lived.
•Every physicist should have an informed opinion on these important
issues!

5. Medical Applications
•Radiotherapy for cancer
–Kill cancer cells.
–Used for 100 years but can be improved by better delivery and dosimetery
–Heavy ion beams can give more localised energy deposition.
•Medical Imaging
–MRI (Nuclear magnetic resonance)
–X-rays (better detectors  lower doses)
–PET
–Many others…see Medical & Environmental short option.

6. Other Applications
•Radioactive Dating
–C
14
/C
12
gives ages for dead plants/animals/people.
– Rb/Sr gives age of earth as 4.5 Gyr.
•Element analysis
–Forenesic (eg date As in hair).
–Biology (eg elements in blood cells)
–Archaeology (eg provenance via isotope ratios).

7. Nuclear Physics Models
•Progress with understanding nuclear physics from QCD=0  use
simple, approximate, phenomenological models.
•Liquid Drop Model: phenomenology + QM + EM.
•Shell Model: look at quantum states of individual nucleons 
understand spin/parity magnetic moments and deviations from SEMF
for binding energy.

8. Nuclear Structure
The atomic nucleus is composed of neutrons
and protons. These particles are called
nucleons.
The atom’s atomic number (Z) gives the number
of protons in its nucleus. It is the atomic number
that determines an atom’s identity.

9. The nucleon number or mass number
is A = Z+N, where N is the number of
neutrons.
Masses of atoms are sometimes give
in terms of atomic mass units. 1u =
1.66053910
-27
kg.

10. The mass quoted for an atom in the periodic
table is a weighted average over all of the
natural isotopes of that element. The weight
factors are determined by using the relative
abundance on Earth of each isotope.
Atoms of the same element with differing
numbers of neutrons are known as isotopes.

11. For an atomic nucleus
.AV
Am


This implies the density of an atomic nucleus is
independent of A.
3
1
3
3
4
Ar
ArV


As an equality
3
1
0Arr
where r
0 = 1.210
-15
m = 1.2 fm

12. Calculate the mass density of nuclear matter.
m 102.1
15
0
3
1

Arr
Consider a nucleus with one nucleon
(A = 1).
The density is
 
.kg/m 103.2
m 102.1
3
4
kg 1066.1
3
4
317
3
15
27
3










r
m

13. Find the radius and volume
of the nucleus.
  m. 1070.5107m 102.1
151/3151/3
0

Arr
.m 107.7
3
4
3433 
rV
The radius is
The volume is
Tc
107
43

14. Notation
•Nuclei are labelled where El is the chemical
symbol of the element, mass number A = number
of neutrons N + number of protons Z. eg
•Excited states labelled by * or m if they are
metastable (long lived).
El
A
Z
Li
7
3

15. Units
•SI units are fine for macroscopic objects like footballs but
are very inconvenient for nuclei and particles  use
natural units.
•Energy: 1 eV = energy gained by electron in being
accelerated by 1V.
–1 eV= e
J.
•Mass: MeV/c
2
(or GeV/c
2
)
–1 eV/c
2 = e/c
2 kg.
–Or use AMU defined by mass of
12C= 12 u
•Momentum: MeV/c (or GeV/c)
–1 eV/c = e/c kg m s
-1
•Cross sections: (as big as a barn door)
–1 barn =10
-28 m
2
•Length: fermi 1 fm = 10
-15
m.

16. Nuclear Masses and Sizes
•Masses and binding energies
–Absolute values measured with mass spectrometers.
–Relative values from reactions and decays.
•Nuclear Sizes
–Measured with scattering experiments (leave discussion until after
we have looked at Rutherford scattering).
–Isotope shifts

17. Nuclei Sizes
•Scattering experiments
determine size
•Measured in femtometers
(aka fermis)
•All nuclei have nearly the
same density
1.29
31
0Arr
Fig. 29.2, p. 959
m 10fm 1
15


18. Nuclear Mass Measurements
•Measure relative masses by energy released in decays or
reactions.
–X  Y +Z + DE
–Mass difference between X and Y+Z is DE/c
2
.
•Absolute mass by mass spectrometers (next transparency).

19. Mass Spectrometer
•Ion Source
•Velocity selector  electric and
magnetic forces equal and
opposite
–qE=qvB  v=E/B
•Momentum selector, circular orbit
satisfies:
–Mv=qBr
–Measurement r gives M.
Ion Source
Velocity
selector
Detector

20. Binding Energy
A nucleus is held together by the strong nuclear
force. This force only acts over distances of a few
fermis.
The binding energy (E
B) of a nucleus is the energy that
must be supplied to separate it into individual protons
and neutrons.
E
B = Total energy of Z protons and N neutrons – total
energy of nucleus.

21. Total energy of Z protons and N neutrons =
(mass of Z protons and N neutrons)c
2.
Total energy of nucleus = (mass of
nucleus)c
2.
These can be used to define the mass defect
Dm = (mass of Z protons and N neutrons) 
(mass of nucleus) so that
.
2
mcE
BD

22. Nucleons also obey the Pauli Exclusion
Principle such that only two protons
(neutrons) can occupy each proton
(neutron) energy level.
Like an atom, a nucleus can be put into
an excited state if it absorbs a photon
of the correct energy. The nucleus can
then emit a photon to go to a lower
energy state.

23. Binding Energy
•Total energy of
nucleus is less than
combined energy of
individual nucleons
•Difference is called
the binding energy
(aka mass defect)
•Energy required to
separate nucleus
into its constituents
Binding Energy vs. Mass Number
 
Aimmm D

24. Binding Energy vs A
•B increases with A up to
56Fe and then slowly
decreases. Why?
•Lower values and not smooth at small A.

25. Binding Energy as function of A
Mass and Binding energy:
B = [Z M
H + N M
n – M(A,Z)]/c
2

26. Find the binding energy of the
16
O nucleus.
 
 
u1370046.0
u9949146.15u0086649.1 1.00782508
atom Oneutral of mass
neutrons 8 of mass atoms H 8 of mass
16



Dm
MeV 8.127
2
DmcE
B

27. (b) What is the average binding energy per
nucleon?
Example continued:
n.MeV/nucleo 986.7
nucleons ofnumber
nucleonper energy Binding


B
E

28. Nuclear Sizes & Isotope Shift
•Coulomb field modified by finite size of nucleus.
•Assume a uniform charge distribution in the nucleus.
Gauss’s law 
integrate and apply boundary conditions
•Difference between actual potential and Coulomb
•Use 1
st
order perturbation theory
3
2
0
)(
4 R
r
r
Ze
E


R
Ze
R
Zer
rV
0
3
0
2
8
3
8
)(


)Rr(
r4
Ze
R8
Ze3
R8
Zer
)r(V
00
3
0
2
D

drrrVerrE
R
)()]()[(4
*
0
2
 DD
2/3
0
0
2/3
0
)(2)/exp()(2)(
a
Z
aZr
a
Z
r 

29. Isotope Shifts
•Isotope shift for optical spectra
•Isotope shift for X-ray spectra (bigger effect because electrons
closer to nucleus)
•Isotope shift for X-ray spectra for muonic atoms. Effect greatly
enhanced because m
m~ 207 m
e and a
0~1/m.
•All data consistent with R=R
0 A
1/3 with R
0=1.25fm.

30. Frequency shift of an optical
transition in Hg at =253.7nm
for different A relative to A=198.
Data obtained by laser
spectroscopy.
The effect is about 1 in 10
7
.
(Note the even/odd structure.)
Bonn et al Z Phys A 276, 203
(1976)
A
2/3
Isotope Shift in Optical Spectra
D
E/h (GHz)

31. Data on the isotope shift of K X ray lines in Hg. The effect is about 1 in
10
6
. Again the data show the R
2
= A
2/3
dependence and the even/odd effect.
Lee et al, Phys Rev C 17, 1859 (1978)

32. Data on Isotope Shift of K Xrays
from muonic atoms [in which a
muon with m=207m
e takes the place
of the atomic electron].
Because a
0 ~ 1/m the effect is ~0.4%,
much larger than for an electron.
The large peak is 2p
3/2 to 1s
1/2. The
small peak is 2p
1/2 to 1s
1/2. The size
comes from the 2j+1 statistical
weight.
Shera et al Phys Rev C 14, 731
(1976)
58
Fe
56
Fe
54
Fe
Energy (keV)

33. SEMF
•Aim: phenomenological understanding of nuclear
binding energies as function of A & Z.
•Nuclear density constant (see lecture 1).
•Model effect of short range attraction due to strong
interaction by liquid drop model.
•Coulomb corrections.
•Fermi gas model  asymmetry term.
•QM pairing term.
•Compare with experiment: success & failure!

34. Liquid Drop Model Nucleus
•Phenomenological model to understand binding energies.
•Consider a liquid drop
–Ignore gravity and assume no rotation
–Intermolecular force repulsive at short distances, attractive at
intermediate distances and negligible at large distances  constant
density.
E=-an + 4R
2T B=an-bn
2/3
•Analogy with nucleus
–Nucleus has constant density
–From nucleon nucleon scattering experiments: Nuclear force has
short range repulsion and attractive at intermediate distances.
–Assume charge independence of nuclear force, neutrons and protons
have same strong interactions check with experiment!

35. Mirror Nuclei
•Compare binding energies of mirror nuclei (nuclei n
p). Eg
7
3Li and
7
4Be.
•Mass difference due to n/p mass and Coulomb energy.
dQ
r
rQ
E
R

0 04
)(

323
/3)/()( RZerdQRrZerQ 
R
Ze
dr
R
r
r
Ze
E
R
0
2
6
5
0 0
2
4
)(
)5/3(
4
)(3


3/1
0
2
;2/~;)]2)(1()1([
45
3
)1,( ARAZZZZZ
R
e
ZZE
c D

3/2
)1,( AZZE
C D

36. nn and pp interaction same
(apart from Coulomb)
“Charge symmetry”

37. Charge Symmetry and Charge Independence
•Mirror nuclei showed that strong interaction is
the same for nn and pp.
•What about np ?
•Compare energy levels in “triplets” with same A,
different number of n and p. e.g.
•Same energy levels for the same spin states  SI
same for np as nn and pp.
MgNaNe
22
12
22
11
22
10

38. Charge Independence
•Is np force is same as nn and pp?
•Compare energy levels in nuclei
with same A.
•Same spin/parity states have
same energy.
•np=nn=pp
23
11Na
23
12 Mg
22
12Mg
22
11Na
22
10Ne

39. Charge Independence of Strong Interaction
•If we correct for n/p mass difference and Coulomb interaction,
then energy levels same under n p.
•Conclusion: strong interaction same for pp, pn and nn if
nucleons are in the same quantum state.
•Beware of Pauli exclusion principle! eg why do we have
bound state of pn but not pp or nn?

40. Asymmetry Term
•Neutrons and protons are spin ½ fermions  obey
Pauli exclusion principle.
•If other factors were equal  ground state would
have equal numbers of n & p.
Illustration
Neutron and proton states with same
spacing D.
Crosses represent initially occupied states
in ground state.
If three protons were turned into neutrons
the extra energy required would be 3×3 D.
In general if there are Z-N excess protons
over neutrons the extra energy is
((Z-N)/2)
2
D. relative to Z=N.

41. Asymmetry Term
•From stat. mech. density of states in 6d phase space = 1/h
3
•Integrate to get total number of protons Z, & Fermi Energy (all states filled
up to this energy level).
•Change variables p  E
3
2
4
h
dpVp
dN


3
F
)h/Vp()3/8(Z
F
E
0
2/1
E
0
2/3
2/1
E)5/3(
dEE
dEE
EAE
dE/dp
dp/dN
dE/dN
F
F




3/1
0
3/1
F
A
Z
R
h
)8/3(P 






3/2
2
0
2
3/2
F
A
Z
mR2
h
)8/3(E 







42. Asymmetry Term
•Binomial expansion keep lowest term in y/A
•Correct functional form but too small by factor of 2. Why?
A
ZN
KE
2
)(

3/2
2
0
2
3/2P
Total
A
Z
mR2
Zh
)8/3(
5
3
E 





 
 
3/53/5
3/2
Total NZ
A
K
E  ZNy
 
3/53/5
3/2
3/5
)/1()/1( AyAy
A
KA
E
Total


43. Pairing Term
•Nuclei with even number of n or even number
of p more tightly bound fig.
•Only 4 stable o-o nuclei cf 153 e-e.
•p and n have different energy levels  small
overlap of wave functions. Two p(n) in same
level with opposite values of j
z have AS spin
state  sym spatial w.f. maximum overlap
maximum binding energy because of short
range attraction.
Neutron number
Neutron separation energy in Ba

44. Pairing Term
•Phenomenological fit to A dependence
•Effect smaller for larger A
2/1
AE

e-e +ive
e-o 0
o-o -ive

45. Semi Empirical Mass Formula
•Put everything together:
•Fit to measured binding energy.
–Fit not too bad (good to <1%).
–Deviations are interesting  shell effects.
–Coulomb term agrees with calculation.
–Asymmetry term larger ?
–Explain valley of stability.
–Explains energetics of radioactive decays, fission and fusion.
2/13/1
22
3/2 )(
),(
AA
Z
d
A
ZN
cbAaAZNB





46. The Binding Energy per
nucleon of beta-stable (odd A)
nuclei.
Fit values in MeV
a15.56
b17.23
c23.285
d0.697
+12 (o-o)
0 (o-e)
-12 (e-e)
A
B/A (MeV)
7.5
9.0

47. Valley of Stability
•SEMF allows us to
understand valley of
stability.
•Low Z, asymmetry
term  Z=N
•Higher Z, Coulomb
term  N>Z.