RFSS and NFSS: Lecture 2Nuclear Properties
Download
Report
Transcript RFSS and NFSS: Lecture 2Nuclear Properties
RFSS and NFSS: Lecture 2
Nuclear Properties
• Readings:
Modern Nuclear Chemistry: Chapter 2
Nuclear Properties
Nuclear and Radiochemistry: Chapter 1
Introduction, Chapter 2 Atomic Nuclei
Nuclear Forensics Analysis, Chapter 2
• Nuclear properties
Masses
Binding energies
Reaction energetics
Nuclear shapes
2-1
Nuclear Properties
• Systematic examination of general nuclear properties
masses
matter distributions
• Size, shape, mass, and relative stability of nuclei follow patterns
that can be understood and interpreted with two models
average size and stability of a nucleus can be described by
average binding of the nucleons in a macroscopic model
detailed energy levels and decay properties evaluated with
a quantum mechanical or microscopic model
Number of stable nuclei based on neutron and proton number
N
Z
Number
even
even
160
odd
even
53
even
odd
49
odd
odd
4
2-2
Mass excess
• Difference between actual mass and atomic
number
Expression of binding energy relative to 12C
By definition 12C = 12 amu
* If mass excess negative, then isotope
has more binding energy the 12C
• Mass excess==M-A
24Na
23.990962782 amu
23.990962782-24 = -0.009037218 amu
• 1 amu = 931.5 MeV
2-3
• -8.41817 = Mass excess=
Masses
• Atomic masses
Nuclei and electrons
• Nuclear mass can be found from atomic mass
m0 is electron rest mass, Be (Z) is the total binding energy of all
the electrons
Be(Z) is small compared to total mass
• Energy (Q) from difference in mass between parent and daughter
• Consider beta decay of 14C
14C14N+ + β- +n + Q
Energy = mass 14C – mass 14N
• Positron decay
• Electron Capture
2-4
Masses
• Electron Capture
Electron comes from parent
orbital
Parent designated as cation to
represent this behavior
• Alpha Decay
241Am237Np + 4He + Q
Use mass excess or Q value
calculator to determine Q
value
Q = 52.937 – (44.874 + 2.425)
Q = 5.638 MeV
Alpha decay energy for 241Am is
5.48 and 5.44 MeV
2-5
Masses
• For a general reaction
Treat Energy (Q) as part of the equation
56Fe+4He59Co+1H+Q
Q= [M56Fe+M4He-(M59Co+M1H)]c2
* M represents mass of isotope
2-6
For coursework please note if 1.022 MeV added to positron Q value
Q value calculation
•
•
•
Find Q value for the Beta decay of 24Na
24Na24Mg+ +b- + n +Q
Q= 24Na-24Mg
M (24Na)-M(24Mg)
23.990962782-23.985041699
0.005921 amu
* 5.5154 MeV
From mass excess
-8.417 - -13.933
5.516 MeV
Q value for the EC of 22Na
22Na+ + e- 22Ne + n +Q
Q= 22Na - 22Ne
M (22Na)-M(22Ne)
21.994436425-21.991385113
0.003051 amu
2.842297 MeV
From mass excess
-5.181 - -8.024
2.843 MeV
Can also use Q-value calculator
http://www.nndc.bnl.gov/qcalc/
2-7
Terms
•
Binding energy
Difference between mass of nucleus
and constituent nucleons
Energy released if nucleons
formed nucleus
Btot (A,Z)=[ZM(1H)+(A-Z)M(n)-M(A,Z)]c2
average binding energy per nucleon
Bave(A,Z)= Btot (A,Z)/A
Measures relative stability
•
Binding Energy of an even-A nucleus is generally higher than values for adjacent odd-A
nuclei
•
Exothermic nature of the fusion of H atoms to form He from the very large binding
energy of 4He
•
Energy released from fission of the heaviest nuclei is large
•
Nuclei near the middle of the periodic table have higher binding energies per
nucleon
Maximum in the nuclear stability curve in the iron-nickel region (A~56 through 59)
2-8
Thought to be responsible for the abnormally high natural abundances of these
elements
Mass Based Energetics Calculations
• Why does 235U undergo neutron
induced fission for thermal
energies while 238U doesn’t?
• Generalized energy equation
AZ + n A+1Z + Q
• For 235U
Q=(40.914+8.071)-42.441
Q=6.544 MeV
• For 238U
Q=(47.304+8.071)-50.569
Q=4.806 MeV
• For 233U
Q=(36.913+8.071)-38.141
Q=6.843 MeV
• Fission requires around 5-6 MeV
Does 233U fission from
thermal neutron
2-9
Binding-Energy Calculation
• Volume of nuclei are nearly proportional to the number of
nucleons present
Nuclear matter is incompressible
Basis of equation for nuclear radius
• Total binding energies of nuclei are nearly proportional to
the numbers of nucleons present
saturation character
Nucleon in a nucleus can apparently interact with
only a small number of other nucleons
Those nucleons on the surface will have different
interactions
• Basis of liquid-drop model of nucleus
Considers number of neutrons and protons in nucleus
2-10
Liquid-Drop Binding Energy:
2
2
N -Z
N -Z
2/3
2 -1/ 3
2 -1
EB c1 A1 - k
- c2 A 1 - k
- c3 Z A + c4 Z A +
A
A
• c1=15.677 MeV, c2=18.56 MeV, c3=0.717 MeV, c4=1.211 MeV,
k=1.79 and =11/A1/2
• 1st Term: Volume Energy
dominant term
in first approximation, binding energy is
proportional to the number of nucleons
(N-Z)2/A represents symmetry energy
binding E due to nuclear forces is greatest for
the nucleus with equal numbers of neutrons
and protons
2-11
Liquid drop model
2
2
N -Z
N -Z
2/3
2 -1/ 3
2 -1
EB c1 A1 - k
- c2 A 1 - k
- c3 Z A + c4 Z A +
A
A
• 2nd Term: Surface Energy
Nucleons at surface of nucleus have unsaturated forces
decreasing importance with increasing nuclear size
• 3rd and 4thTerms: Coulomb Energy
3rd term represents the electrostatic energy that arises
from the Coulomb repulsion between the protons
lowers binding energy
4th term represents correction term for charge distribution
with diffuse boundary
• term: Pairing Energy
binding energies for a given A depend on whether N and Z
are even or odd
even-even nuclei, where =11/A1/2, are the stablest
two like particles tend to complete an energy level by
pairing opposite spins
2-12
Mass Parabolas
• Method of
demonstrating
stability for given
mass constructed
from binding energy
Values given in
difference, can
use energy
difference
• For odd A there is
only one b-stable
nuclide
nearest the
minimum of the
parabola
2-13
Friedlander & Kennedy, p.47
Even A mass parabola
• For even A there are usually two or three possible b-stable isobars
Stable tend to be even-even nuclei
Even number of protons, even number of neutron for
these cases
2-14
Magic Numbers
•
Certain values of N
and Z--2, 8, 20, 28,
50, 82, and 126 -exhibit unusual
stability
• Evidence from
masses, binding
energies, elemental
and isotopic
abundances,
numbers of species
with given N or Z,
and -particle
energies
• Concept of closed
shells in nuclei
• Demonstrates
limitation in liquid
drop model
2-15
Nuclear Shapes: Radii
R=roA1/3
• Nuclear volumes are about proportional to nuclear
masses
nuclei have approximately same density
• nuclei are not densely packed with nucleons
Density varies
• ro~1.1 to 1.6 fm
• Nuclear radii can mean different things
nuclear force field
distribution of charges
nuclear mass distribution
2-16
Nuclear Force Radii
•
•
The radius of the nuclear force field must be less than
the distance of closest approach (do)
d = distance from center of nucleus
T’ = particle’s kinetic energy
T = particle’s initial kinetic energy
do = distance of closest approach in a head on
collision when T’=0
do~10-20 fm for Cu and 30-60 fm for U
2Ze 2
T' T do
2 Ze 2
do
T
2-17
http://hyperphysics.phy-astr.gsu.edu/hbase/rutsca.html#c1
Measurement of Nuclear Radii
• Any positively charged particle can be used to
probe the distance
nuclear (attractive) forces become significant
relative to the Coulombic (repulsive force)
• Neutrons can be used but require high energy
neutrons are not subject to Coulomb forces
high energy needed for de Broglie
wavelengths small compared to nuclear
dimensions
At high energies, nuclei become transparent to
neutrons
Small cross sections
2-18
Electron Scattering
• Using moderate energies of electrons, data is compatible
with nuclei being spheres of uniformly distributed charges
• High energy electrons yield more detailed information about
the charge distribution
no longer uniformly charged spheres)
• Radii distinctly smaller than indicated by methods that
determine nuclear force radii
• Re (half-density radius)~1.07 fm
• de (“skin thickness”)~2.4 fm
2-19
Nuclear potentials
• Scattering experiments has approximate
agreement the Square-Well potential
Woods-Saxon equation better fit
Vo
V
(r -R) / A
1+ e
Vo=potential at center of nucleus
a=constant~0.5 fm
R=distance from center at which V=0.5Vo (for
half-potential radii) or V=0.9Vo and V=0.1Vo
for a drop-off from 90 to 10% of the full
potential
2-20
Square-Well and Woods-Saxon Potentials
• ro~1.35 to 1.6 fm for Square-Well
• ro~1.25 fm for Woods-Saxon with half-potential radii,
• ro~2.2 fm for Woods-Saxon with drop-off from 90 to
10%, skin thickness, of the full potential
2-21
Nuclear Skin
• charge density give information on protons distribution in
nuclei
no experimental techniques exist for determining total
nucleon distribution
generally assumed that neutrons and protons are
distributed in same way
nuclear-potential radii are about 0.2 fm larger than radii
of the charge distributions
Nucleus Fraction of nucleons in the “skin”
o
12C
(r )
0.90
[( r - R ) / a ]
1
+
e
24Mg
0.79
56Fe
0.65
107Ag
0.55
139Ba
0.51
208Pb
0.46
2-22
238U
0.44
e
e
e
Spin
• Nuclei possess angular momenta Ih/2
I is an integral or half-integral number known as the
nuclear spin
For electrons, generally distinguish between electron
spin and orbital angular momentum
• Protons and neutrons have I=1/2
• Nucleons in the nucleus contribute orbital angular
momentum (integral multiple of h/2 ) and their intrinsic
spins (1/2)
Protons and neutrons can fill shell (shell model)
Shells have orbital angular momentum like electron
orbitals (s,p,d,f,g,h,i,….)
spin of even-A nucleus is zero or integral
spin of odd-A nucleus is half-integral
• All nuclei of even A and even Z have I=0 in ground state2-23
Magnetic Moments
• Nuclei with nonzero angular momenta have magnetic
moments
From spin of protons and neutrons
• Bme/Mp is used as the unit of nuclear magnetic moments
and called a nuclear magneton
• Magnetic moments are often expressed in terms of
gyromagnetic ratios
g*I nuclear magnetons, where g is + or - depending
upon whether spin and magnetic moment are in the
same direction
• Measured magnetic moments tend to differ from calculated
values
Proton and neutron not simple structures
Neutron has charge distribution
* Negative (from negative mesons) near edge
2-24
Methods of measurements
•
Hyperfine structure in atomic spectra
•
Atomic Beam method
Element beam split into 2I+1 components in magnetic field
•
Resonance techniques
•
2I+1 different orientations
Quadrupole Moments: q=(2/5)Z(a2-c2), R2 = (1/2)(a2 + c2)= (roA1/3)2
Data in barns, can solve for a and c
•
Only nuclei with I1/2 have quadrupole moments
Non-spherical nuclei
Interactions of nuclear quadrupole moments with the electric fields produced by electrons in
atoms and molecules give rise to abnormal hyperfine splittings in spectra
•
Methods of measurement: optical spectroscopy, microwave spectroscopy, nuclear resonance
absorption, and modified molecular-beam techniques
2-25
Parity
•
•
•
System wave function sign change if the sign of the space coordinates change
system has odd or even parity
Parity is conserved
even+odd=odd, even+even=even, odd+odd=odd
allowed transitions in atoms occur only between an atomic state of
even and one of odd parity
•
Parity is connected with the angular-momentum quantum number l
states with even l have even parity
states with odd l have odd parity
2-26
Topic review
• Understand role of nuclear mass in
reactions
Use mass defect to determine energetics
Binding energies, mass parabola, models
• Determine Q values
• How are nuclear shapes described and
determined
Potentials
Nucleon distribution
• Quantum mechanical terms
2-27
Study Questions
• What do binding energetic predict about
abundance and energy release?
• Determine and compare the alpha decay Q
values for 2 even and 2 odd Np isotopes.
Compare to a similar set of Pu isotopes.
• What are some descriptions of nuclear shape?
• Construct a mass parabola for A=117 and
A=50
• Describe nuclear spin, parity, and magnetic
moment
2-28
Pop Quiz
• Using the appropriate mass excess calculate the
following Q values for 212Bi. Show the reaction
b- decay
b+ decay
EC
Alpha decay
• Which decay modes are likely?
2-29