advanced nuclear physics by imran aziz
DESCRIPTION
Advanced Nuclear PhysicsMOHAMMAD IMRAN AZIZ Assistant Professor PHYSICS DEPARTMENT SHIBLI NATIONAL COLLEGE, AZAMGARH (India)[email protected] total number of protons in the nucleus of an atom is called the atomic number of the atom and is given the symbol Z. The number of electrons in an electrically-neutral atom is the same as the number of protons in the nucleus. The number of neutrons in a nucleus is known as the neutron number and is given the symbol N. The mass numbTRANSCRIPT
Advanced Nuclear Physics
MOHAMMAD IMRAN AZIZAssistant Professor
PHYSICS DEPARTMENTSHIBLI NATIONAL COLLEGE, AZAMGARH
(India).
The total number of protons in the nucleus of an atom is called the atomic number of the atom and is given the symbol Z. The number of electrons in an electrically-neutral atom is the same as the number of protons in the nucleus. The number of neutrons in a nucleus is known as the neutron number and is given the symbol N. The mass number of the nucleus is the total number of nucleons, that is, protons and neutrons in the nucleus. The mass number is given the symbol A and can be found by the equation Z + N = A..
Nuclides
Nuclear constituents and their properties
why electrons cannot exist inside a nucleus:This can be explained mathematically also using Heisenberg's uncertainty principle as follows
Confinement Calculation
Electtron is not found in the nucleus, it means we are talking about free electron. That is free electron does not exist.n ....> p + e, here this is beta deacy and electron that becomes free is
emitted out of the nucleus as free electron as can not exist due to Heisenberg uncertainty.
Nuclear Spin
The nuclear spins for individual protons and neutrons parallels the treatment of electron spin, with spin 1/2 and an associated magnetic moment. The magnetic moment is much smaller than that of the electron. For the combination neutrons and protons into nuclei, the situation is more complicated.
It is common practice to represent the total angular momentum of a nucleus by the symbol I and to call it "nuclear spin". For electrons in atoms we make a clear distinction between electron spin and electron orbital angular momentum, and then combine them to give the total angular momentum. But nuclei often act as if they are a single entity with intrinsic angular momentum I. Associated with each nuclear spin is a nuclear magnetic moment which produces magnetic interactions with its environment.
• A characteristic of the collection of protons and neutrons (which are fermions) is that a nucleus of odd mass number A will have a half-integer spin and a nucleus of even A will have integer spin. The suggestion that the angular momenta of nucleons tend to form pairs is supported by the fact that all nuclei with even Z and even N have nuclear spin I=0. The half-integer spins of the odd-A nuclides suggests that this is the nuclear spin contributed by the odd neutron
Nuclear Magnetic Moments
• Associated with each nuclear spin is a magnetic moment which is associated with the angular momentum of the nucleus. It is common practice to express these magnetic moments in terms of the nuclear spin in a manner parallel to the treatment of the magnetic moments of electron spin and electron orbital angular momentum.
• For the electron spin and orbital cases, the magnetic moments are expressed in terms of a unit called a Bohr magneton which arises naturally in the treatment of quantized angular momentum
Electric Quadrupole Moments of Nuclei
The nuclear electric quadrupole moment is a parameter which describes the effective shape of the ellipsoid of nuclear charge distribution. A non-zero quadrupole moment Q indicates that the charge distribution is not spherically symmetric. By convention, the value of Q is taken to be positive if the ellipsoid is prolate and negative if it is oblate.
The quantity Q0 is the classical form of the calculation represents the departure from spherical symmetry in the rest frame of the nucleus.
Generally, the measured quantity is proportional to the z-component of the magnetic moment (the component along the experimentally determined direction such as the direction of an applied magnetic field, etc. ). In this treatment, the use of a "gyromagnetic ratio" or "g-factor" is introduced. The g-factor for orbital is just gL = 1, but the electron spin g-factor is approximately gS = 2
For free protons and neutrons with spin I =1/2, the magnetic moments are of the form
The proton g-factor is far from the gS = 2 for the electron, and even the uncharged neutron has a sizable magnetic moment! For the neutron, this suggests that there is internal structure involving the movement of charged particles, even though the net charge of the neutron is zero. If g=2 were an expected value for the proton and g=0 were expected for the neutron, then it was noted by early researchers that the the proton g-factor is 3.6 units above its expected value and the neutron value is 3.8 units below its expected value. This approximate symmetry was used in trial models of the magnetic moment, and in retrospect is taken as an indication of the internal structure of quarks in the standard model of the proton and neutron
whereProton: g = 5.5856912 +/- 0.0000022Neutron: g = -3.8260837 +/- 0.0000018
NuclideNuclear
spin I
Magnetic moment
in N
n 1/2 -1.9130418
p 1/2 +2.7928456
2H(D) 1 +0.8574376
17O 5/2 -1.89279
57Fe 1/2 +0.09062293
57Co 7/2 +4.733
93Nb 9/2 +6.1705
Note that the maximum effective magnetic moment of a nucleus in nuclear magnetons will be the g-factor multiplied by the nuclear spin. For a proton with g = 5.5857 the quoted magnetic moment is = 2.7928 nuclear magnetons.
Nuclei
• Parameters of nuclei• Strong Interaction• Binding Energy• Stable and Unstable Nuclei• Liquid-Drop Model
Numerous Applications:• nuclear power • applications in medicine, biology and chemistry• evolution of stars and the Universe• nuclear weapons
Size of Nuclei and Rutherford Scattering
1/30R R A
150 1.2 10 1.2R m fm
R – the fitting parameter
Calculations were strictly classical. However, because of the Coulomb interaction between alpha-particles and nucleus, the result miraculously coincides with the exact quantum-mechanical one (recall the success of the Bohr model for atoms).
1/323892 1.2 238 7.4R U fm fm
Geiger, Marsden, Rutherford,1910
- depends weakly on A (number of nucleons in
the nucleus)
-particles: bare He nuclei
Scattering pattern was consistent with that expected for scattering of particles by pointlike objects having a charge of +79e (the charge of the gold nucleus). This allowed Rutherford to put an upper limit on the size of the nucleus (<310-14m for gold).
To measure the size of a nucleus, one has to use more energetic particles (or electrons, which are more commonly used these days) that get close enough to get inside the nucleus.
Nuclear Mass
AZ X
6 protons + 6 neutrons
126 C
Atomic Mass (the mass of a neutral atom):2 2
. ./ /p n nucleus e el nuclM Z m N m U c Z m U c
. ., 0nucleus el nuclU U - attraction between nucleons in the nucleus and between electrons and the nucleus
Mass Unit126 27 21 1.66054 10 931.49 /
12C
Mu kg MeV c
chemical symbol for the element
number of protons in the nucleus(atomic number of the element)
number of nucleons in the nucleus(mass number of the nucleus)
The neutron number: N A Z
2. .el nucl nucleus pU U m c - we can neglect Uel.-nucl. and introduce a
convenient mass unit:
proton mass 1.007277pm u
neutron mass 1.008665nm u
electron mass 0.000549em u [email protected]
Nuclear Density
3 3 30 0
2717 3
315
1 13 3 34 4 4
1.66 102.4 10 /
31.2 10
4
A auM au
R R A R
kgkg m
m
The density of neutron stars is comparable with that of nuclei.
(Unstable) isotopes of tin and zinc.
!
The Need for a “Strong Force”Which interaction controls the size of nucleons? This cannot be electromagnetic interaction: protons have the same electric charge (they would repel each other) and also there are attractive forces between protons and electrically neutral neutrons.
Strong Interaction:
binds protons to protons, neutrons to neutrons, and protons to neutrons with roughly the same force
does not affect certain other kinds of particles (specifically electrons) is short-ranged (the range ~ 2 fm). Nucleons separated by a larger distance exert no strong
forces on each other.
These observations are explained by the quark model of nucleons. Nucleons are the combination of quarks that are strong-interaction-neutral (like an electrically-neutral atom). Two nucleons interact only if they are close enough that the distances between various pairs of quarks are significantly different.
g
u d
u
quarks and
gluons
~ 10-15 m
Structure of Matter
Atom is almost an empty space (the
nuclear volume is ~10-15 of the atomic volume)
g
u d
u
quarks and
gluons
~ 10-10 - 10-9 m
~ 10-15 - 10-14 m
Protons & Neutrons (nucleons) are almost an empty space (the
quark size is <10-18 m)
El.-mag. interaction determines the size of
atoms
~ 10-15 m
Strong interaction determines the size of
nuclei and nucleons
Binding Energy
Binding energy:
11
2
2
mass of all particles consituting the atom mass of the atom
AZ
B
nH X
E c
Z m N m M c
2BE M c
mass deficit
sys B parts B parts sysE E E E E E -the binding energy is positive for a bound system
Recall a H atom: the binding energy is 13.6 eV (the ground state energy with sign “minus”).
We can compute the binding energy if we know masses of a system and its constituents:
add EB
p
np
p
n
nn
sysE2
parts iE m c
Binding Energy curveBecause of the short-range character of strong interaction (basically, between nearest and next-to-nearest neighbors), the interaction energy per nucleon with increasing Z saturates at the level ~ (Z/2)(# of neighbors).
The binding energy ~ 10MeV/nucleon is ~1% of the nucleon’s rest energy: we can consider the nucleus as a system of individual nucleons
The decrease of the binding energy with increasing Z is caused by the long-range Coulomb
repulsion of protons:
2B Z
Bind
ing
ener
gy p
er n
ucle
on (E
B/A)
, MeV
Mass number, A
Liquid-Drop Model A “semi-classical” model of the nucleus: describes reasonably well the dependence EB(A):
222/3
1/3
2B V s c a
A ZZE a A a A a a
A A
0
3 2
000
44
34
R r r drU R
r
q r dq r
q r
dq r - charge density
Stable Nuclei
Isotopes: all nuclei that have the same number of protons (Z) but different number of neutrons (N). Since the chemical properties of an atom are determined by the number of its electrons, isotopes of the same element have almost identical chemical properties.
Example: naturally occurring isotopes of oxygen168 0 17
8 0 198 0
Related questions: - What makes unstable nuclei unstable?- What are the mechanisms by which they transform themselves into stable nuclei?-Why do light stable nuclei tend to have N Z?- Why do heavier nuclei tend to have more neutrons than protons?- Why are there no stable nuclei with Z>83?
What makes unstable nuclei unstable?-
If a nucleus is allowed to decrease its energy by transforming “excessive” protons (neutrons) into neutrons (protons), it will do it!
The potential experienced by nucleons is a 3D potential well. The ground-state configuration of the carbon-16 nucleus :
126 C
protons neutrons
energyBoth protons and neutrons are fermions (they obey the exclusion principle). Nuclei are two-component Fermi systems. Each nuclear energy level can contain four particles: two protons (s=½) and two neutrons (s=½).
The processes responsible for these transformations are driven by weak interaction (the fourth fundamental interaction):
The weak interaction (unlike the strong interaction) affects both quarks and leptons, (unlike the el.-mag. interaction) can affect electrically neutral particles, and (unlike gravity) does not affect photons.
The effective range of the weak interaction is ~ 10-18m.
Some important transformation processes driven by weak interaction:
n e p p n e p e n
0r
Why N Z for light nuclei
If the electrostatic repulsion of protons can be neglected (this is the case of light nuclei: recall that the positive electrostatic energy Z2), the nucleus tends to keep approximately equal numbers of protons and neutrons.
protons neutrons
energy138 O
1/2 8.9t ms
protons neutrons
energy137 N
1/2 5730t y
protons neutrons
energy147 N
protons neutrons
energy146 C
Even in this case, the nucleus can still lower its total energy: the rest energy of neutron is slightly more than the rest energy of a proton and an electron.
Why N > Z for heavy nuclei
In the heavy nuclei, the electrostatic energy cannot be neglected. As a result, the protons’ energy levels are “pushed up” with respect to the neutrons’ levels. In the “otherwise stable” 44Ti, two protons undergo the transformation into neutrons, the end product is stable 44Ca.
protons neutrons
energy4422Ti
.
.
...
protons neutrons
energy4420Ca
.
.
...
The proton-neutron disbalance becomes more pronounced with increasing Z.
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
Nuclear Mass Measurements
• Measure relative masses by energy released in decays or reactions.– X Y +Z + E – Mass difference between X and Y+Z is E/c2.
• Absolute mass by mass spectrometers (next transparency).
• Mass and Binding energy:• B = [Z MH + N Mn – M(A,Z)]/c2
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
Binding Energy vs A• B increases with A up to 56Fe and then slowly
decreases. Why?• Lower values and not smooth at small A.
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 1st order perturbation theory
32
0
)(4 R
r
r
ZeE
R
Ze
R
ZerrV
03
0
2
8
3
8)(
)Rr(r4
Ze
R8
Ze3
R8
Zer)r(V
003
0
2
drrrVerrER
)()]()[(4 *
0
2 2/3
00
2/3
0)(2)/exp()(2)(
a
ZaZr
a
Zr
Isotope Shifts
2 23
00
2( / )
5
Ze RE Z a
dr]r4
Ze
R8
Ze3
R8
Zer)[e()a/Z(4r4E
003
0
23
R
0
2
5
R4drrr4
522
R
0
3
R4drr4
32
R
0
22R
0R2dr
r
1r4
]22
3
3
4
10
4[R
4
Ze)a/Z)(e4(E 2
0
3
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~ 207 me and a0~1/m.
• All data consistent with R=R0 A1/3 with R0=1.25fm.
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=-n + 4R2T B=n-n2/3
• 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!
Mirror Nuclei• Compare binding energies of mirror nuclei (nuclei n p). Eg 7
3Li and 74Be.
• Mass difference due to n/p mass and Coulomb energy.
dQr
rQE
R
0 04
)(
323 /3)/()( RZerdQRrZerQ
R
Zedr
R
r
r
ZeE
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
eZZEc
3/2)1,( AZZEC
surface area ~ n2/3
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.– n=number of molecules, T=surface tension, B=binding energy
E=total energy of the drop, ,=free constantsE=-n + 4R2T B=n-n2/3
Analogy with nucleus Nucleus has constant density From nucleon-nucleon scattering experiments we know:
Nuclear force has short range repulsion and is attractive at intermediate distances.
Assume charge independence of nuclear force, neutrons and protons have same strong interactions check with experiment (Mirror Nuclei!)
Coulomb Term• The nucleus is electrically charged with total charge Ze• Assume that the charge distribution is spherical and compute the
reduction in binding energy due to the Coulomb interaction
00
( )
4
Ze
Coulomb
Q rE dQ
r 3 2 3( ) ( / ) 3 /Q r Ze r R dQ Zer R dr
2 5 2
60 00
3( ) ( )(3 / 5)
4 4
R
Coulomb
Ze r ZeE dr
r R R
to change the integral to dr ; R=outer radius of nucleus
includes self interaction of last proton with itself. To correct this replace Z2 with Z*(Z-1)
1/3
*( 1)( , )Coulomb
Z ZB Z A d
A
in principle you could take d from this calculation but it is more accurate to take it from the overall fit of the SEMF to data (nuclei not totally spherical or homogeneous)
… and remember R=R0A-1/3
Mirror Nuclei• Does the assumption of the drop model of constant binding energy for every
constituent of the drop acatually hold for nuclei?• Compare binding energies of mirror nuclei (nuclei with np). Eg 7
3Li and 74Be.
• If the assumption holds the mass difference should be due to n/p mass difference and Coulomb energy alone.
• Let’s compute the Coulomb energy correction from results on previous page
2 2
0 0
3 3( , 1) [ ( 1) ( 1)( 2)] 2( 1)
5 4 5 4coulomb
e eE Z Z Z Z Z Z Z
R R
2/3( , 1)CE Z Z A 1/30~ / 2 ;Z A R R A to find that
Now lets measure mirror nuclei masse, assume that the model holds and derive ECoulomb from the measurement.
This should show an A2/3 dependence And the scaling factor should yield the correct R0 of 1.2 fm if the assumptions were [email protected]
More charge symmetry• Energy Levels of two mirror nuclei for a number of excited states• Corrected for n/p mass difference and Coulomb Energy
Ecorrected
From Charge Symmetry to 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.
• If we find the same energy levels for the same spin states Strong interaction is the same for np as nn and pp.
MgNaNe 2212
2211
2210
Charge Independence
2212Mg
2211Na
2210Ne
Ecorrected• Same spin/parity states should have the same energy.
• Yes: np=nn=pp• Note: Far more states in
2211Na. Why?
• Because it has more np pairs then the others
• np pairs can be in any Spin-Space configuration
• pp or nn pairs are excluded from the totally symmetric ones by Herr Pauli
• Note also that 2211Na has the
lowest (most bound) state, remember for the deuteron on next page
Charge Independence
• We have shown by measurement that:– If we correct for n/p mass difference and Coulomb interaction, then
energy levels in nuclei are unchanged under n p – and we must change nothing else! I.e. spin and space wavefunctions
must remain the same!• Conclusion: strong two-body interaction same for pp, pn and
nn if nucleons are in the same quantum state.• Beware of the Pauli exclusion principle! eg why do we have
bound state of pn but not pp or nn?– because the strong force is spin dependent and the most strongly
bound spin-space configurations (deuteron) are not available to nn or pp. It’s Herr Pauli again!
– Just like 2211Na on the previous triplet level schema
Volume and Surface Term• We now have all we need to trust that we can apply the
liquid drop model to a nucleus– constant density– same binding energy for all constituents
• Volume term:• Surface term:
• Since we are building a phenomenological model in which the coefficients a and b will be determined by a fit to measured nuclear binding energies we must inlcude any further terms we may find with the same A dependence together with the above
( )VolumeB A aA2/3( )SurfaceB A bA
Asymmetry Term• Neutrons and protons are spin ½ fermions obey
Pauli exclusion principle.• If all other factors were equal nuclear ground state
would have equal numbers of n & p.neutrons protons Illustration
n and p states with same spacing . Crosses represent initially occupied states in
ground state. If three protons were turned into neutrons the extra energy required would be 3×3 . In general if there are Z-N excess protons over
neutrons the extra energy is ((Z-N)/2)2 . relative to Z=N.
But how big is ?
Asymmetry Term
• Assume:– p and n form two independent, non-interacting
gases occupying their own square Fermi wells– kT << – so we can neglect kT and assume T=0– This ought to be obvious as nuclei don’t suddenly
change state on a warm summers day!– Nucleons move non-relativistically (check later if
this makes sense)
Asymmetry Term• From stat. mech. density of states in 6d phase space = 1/h3
• Integrate up to pf to get total number of protons Z (or Neutrons N), & Fermi Energy (all states filled up to this energy level).
• Change variables p E to find avg. E
2
3
4particle
p dpVdN
h
3 3 30
4(4 / 3 ) and
3FZ V h p V R A
3/ 2
1/ 2 0
1/ 2
0
// (3 / 5)
/
F
F
E
FE
E dEdN dp
dN dE const E E Edp dE
E dE
1/3 22/3
0
(3 / 4 ) and 2F
h Z pP E mR A
2/324/3
20
(3 / 4 )2F
h ZE
mR A
here Nparticle could be the number of protons or neutrons
These are all standard stat. mech. [email protected]
comes from a fit of the SEMF to measurementsanalytical ≈ 24 MeV
This terms is only proportional to volume (A). It has already been captured by the Volume term of the liquid drop model
call this K
Asymmetry Term
• Binomial expansion keep lowest term in y/A1 3 2
2 3 2 5 ( )2
9Total
N ZE KA K
A
2/324/3
20
3(3/ 4 )
5 2P PTotal
h ZE Z E Z
mR A
5/3 5/32 /3Total
KE Z N
A
5/35/3 5/3
2 /3 5 3(1 / ) (1 / )
2Total
KAE y A y A
A
Compute total energy of all protons by Z*<E>
Use the above to compute total energy of Z protons and N neutrons
change variables from (Z,N,A) to (y,A) with y=N-Z
where y/A is a small number ()
note! linear terms cancel
5/3 5/35 3
(1 ) (1 )2Total
KAE
2( )( ) *Total
N ZE Fermi Gas const
A
Asymmetry term
• From the Fermi Gas model we learn that– due to the fermionic nature of p and n we loose in
binding energy if the nucleus deviates from N=Z
• The Asymmetry term: 2( )
( , )Asymmetry
N ZB N Z c
A
Note: this only holds for nn and pp, not fornp. We don’t have a preference for even A
Pairing Term• Observations:• Nuclei with even number of n or
even number of p more tightly bound then with odd numbers. See figure
• Only 4 stable o-o nuclei but 153 stable e-e nuclei.
• p energy levels are Coulomb shifted wrt n small overlap of wave functions between n and p.
• Two p or two n in same energy level with opposite values of jz have AS spin state forced into sym spatial w.f. maximum overlap maximum binding energy
because of short range attraction. Neutron number
Neutron separation energy [MeV] in Ba isotopes
56+N56Ba
Pairing Term• Measure that the Pairing effect smaller for larger A • Phenomenological*) fit to A dependence gives A-1/2
e-e +ive
e-o 0
o-o -ive
1/ 2( )PairingB A
A
*) For an even more insightful explanation of the A dependence read the book by Jelley
Note: If you want to plot binding energies versus A it is often best to use odd A only as for these the pairing term does not appear
Semi Empirical Mass Formula
• Put everything together:2 2
2/31/3 1/ 2
( )( , )
N Z ZB N Z aA bA c d
A A A
Volume Term
Surface Term
AsymmetryTerm
CoulombTerm
PairingTerm
Lets see how all of these assumptions fit reality And find out what the constants are
Note: we went back to the simpler Z2 instead of Z*(Z-1)
Semi Empirical Mass Formula Binding Energy vs. A for beta-stable odd-A nuclei
Iron
Not smooth because Z not smooth function of A
Fit parameters in
MeV
a 15.56
b 17.23
c 23.285
d 0.697
+12 (o-o)
0 (o-e)
-12 (e-e)
Semi Empirical Mass Formula
• Conclusions– Only makes sense for A≥20– Good fit for large A (good to <1%) in most places.– Deviations are interesting shell effects.– Coulomb term constant agrees with calculation.– Explains the valley of stability (see next lecture).– Explains energetics of radioactive decays, fission and fusion.
Nuclear Shell Model• Potential between nucleons can be studied by studying
bound states (pn, ppn, pnn, ppnn) or by scattering cross sections: np -> np pp -> pp nD -> nD pD -> pD
• If had potential could solve Schrod. Eq. Don’t know precise form but can make general approximation
• 3d Finite Well with little r-dependence (except at edge of well)
• Almost spherically symmetric (fusion can be modeled as deformations but we’ll skip)
• N-N interactions are limited (at high A) due to Pauli exclusion. p + n -> p’ + n’ only if state is available
Infinite Radial Well • Radial part of Scrod Eq
• Easy to solve if l=0
• For L>0, angular momentum term goes to infinity at r=0. Reduces effective wavelength, giving higher energy
• Go to finite well. Wave function extends a bit outside well giving longer effective wavelength and lower energy (ala 1D square wells)
• In nuceli, potential goes to infinity at r=0 (even with L=0) as that would be equivalent to nucleon “inside” other nucleon
2 2
2
2
2
2
2 2
1
4
m
d u
drV r
m
l l
ru E u
u r rR r P r u
( )( )
( ) ( ) ( )
u kr kn
aE
p
m
k
m
hn
m a sin
( ) ( )2
2 2 2 8
2 2 2
2
Angular part• If V(r) then can separate variables (r, = R(r)Y( have spherical
harmonics for angular wave function • Angular momentum then quantized like in Hydrogen (except that L>0 for
n=1, etc)
• Energy doesn’t depend on m• Energy increases with increasing n (same l)• Energy increases with increasing l (same n)• If both n,l vary then use experimental observation to determine lower
energy• Energy will also depend on strong magnetic coupling between nucleons• Fill up states separately for p,n
L l l L m
l m l l n r quan tumZ
2 21
0 1 2
( )
, , #
L,S,J Coupling: Atoms vs Nuclei• ATOMS: If 2 or more electrons, Hund’s rules:• Maximise total S for lowest E (S=1 if two)• Maximise total L for lowest E (L=2 if 2 P)• Energy split by total J (J=3,2,1 for S=1,L=2)
• NUCLEI: large self-coupling. Plus if 2 p (or 2 n) then will anti-align giving a state with J=0, S=0, L=0
leftover “odd” p (or n) will have two possible J = L + ½ or J = L – ½ higher J has lower energy if there are both an odd P and an odd n (which is very rare in stable) then
add up Jn + Jp• Atom called LS coupling nuclei called jj • Note that magnetic moments add differently as different g-factor for p,n
Spin Coupling in Nuclei• All nucleons in valence shell have same J• Strong pairing causes Jz antiparallel (3 and -3) spin
wavefunction = antisymmetric space wavefunction = symmetric
• This causes the N-N to be closer together and increases the attractive force between them
• e-e in atoms opposite as repulsive force• Can also see in scattering of polarized particles• Even N, even Z nuclei. Total J=S=L=0 as all n,p paired off• Even N, odd Z or odd N, even Z. nuclear spin and parity determined
by unpaired nucleon• Odd N, odd Z. add together unpaired n,p• Explains ad hoc pairing term in mass formula
Energy Levels in Nuclei• Levels in ascending order (both p,n) State n L degeneracy(2j+1) sum 1S1/2 1 0 2 2*** 1P3/2 1 1 4 6 1P1/2 1 1 2 8*** 1D5/2 1 2 6 14 2S1/2 2 0 2 16 1D3/2 1 2 4 20*** 1F7/2 1 3 8 28*** 2P3/2 2 1 4 32 1F5/2 1 3 6 38 2P1/2 2 1 2 40 1G9/2 1 4 10 50***
*** “magic” number is where there is a large energy gap between a filled shell and the next level. More tightly bound nuclei. (all filled subshells are slightly “magic”)
Magic Numbers• Large energy gaps between some filled shells and next (unfilled) shell give
larger dE/A and more made during nucleosnthesis in stars # protons #neutrons 2 He 2 He-4 6 C 6 C-12 8 O 8 O-16 20 Ca 20 28 Ni 28 Cr-52(24,28) 50 Sn 50 Ni-78 82 Pb 82 126 136 • Ni-78 (2005) doubly magic. While it is unstable, it is the much neutron rich. • Usually more isotopes if p or n are magic. Sn has 20 isotopes, 10 of which
are stable
Nuclear Magnetic Moments• Protons and neutrons are made from quarks and gluons. Their magnetic
moment is due to their spin and orbital angular momentum
• The g-factors are different than electrons. orbital, p=1 and n=0 as the neutron doesn’t have charge
• spin, g for proton is 5.6 and for neutron is -3.8 (compared to -2 for the electron; sometimes just 2).
• A proton is made from 2 up and 1 down quark which have charge 2/3 and -1/3
• A neutron is made from 1 up and 2 down and has “more” negative charge/moments
• No theory which explains hadronic magnetic moments • orbital and spin magnetic moments aren’t aligned, need to repeat the
exercise in atoms (Zeeman effect) to get values for the z-component of the moment
L SN
l S Np
g L g Se
m( )
2
Nuclear Cross Sections
• Definition of Cross Section– Why its useful.
• Breit-Wigner Resonances• Rutherford Scattering
Cross-Sections
• Why concept is important– Learn about dynamics of interaction and/or constituents
(cf Feynman’s watches).– Needed for practical calculations.
• Experimental Definition• How to calculate
– Fermi Golden Rule– Breit-Wigner Resonances– QM calculation of Rutherford Scattering
Definition of • a+bx• Effective area for reaction to occur is
Beam a
dx
Na
Na(0) particles type a/unit time hit target b
Nb atoms b/unit volume
Number /unit area= Nb dx
Probability interaction = Nbdx
dNa=-Na Nb dx
Na(x)=Na(0) exp(-x/) ; =1/(Nb )
Reaction Rates• Na beam particles/unit volume, speed v
• Flux F= Na v
• Rate/target b atom R=F• Thin target x<<: R=(Nb
T) F Total
• This is total cross section. Can also define differential cross sections, as a function of reaction product, energy, transverse momentum, angle etc.
• dR(a+bc+d)/dE=(NbT) F d(a+bc+d) /dE
Breit-Wigner Line Shape
• Start with NR Schrödinger equation:
nn0nn )/tiEexp()t(a)t(;H
dti
nm
mmnnnnnnn )/tiEexp(Ha)/tiEexp(Ea)/tiEexp(ai
X by n and integrate rdHHrd nmmnnmnm3*3* ;
m
mmnmnnnnn tiEHatiEEatiEai )/exp()/exp()/exp(
Start in state m exponential decay )2/texp()t(am )/texp()t(a 2
Breit-Wigner Line Shape - 2
}/t]2/)EE(iexp{[Hai mnmnn
/t]2/)EE(iexp[Hdt)t(ai mnmn
t
0n
t
0mn
mnmnn 2/)EE(i
/t]2/)EE(iexp[H)t(a
For
2/)EE(i
H)t(a
mn
mnn
/t
Breit-Wigner Line Shape -3
)EE(PH2
)t(a
4/)EE(
H)t(a
nm2
mn2
n
22nm
2mn2
n
4)EE(
1
2)EE(P
22nm
nm
Normalised Breit-Wigner line shape
Q: where have you seen this shape before?
We will see this many times in NP and PP.
Breit-Wigner Resonance
• Important in atomic, nuclear and particle physics.
• Uncertainty relationship
• Determine lifetimes of states from width.
• =FWHM;
~tE
~tE
~tE
Fermi Golden Rule
• Want to be able to calculate reaction rates in terms of matrix elements of H.
• Warning: We will use this many times to calculate but derivation not required for exams, given here for completeness.
Discrete Continuum • Decays to a channel i (range of states n). Density of
states ni(E). Assume narrow resonance
dE)EE(P)E(nH2
P 0i2
0ii
)E(nH2
P 0i2
0ii
TotaliiTotali
i RPR;R;P
)E(nH2
R 0i2
0ii
i
Cross Section• Breit Wigner cross section.• Definition of and flux F:
)r.kiexp(V 2/1 FR
vVF 12
3k4
)2(
V)k(n
vdk
dE;
m2
)k(E
2
v
k4
)2(
V)E(n
2
3
Breit-Wigner Cross Section
• Combine rate, flux & density states
4/)EE(
E
2
1
k4V
v)2(
v
V22
01
f1i2
3
f
2201
201f2
o4/)EE(
H)t(aR
)E(nH2)E( 210i
4/)EE(
E
)E(n
1
2
1R
2201
f1i
Low Energy Resonances
• n + Cd total cross section.
• Cross section scales ~ 1/E1/2 at low E.
• B-W: 1/k2 and ~n(E)~k
Rutherford Scattering 1
1c;c4
e;
r
ZZ)r(V
0
221
)r.kiexp(V;)r.kiexp(V f2/1
fi2/1
i
rd)r.kiexp(r
ZZ)r.kiexp(VH 3
f21
i1
fi
fi kkq
rdr
)r.qiexp(ZZVH 321
1fi
cosddrrr
)cosiqrexp(2ZZVH 2
211
fi
Rutherford Scattering 2
drriqr
)iqrexp()iqrexp(2ZZVH 2
2211
fi
a)a/rexp();r(xV
drriqariqaiq
ZZVH fi )/1exp()/1exp(
2211
iqa/1
1
iqa/1
1
iq
2ZZVH 211
fi
Rutherford Scattering 3• Fermi Golden Rule:
f
2fi dE
dnH
2R
v/1dE
dp;
dE
dp
dp
dn
dE
dn;
4
d
h
Vp4
dp
dn3
2
)22(;)2(
)( 3
2
hdv
VpEn
vVF;F/R 1
v
V
)2(v
Vp
Vq
4ZZ2
d
d3
22
221
42
221
2 )(4
qv
ZZp
d
d
Rutherford Scattering 4
)2/(sinp4)cos1(p2)pp(q 2222fi
2
)2/(sin4
)(422
221
vp
ZZ
d
d
Compare with experimental data at low energy
Q: what changes at high energy ?
pipf
Low Energy Experiment• Scattering of on Au & Ag agree with calculation
assuming point nucleus
Sin4(/2)
dN/d
cos
Higher Energy
• Deviation from Rutherford scattering at higher energy determine charge distribution in the nucleus.
• Form factors is F.T. of charge distribution.
Electron - Gold
Induced Fission(required energy)
Neutrons
Ef=Energy needed to penetrate fission barrier immediately ≈6-8MeV
A=238
Neu
tron
Nucleus Potential Energy during fission [MeV]
Esep≈6MeV per nucleon for heavy nuclei
Very slow n
Induced Fission(required energy)
• Spontaneous fission rates low due to high coulomb barrier (6-8 MeV @ A≈240)
• Slow neutron releases Esep as excitation into nucleus• Excited nucleus has enough energy for immediate fission if
Ef - Esep >0• We call this “thermal fission” (slow, thermal neutron
needed)• But due to pairing term …• even N nuclei have low Esep for additional n
• odd N nuclei have high Esep for additional n
• Fission yield in n -absorption varies dramatically between odd and even N
Induced Fission(fissile nuclei)
• Esep(n,23892U) = 4.78 MeV only
• Fission of 238U needs additional kinetic energy from neutron En,kin>Ef-Esep≈1.4 MeV
• We call this “fast fission” (fast neutrons needed)• Thermally fissile nuclei, En,kin
thermal=0.1eV @ 1160K– 233
92U, 23592U, 239
94Pu, 24194Pu
• Fast fissile nuclei En,kin=O(MeV)– 232
90Th, 23892U, 240
94Pu, 24294Pu
• Note: all Pu isotopes on earth are man made• Note: only 0.72% of natural U is 235U
Induced Fission (Reminder: stages of the process up to a few seconds after fission event)
t=0
t≈10-14 s
t>10-10 s
<# prompt n>prompt=2.5
<n-delay>d=few s
<# delayed n>d=0.006
Induced Fission (the fission process)
Energy balance of 23592U induced thermal fission MeV:
– Prompt (t<10-10s):• Ekin( fragments) 167• Ekin(prompt n) 5 3-12 from X+nY+• E(prompt ) 6• Subtotal: 178 (good for power production)
– Delayed (10-10<t<):• Ekin(e from -decays) 8• E( following -decay) 7• Subtotal: 15 (bad, spent fuel heats up)
– Neutrinos: 12 (invisible)
– Grand total: 205
Induced Fission(n -induced fission crossections (n,f) )
• 23892U does nearly no n -induced fission below En,kin≈1.4 MeV
• 23592U does O(85%) fission starting at very low En,kin
• Consistent with SEMF-pairing term of 12MeV/√A≈0.8 MeV between– odd-even= 235
92U and even-even= 23892U
unresolved, narrowresonances
unresolved, narrowresonances
238U 235U
Induced Fission((n,f) and (n,) probabilities in natural Uranium)
23592U(n,f)
23592U(n,)
23892U(n,) 238
92U(n,f)235
92U(n,f)
23592U(n,)
23892U(n,)
23892U(n,)
ener
gy ra
nge
of fi
ssio
n ne
utro
ns
fastthermal
neut
ron
abso
rbtio
n pr
obab
ilit p
er 1
m
“bad-238”
“good 238 ”
“bad-235”
“good 235 ”
Induced Fission(a simple bomb)
– mean free path for fission n:
235 238(1 )tot tot totc c 1 ( ) 3 cmnucl tot
Simplify to c=1 (the bomb mixture) prob(235U(nprompt ,f)) @ 2MeV ≈ 18% (see slide 8) rest of n scatter, loosing Ekin prob(235U(n,f)) grows most probable #collisions before 235U(n,f) = 6 (work it out!) 6 random steps of =3cm lmp=√6*3cm≈7cm in tmp=10-8 s
Uranium mix 235U:238U =c:(1-c) nucl(U)=4.8*1028 nuclei m-3
average n crossection:
mean time between collisions =1.5*10-9 s @ Ekin(n)=2MeV
Induced Fission(a simple bomb)
• After 10-8 s 1n is replaced with =2.5 n, =average prompt neutron yield of this fission process
• Let probability of new n inducing fission before it is lost = q • (others escape or give radiative capture)• Each n produces on average (q-1) new such n in tp=10-8 s (ignoring
delayed n as bombs don’t last for seconds!)
0
( 1)
( ) ( ) ( 1) ( ) ( )
( ) 1lim ( )
solved by: ( ) (0) mp
mp
tmp
tq t
n t t n t q n t t t
dn t qn t
dt t
n t n e if q>1 exponential growths of neutron number For 235U, =2.5 if q>0.4 you get a bomb
Induced Fission(a simple bomb)
• If object dimensions << lmp=7 cm most n escape through surface q << 1
• If Rsphere(235U)≥8.7cm M(235U)≥52 kg q = 1 explosion in < tp=10-8 s little time for sphere to blow apart significant fraction of 235U will do fission
Fission Reactors(not so simple)
• Q: What happens to a 2 MeV fission neutron in a block of natural Uranium (c=0.72%)?
• A: In order of probability – Inelastic 238U scatter (slide 8)– Fission of 238U (5%)– rest is negligible
• as Eneutron decreases via inelastic scattering – (238
92U(n,)) increases and becomes resonant– (238
92U(n,f)) decreases rapidly and vanishes below 1.4 MeV– only remaining chance for fission is (235
92U(n,f)) which is much smaller then (238
92U(n,)) • Conclusion: piling up natural U won’t make a reactor because n get
“eaten” by (n,) resonances. I said it is not SO simple
23592U(n,f)
23592U(n,)
23892U(n,) 238
92U(n,f)235
92U(n,f)
23592U(n,)
23892U(n,)
23892U(n,)
Fission Reactors(two ways out)
• Way 1: Thermal Reactors– bring neutrons to thermal energies without
absorbing them = moderate them– use low mass nuclei with low n-capture
crossection as moderator. (Why low mass?)– sandwich fuel rods with moderator and coolant
layers– when n returns from moderator its energy is so
low that it will predominantly cause fission in 235U
Fission Reactors(two ways out)
• Way 2: Fast Reactors– Use fast neutrons for fission– Use higher fraction of fissile material, typically
20% of 239Pu + 80% 238U– This is self refuelling (fast breeding) via:
• 23892U+n 239
92U +
• 23993Np + e- + e
• 23994Pu + e¯ + e
– Details about fast reactors later
Fission Reactors (Pu fuel)
• 239Pu fission crossection slightly “better” then 235U• Chemically separable from 238U (no centrifuges)• More prompt neutrons (239Pu)=2.96• Fewer delayed n & higher n-absorbtion, more later
Fission Reactors (Reactor control)
• For bomb we found:– “boom” if: q > 1 where was number of prompt n– we don’t want “boom” need to get rid of most prompt
n• Reactors use control rods with large n-capture
crossection nc like B or Cd to regulate q• Lifetime of prompt n:
– O(10-8 s) in pure 235U– O(10-3 s) in thermal reactor (“long” time in moderator)
• not “long” enough Far too fast to control• … but there are also delayed neutrons
Fission Reactors (Reactor control)
• Fission products all n -rich all - active• Some - decays have excited states as daughters• These can directly emit n (see table of nuclides, green at bottom of curve)
Group Half-Life
(sec)
Delayed Neutron Fraction
Average Energy (MeV)
1 55.7 0.00021 0.252 22.7 0.00142 0.463 6.2 0.00127 0.414 2.3 0.0026 0.455 0.61 0.00075 0.416 0.23 0.00027 -
Total - 0.0065 -
Delayed Neutron Precursor Groups for Thermal Fission in 235-U
several sources of delayed n typical lifetimes ≈O(1 sec) Fraction d ≈ 0.6%
Ener
gy
off s
ylla
bus
Fission Reactors (Reactor control)
• Since fuel rods “hopefully” remain in reactor longer then 10-2 s must include delayed n fraction d into our calculations
• New control problem:– keep (+d)q = 1
– to accuracy of < 0.6%– at time scale of a few seconds
• Doable with mechanical systems but not easy
Fission Reactors (Reactor cooling)
• As q rises during control, power produced in reactor rises – we cool reactor and drive “heat engine” with coolant – coolant will often also act as moderator
• Coolant/Moderator choices:Material
State n-abs reduce En
chemistry
other coolant
H2O liquid
small
best reactive cheap good
D2O liquid
none
2nd best reactive rare good
C solid mild medium reactive cheap medium
CO2press. gas mild medium passive cheap ok
He gas mild 3rd best very passi.
leaks ok
Na liquid
small
medium very react.
difficult excellent
off s
ylla
bus
Fission Reactors (Thermal Stability)
• Want dq/dT < 0– Many mechanical influences via thermal
expansion– Change in n-energy spectrum– Doppler broadening of 238U(n,) resonances
large negative contribution to dq/dT due to increased n -absorbtion in broadened spectrum
– Doppler broadening of 239Pu(n,f) in fast reactors gives positive contribution to dq/dt
– Chernobyl No 4. had dq/dT >0 at low power– … which proved that you really want dq/dT < 0
Fission Bombs (fission fuel properties)
• ideal bomb fuel = pure 239Pu
Isotope Half-lifea Bare critical mass
Spontaneousfission neutrons
Decay heat
yearskg, Alpha-phase
(gm-sec)-1 watts kg-1
Pu-238 87.7 10 2.6x103 560
Pu-239 24,100 10 22x10-3 1.9
Pu-240 6,560 40 0.91x103 6.8
Pu-241 14.4 10 49x10-3 4.2
Pu-242 376,000 100 1.7x103 0.1
Am-241 430 100 1.2 114
a. By Alpha-decay, except Pu-241, which is by Beta-decay to Am-241.
Fission Bombs (where to get Pu from? Sainsbury’s?)
Grade Isotope
Pu-238 Pu-239 Pu-240 Pu-241a Pu-242
Super-grade - .98 .02 - -
Weapons-gradeb .00012 .938 .058 .0035 .00022
Reactor-gradec .013 .603 .243 .091 .050
MOX-graded .019 .404 .321 .178 .078
FBR blankete - .96 .04 - -
c. Plutonium recovered from low-enriched uranium pressurized-water reactor fuel that has released 33 megawatt-days/kg fission energy and has been stored for ten years prior to reprocessing (Plutonium Fuel: An Assessment (Paris:OECD/NEA, 1989) Table 12A).
a. Pu-241 plus Am-241.
d. Plutonium recovered from 3.64% fissile plutonium MOX fuel produced from reactor-grade plutonium and which has released 33 MWd/kg fission energy and has been stored for ten years prior to reprocessing (Plutonium Fuel: An Assessment(Paris:OECD/NEA, 1989) Table 12A).
Fission Bombs (drawbacks of various Pu isotopes)
• 241Pu : decays to 241Am which gives very high energy -rays shielding problem
• 240Pu : lots of n from spontaneous fission • 238Pu : -decays quickly (= 88 years) lots of heat conventional ignition explosives don’t like that!
• in pure 239Pu bomb, the nuclear ignition is timed optimally during compression using a burst of external n maximum explosion yield
• … but using reactor grade Pu, n from 240Pu decays can ignite bomb prematurely lower explosion yield but still very bad if you are holding it in your hand
• Reactor grade Pu mix has “drawbacks” but can “readily” be made into a bomb.
Plutonium isotope composition as a function of fuel exposure in a pressurized-water reactor, upon discharge.
Fission Bombs (suspicious behaviour)
• Early removal of fission fuel rods need control of reactor fuel changing cycle!
• Building fast breaders if you have no fuel recycling plants
• Large high-E sources from 241Am outside a reactor
• large n fluxes from 240Pu outside reactors very penetrating easy to spot over long range
Fission Reactors (Thermal vs. Fast)
• Fast reactors– need very high 239Pu concentration Bombs– very compact core hard to cool need high Cp coolant
like liq.Na or liq. NaK-mix don’t like water & air & must keep coolant circuit molten & high activation of Na
– High coolant temperature (550C) good thermal efficiency– Low pressure in vessel better safety– can utilise all 238U via breeding 141 times more fuel– High fuel concentration + breading Can operate for long
time without rod changes– Designs for 4th generation molten Pb or gas cooled fast reactors
exist. Could overcome the Na problems
Fission Reactors (Thermal vs. Fast)
• Thermal Reactors– Many different types exist
• BWR = Boiling Water Reactor• PWR = Pressure Water Reactor • BWP/PWR exist as
– LWR = Light Water Reactors (H2O)– HWR = Heavy Water Reactors (D2O)
• (HT)GCR = (High Temperature) Gas Cooled Reactor exist as
– PBR = Pebble Bed Reactor– other more conventional geometries
Fission Reactors (Thermal vs. Fast)
• Thermal Reactors (general features)– If moderated with D2O (low n-capture) can
burn natural U now need for enrichment (saves lots of energy!)
– Larger reactor cores needed more activation– If natural U used small burn-up time
often need continuous fuel exchange hard to control
Fission Reactors (Light vs. Heavy water thermal reactors)
• Light Water – it is cheap – very well understood chemistry– compatible with steam part of plant – can not use natural uranium (too much n-capture) must have enrichment plant bombs
– need larger moderator volume larger core with more activation
– enriched U has bigger n-margin easier to control
Fission Reactors (Light vs. Heavy water thermal reactors)
• Heavy Water– it is expensive– allows use of natural U – natural U has smaller n-margin harder to
control– smaller moderator volume less activation– CANDU PWR designs (pressure tube reactors)
allow D2O moderation with different coolants to save D2O
Fission Reactors (PWR = most common power reactor)
• Avoid boiling better control of moderation• Higher coolant temperature higher thermal efficiency• If pressure fails (140 bar) risk of cooling failure via boiling Steam raised in secondary
circuit no activity in turbine and generator
Usually used with H2O need enriched U
Difficult fuel access long fuel cycle (1yr) need highly enriched U
Large fuel reactivity variation over life cycle need variale “n-poison” dose in coolant
Fission Reactors (BWR = second most common power reactor)
• lower pressure then PWR (70 bar) safer pressure vessel• simpler design of vessel and heat steam circuit• primary water enters turbine activation of tubine no access during
operation (½(16N)=7s, main contaminant) lower temperature lower efficiency if steam fraction too large (norm. 18%) Boiling crisis =loss of cooling
Fission Reactors (“cool” reactors)
• no boiling crisis• no steam handling• high efficiency 44%• compact core• low coolant mass
Fission Reactors (enrichment)
• Two main techniques to separate 235U from 238U in gas form UF6 @ T>56C, P=1bar– centrifugal separation
• high separation power per centrifugal step• low volume capacity per centrifuge• total 10-20 stages to get to O(4%) enrichment• energy requirement: 5GWh to supply a 1GW reactor with 1 year of fuel
– diffusive separation• low separation power per diffusion step• high volume capacity per diffusion element• total 1400 stages to get O(4%) enrichment• energy requirement: 240GWh = 10 GWdays to supply a 1GW reactor
with 1 year of fuel
15-20 cm
1-2
m
O(70,000) rpm Vmax≈1,800 km/h = supersonic! & gmax=106g difficult to [email protected]
Nuclear Fusion,as a source of stellar energy In stars 12C formation sets the stage for the entire nucleosynthesis of heavy elements:
T ~ 6*108 K and ~ 2*105 gcm-3
4He + 4He 8Be8Be unstable ( ~ 10-16 s)
8Be + 4He 12C Large density helps to overcome the bottleneck caused by the absence of stable nuclei with 8
nucleons.
Example: show that the nucleus 8be has a positive binding energy but is unstable against the decay into two alpha particles.
The binding energy of 8Be:
1 81 4
8 24 4 4
4 1.008665 4 1.007825 8.005304 931.5 / 56.5
B n H BeE Be m m m c
u u u MeV u MeV
/ 7.06BE A MeV
The energy of the decay 8Be two alpha particles:
8 44 2
8 24 2 8.005304 2 4.002602 931.5 / 0.093B Be He
E Be m m c u u MeV u MeV Because the energy of the decay 8Be two alpha particles is positive, 8Be is unstable (an important factor for the nucleosynthesis in the Universe).
Stellar Nucleosynthesis
A massive star near the end of its lifetime has “onion ring” structure.
C burning T ~ 6*108 K ~ 2*105 gcm-3
Ne burning T ~ 1.2*109 K ~ 4*106 gcm-3
O burning T ~ 1.5*109 K ~ 107 gcm-3
Si burning T ~ 3*109 K ~ 108 gcm-3
major ash: Fe - the end of exothermic processes
Multi-step processes of the formation of heavier elements up to Fe.
Two key parameters: temperature (thermal energy is sufficiently large to overcome Coulomb repulsion ) and density (controls the frequency of collisions). With increasing Z, the temperatures should also increase to facilitate the reactions.
Stability Issues (Stable Stars vs. Unstable Bombs)Why are the stars stable (in contrast to the hydrogen bomb)?In stars, the increase of temperature results in the increase of the pressure and the subsequent increase of its size (think the ideal gas law).The density becomes smaller, and the rate of thermonuclear reactions decreases. This is the build-in negative feedback.
Sunred
dwarf
Sirius A
proton cycle
carbon-nitrogen
cycle
T, K
107 109105
lum
inos
ity
The negative feedback works well for young stars. For more dense and old stars, the pressure increase is not sufficient to produce a significant increase of volume (the matter in such stars is not described by gas laws) – and the thermonuclear explosion occurs! This is the star explosion (supernova: “carbon-nitrogen” bomb).
12 13 76 7 10C p N y The carbon-nitrogen cycle:
13 137 6 7 minN C e 13 14 66 7 10C p N y 14 13 15 87 6 8 10N p O y 15 158 7 82O N e s 15 12 4 57 6 2 10N p C He y
a “catalyst”
Explosive Nucleosynthesis (Elements Heavier than Iron)
Elements heavier than iron are created (mostly) by neutron capture.
Explosive nucleosynthesis Endothermic fusion
s-process (slow neutron
capture): n 5626 Fe 57
26 Fe 5727Co
e-
n
5626 Fe 60
26 Fe 6127Co
n
nr-process
(a succession of rapid neutron
captures on iron seed nuclei):
The neutron is added to the nucleus and (later) converted into a proton by decay; this increases the atomic number by 1. Repetition of this process – progress up the valley of stability.
e-
These processes require energy, occur only at high densities & temperatures (e.g., r-processes occur in core-collapse supernovae).
High n flux: fast neutron capture until the nuclear force is unable to bind an extra neutron. Then, a beta decay occurs, and in the new chain the neutron capture continues. This process is responsible
for the creation of about half of neutron-rich nuclei heavier than Fe.
Summary
charged-particle induced reaction
mainly neutron capture reaction
Both occur during quiescent and explosive stages of stellar evolution
involve mainly STABLE NUCLEI involve mainly UNSTABLE NUCLEI
10-1
Abun
danc
e re
lativ
e to
Sili
con
(=10
6 )
Decay Theory• Consider 232Th, Z=90, with radius of R=7.6 fm • It alpha decays with Ea=4.08 MeV at r=• But at R=7.6 fm the potential energy of the alpha would be
E,pot=34 MeV if we believe:
• Question: How does the escape from the Th nucleus?• Answer: by QM tunnelling
2
1 204
e cE Z Z
c R
which we really should!
r
nucleus inside barrier (negative KE) small flux of real α
Decay Theory
I II III
potential energy of
total energy of
Exponential decay of
radial wave function in alpha decay in 3 regions oscillatory oscillatory
r=t
r=R •see also Williams, p.85 to [email protected]
QM Tunnelling through a square well (the easy bit)
• Boundary condition for and d/dx at r=0 and r=t give 4 equations • for times such that Kt>>1 and approximating k≈K we get transmission
probability: T=|D|2~exp(-2Kt) [Williams, p.85]
exp( ) exp( )
exp( ) exp( )
exp( )
I
II
III
ikr A ikr
B Kr C Kr
D ikr
0
2
2 ( )
kin
kin
k mE
K m V E
in regions I and III
in region II
unit incoming oscillatory wavereflected wave of amplitude A
two exponential decaying waves of amplitude B and C
transmitted oscillatory wave of amplitude D
Wave vector Ansatz:
Stationary Wavefunction Ansatz:
Etot
Potential :V
r=0 r=t
V=V0
I II IIIrV=0
4 unknowns !
-decay
Neutrons Protons Alphas
Ebind(42)=28.3 MeV > 4*6MeV
Esep≈6MeV per nucleon for heavy nuclei
Tunnelling in -decay• Assume there is no recoil in the remnant nucleus• Assume we can approximate the Coulomb potential by sequence of many
square wells of thickness r with variable height Vi
• Transmission probability is then product of many T factors where the K inside T is a function of the potential:
• The region between R and Rexit is defined via: V(r)>Ekin
• Inserting K into the above gives:
• We call G the Gamov factor
22 ( )
2
00
lim
Rexit
iR
k r drN K r
trans ri
P T e e
2 2exp 2 ( ( ) ) exp( )
exitR
kin
R
T m V r E dr G
2 21 2 1 2
0 0
( )4 4kin
exit
Z Z e Z Z eV r E
r R
1/ 21/ 22
2
0
2
0
21/ 1/
2
cos 2 cos sin
0
exitR
exit
R
exit exit
exit
mZ eG r R dr
r R dr R d
r R r R
substituting
and and
0 0
1/ 221/ 2 2 22
0 0 00 0 0
4sin sin (1/ 2) sin cos
2 exit
mZ eG R d d
where
Tunnelling in -decay• Use the Coulomb potential for an a particle of charge Z1 and a nucleus of charge Z2 for V(r)
the latter defines the relation between the exit radius and the alpha particles kinetic energy
2exp( ) exp 2 ( ( ) )
exitR
kin
R
G m V r E dr inserted into: and Z1=2 gives
Tunnelling in -decay• How can we simplify this ?
– for nuclei that actually do a-decay we know typical decay energies and sizes– Rtyp≈10 fm, Etyp ≈ 5 MeV, Ztyp ≈ 80
– Rexit,typ ≈ 60 fm >>Rtyp
– since
• Inserting all this into G gives:
• And further expressing Rexit via Ekin gives:
1/ 22
04exitmZe R
G
0 0 0cos / cos 0 / 2exitR R
1/ 22 2 2
0 0
2
4 4 2exitkin
Ze e mZR G
E E
-decay Rates• How can we turn the tunnelling probability into a decay rate?• We need to estimate the “number of hits” that an makes onto the
inside surface of a nucleus.• Assume:
– the a already exists in the nucleus– it has a velocity v0=(2Ekin/m)1/2
– it will cross the nucleus in t=2R/v0
– it will hit the surface with a rate of 0=v0/2R
• Decay rate is then “rate of hits” x tunnelling probability
• Note: 0 is a very rough plausibility estimate! Williams tells you how to do it better but he can’t do it either!
1/ 22 2
0
(2 / ) (2 / )exp( ) exp
2 2 4 2kin kin
kin
E m E m e mZG
R R E
-decay experimental tests• Predict exponential decay rate proportional to (Ekin)1/2
• Agrees approximately with data for even-even nuclei.• But angular momentum effects complicate the picture:
– Additional angular momentum barrier (as in atomic physics)
– El is small compared to ECoulomb – E.g. l=1, R=15 fm El~0.05 MeV compared to – Z=90 Ecoulomb~17 MeV.– but still generates noticeable extra exponential suppression.
• Spin (J) and parity (P) change from parent to daughterJ=L P=(-1)L
2
2 2
( 1)( )
2l
l l cE
mc r
-decay experimental tests
• We expect:
ln(d
ecay
rate
)
.,,
,/ /red daughter
daughter daughter kintot kin
mZ Z EE m
1/ 22 2
0
(2 / )exp
2 4 2kin
kin
E m e mZ
R E
Fermi Decay Theory• Consider simplest case: of -
decay, i.e. n decay• At quark level: du+W
followed by decay of virtual W to electron + anti-neutrino
• this section is close to Cottingham & Greenwood p.166 - ff
• but also check that you understand Williams p. 292 - ff
(782 )e
e
n pe Q keV
d ue
or at quark level
W-
e-
( ) e
d
u u d
u d
n
p
Fermi Theory• 4 point interaction
– Energy of virtual W << mW life time is negligible– assume interaction is described by only a single number– we call this number the Fermi constant of beta decay G
– also assume that p is heavy and does not recoil (it is often bound into an even heavier nucleus for other -decays)
– We ignore parity non-conservation* * * 3( ) ( ) ( ) ( )fi e p nH r r r G r d r
1/ 2 1/ 2( ) exp( . ) ; ( ) exp( . ) ;e e er V ik r r V ik r q k k
* 3
~ 1 / ~ 5 ~ 1/ 40
exp( . ) 1
0
( ) ( )
n p
fi p n f
q MeV c R fm q r
iq r
L R L L L
H G r r d r G H
From nuclear observations we know :
which is only applicable for as otherwise can be larger
which is just a numbe
n pr since and are at rest
e- ( ) e
d
u u d
u d
np
Fermi Theory
2 3 2 34 / 4 /e e e edn p dp h dn p dp h n n n ; ;
2 2 3 2 34 / 4 /e ed n p dp h p dp h
0
/ ( ) / ; / 1/
f
f e
f e f
E
E
E E E
p E c E E c p E c
= total energy released in the decay =
= total energy of the final state
= mass deficit
= total kinetic energy + rest masses of the final state
as we neglect nuclear recoil energy electron energy distribution is determined by density of states
but pe and p or Ee and E are correlated to conserve energy we can not leave them both variable
2 22 2
6 3
16( )e f e
f e
d np E E
dE dp h c
Fermi Theory Kurie Plot
2 2
22
( ) ( )
( )( )
e e f ee
ef e
e
dRI p Ap E E
dp
I pA E E
p
FGR to get a decay rate and insert previous results:2
22
22 2 2
6 3
42 2
7 3
2
2
2 16( )
64( )
fif
fie f e
fi e f e
f e f e
dnR H
dE
dR d nH
dP dE dP
H p E Eh c
G H p E Eh c
* 3( ) ( )fi p n fH G r r d r G H
2 22 2
6 3
16( )e f e
f e
d np E E
dE dp h c
A
let’s plot that from real data
Electron Spectrum• Observe electron kinetic energy spectrum in
tritium decay• Implant tritium directly into a biased silicon
detector• Observe internal ionisation (electron hole pairs)
generated from the emerging electron as current pulse in the detector
• number of pairs proportional to electron energy• Observe continuous spectrum neutrino has to
carrie the rest of the energy• End point of this spectrum is function of neutrino
mass • But this form of spectrum is bad for determining
the endpoint accurately
Ekin,e (keV)
Rel
ativ
e In
tens
ity
Simple Spectrum
Kurie Plot• A plot of: should be linear
• …but it does not! Why?• …because that’s off syllabus! • But if you really must know …• Electron notices Coulomb field of nucleus • e gets enhanced near to proton (nucleus)
• The lower Ee the bigger this effect
• We compensate with a “Fudge Factor” scientifically aka “Fermi Function” K(Z,pe)
• Can be calculated but we don’t have means to do so
• We can’t integrate I(pe) to give a total rate
2
( )ee
e
I pE
p vs.
(I(p
)/p2 K
(Z,p
))1/
2
Ekin,e (keV)
Kurie-Plot
Selection Rules• Fermi Transitions:
– ecouple to give spin Se=0
– “Allowed transitions” Le=0 Jnp=0.
• Gamow-Teller transitions:– ecouple to give spin Se=1
– “Allowed transitions” Le=0 Jnp=0 or ±1
• “Forbidden” transitions
– See arguments on slide 15– Higher order terms correspond to non-zero L. Therefore suppressed
depending on (q.r)2L
– Usual QM rules give: Jnp=Le+Se
...).().(1).exp( 2 rqOrqirqi
Electron Capture• capture atomic electron• Can compete with + decay.• Use FGR again and first look at matrix element
* * 3
0
( ) ( ) ( ) ( )R
fi e p nH G r r r r d r
* 3
0
(0) (0) ( ) ( )R
fi e p nH G r r d r
3/ 221/ 2
20
exp( . ) 1(0) ; ( ) (0)
4e
e
Zm e ik rr
V V
32 22 2
204e
fi f
G Zm eH H
V
* 3
0
( ) ( )R
F n pH r r d r
ee p n
For “allowed” transitions we consider e and const.
Only le=0 has non vanishing e(r=0) and for ne=1 this is largest.
Electron Capture• Density of states easier now
– only a 2-body final state (n,)– n is assumed approximately stationary only matters– final state energy = E
2
3
2
3
4; ;
4
dN dN dN dqqV E q c
dq h dE dq dE
dN qV
dE h c
2
32 2 22 2
4 3 20
2
16
4
ffi
f
eF
dNH
dE
E Zm eG M
h c
apply Fermi’s Golden Rule AGAIN:
Anti-neutrino Discovery• Inverse Beta Decay
• Assume again no recoil on n• But have to treat positron fully relativistic• Same matrix elements as -decay because all
wave functions assume to be plane waves
• Fermi’s Golden Rule (only positron moves in final state!)
2 222 2e efi F
e e
dN dNH G H
dE dE
;e en pe p ne - decay : inverse - decay :
2 22 2fi FH G H V
Anti-neutrino Discovery
• Phase space factor:
• Neglect neutron recoil:
• Combine with FGR
223 2
42 e eF
p EG H
h c V
2
3
4e e
e e
dN dpp V
dE h dE
2 2 2 2 4 2; /ee e e e
e
dpE p c m c E pc
dE
/ ;F c V R F
322
4 3
16 e eF
p EG H
h c
The Cowan & Reines Experiment• for inverse -decay @ E ~ 1MeV ~10-47 cm2
• Pauli’s prediction verified by Cowan and Reines.
1 GW Nuclear Reactor
PMT
H20+CdCl2
Liquid Scint.
2
e p ne
e e
n Cd several
(on protons from the water)
(prompt : shortly after inverse beta decay)
(9MeV,delayed coincidence)
Shielding
original proposalwanted to use abomb instead!
Liquid Scint. PMT
-beam
all this well under groundto reduce cosmic rays!
Parity Definitions
• Parity transforms from a left to a right handed co-ordinate system and vice versa
• Eigenvalues of parity are +/- 1.• If parity is conserved: [H,P]=0 eigenstates of H are eigenstates of
parity all observables have a defined parity• If Parity is conserved all result of an experiment should be unchanged by
parity operation• If parity is violated we can measure observables with mixed parity, i.e. not
eigenstates of parity• best read Bowler, Nuclear Physics, chapter 2.3 on parity!
2
1 2
1
; [ ( )] ( )
[ ( )] ( ) 1
( ) ;
( )
. ( )
.
i
r r P r r
P r r P Eignevalues
v P v v
L r x p P L L
s v v P s s
O v L
let be a true vector:
let be an axial vector:
let be a true scalar:
let be a pseudo scal
;;;;;;;;;;;;;; ( )P O Oar:
Parity Conservation• If parity is conserved for reaction a + b c + d.
• Absolute parity of states that can be singly produced from vacuum (e.g. photons = -1) can be defined wrt. vacuum
• For other particles we can define relative parity. e.g. arbitrarily define p=+1, n=+1 then we can determine parity of other nuclei wrt. this definition
• parity of anti-particle is opposite particle’s parity• Parity is a hermitian operator as it has real eigenvalues!• If parity is conserved <pseudo-scalar>=0 (see next
transparency).• Nuclei are Eigenstates of parity
( 1) ( 1) finalinitialLL
a b c d
x x
where are intrinsic parities of particle
* 3 * 2 3p p pO O d r P O d r
* 3p pO PO P d r
2 * 3( )p p pO O d r * 3
p pO O d r
Parity Conservation Let Op be an observable pseudo scalar operator, i.e. [H, Op]=0
Let parity be conserved [H, P]=0 [P, Op]=0
Let be Eigenfunctions of P and H with intrinsic parity p
<Op> = - <Op> = 0 QED it is often useful to think of parity violation as a non vanishing expectation
value of a pseudo scalar operator
insert Unity
as POp=-OpP since [P, Op]=0
use E.V. of under parity
Q: Is Parity Conserved In Nature?• A1: Yes for all electromagnetic and strong interactions.• Feynman lost his 100$ bet that parity was conserved
everywhere. In 1956 that was a lot of money!• A2: Big surprise was that parity is violated in weak
interactions.• How was this found out?
– can’t find this by just looking at nuclei. They are parity eigenstates (defined via their nuclear and EM interactions)
– must look at properties of leptons in beta decay which are born in the weak interaction
– see Bowler, Nuclear Physics, chapter 3.13
Mme. Wu’s “Cool” Experiment
• Adiabatic demagnetisation to get T ~ 10 mK• Align spins of 60Co with magnetic field.• Measure angular distribution of electrons and photons relative
to B field.• Clear forward-backward asymmetry of the electron direction
(forward=direction of B) Parity violation. • Note:
– Spin S= axial vector– Magnetic field B = axial vector– Momentum p = real vector– Parity will only flip p not B and S
60 60 * 60 * 60( 5) ( 4) ;eCo J Ni J e Ni Ni
5+
0+
4+
2+
- allowedGamov Teller decay J=1
2.51 MeV
1.33 MeV
0 MeV
ExcitationEnergy
60Ni60Co
~100%
The Wu Experiment
’s from late cascadedecays of Ni* measure degree of polarisationof Ni* and thus of Cogamma det. signals
summed over bothB orientations!
scintillator signal
electron signal showsasymmetry of theelectron distribution
see also Burcham & Jobes, P.370
sample warms up asymmetry disappears
Interpreting the Wu Experiment
• Let’s make an observable pseudo scalar Op:– Op=JCo * pe = Polarisation (axial vector dot real
vector)
• If parity were conserved this would have a vanishing expectation value
• But we see that pe prefers to be anti-parallel to B and thus to JCo
• Thus: parity is [email protected]
Improved Wu-Experiment
• Polar diagram of angular dependence of electron intensity
• is angle of electron momentum wrt spin of 60Co or B
• using many detectors at many angles
• points indicate measurements• if P conserved this would have
been a circle centred on the origin
decays• When do they occur?
– Nuclei have excited states similar to atoms. Don’t worry about details E,JP (need a proper shell model to understand).
– EM interaction less strong then the strong (nuclear) interaction– Low energy excited states E<6 MeV above ground state can’t usually
decay by nuclear interaction -decays• -decays important in cascade decays following and decays.
• Practical consequences– Fission. Significant energy released in decays (see later lectures)– Radiotherapy: from Co60 decays– Medical imaging eg Tc (see next slide)
Energy Levels for Mo and Tc• Make Mo-99 in an accelerator• attach it to a bio-compatible molecule• inject that into a patient and observe where the patient emits -rays• don’t need to “eat” the detector as ’s penetrate the body • call this substance a tracer
both decay leaves Tc in excited state.
MeV
MeV
interesting metastable state
Introduction
• Particle Rangesa) If smooth energy loss via many steps
(i.e. ionisation from light ions) sharply defined range, useful for rough
energy measurement
a)
b)
c)c) Sometimes several types of processes happen (i.e. high energy electrons)
mixed curves, extrapolated maximum range
b) If a few or a single event can stop the particle (i.e. photo-effect)
exponential decay of particle beam intensity,
decay constant can have useful energy dependence
No range but mean free path defined
Introduction(classification of interactions)
Particles we are interested in photons
exponential attenuation at low E, often get absorbed in single events detect secondary electrons and ions liberated in absorption process.
charged particles sharper range (continuously loose energy via ionisation) leave tracks of ionisation in matter measure momentum in B sometimes radiate photons can be used to identify particle type
neutrons electrically neutral no first-order em-interaction devils to detect react only via strong force (at nuclear energies!) long exponential range (lots of nuclear scattering events followed by
absorption or decay) need specific nuclear reactions to convert them into photons and/or charged
particles when captured by a target nucleus if stopped, measure decay products, e- + p +
Charged particles in matter(non radiating interactions, what to collide with)
• What could a charged particle collide with– Atomic electrons (“free”)
large energy loss E≈q2/2me (small me, q=momentum transfer)
small scattering angle
– Nuclei small energy loss (E=q2/2mnucleus)
large scattering angle
– Unresolved atoms (predominant at low energies) medium energy loss E<q2/2me
eff because: meeff(bound)>me(free)
medium scattering angle atoms get excited and will later emit photons (scintillation)
Charged particles in matter(Ionisation and the Bethe-Bloch Formula)
• Deal with collisions with electrons first since these give biggest energy loss.
• Task: compute rate of energy loss per path length, dE/dx due to scattering of a charged particle from electrons in matter.
• Remember a similar problem?• Scatter alpha particles of nuclei = Rutherford
scattering
Charged particles in matter(Comparison between Rutherford Scattering and EM-scattering of free electrons)
• Rutherford Scattering– any charged particle X (original
used ’s) scatters of nucleus– Charge(X)=Ze – Charge(nucleus)=Z’e – Mnucl >> MX no nuclear-recoil– first order perturbation theory
(Z*Z’*em<<1)– point point scattering no form-factors
• Bethe-Bloch situation– any charged particle X scatters of
electron (in matter)– Charge(X)=Ze – Charge(electron)=1e – MX >> Me no X-recoil (not true
for X=e-)– first order perturbation theory
(Z*1*em<<1)– point point scattering no form-factors
com
mon
aliti
esdi
ffere
nces spin-0 scatters of spin-0
non-relativistic nucleus assumed unbound
spin-0 scatters of spin-½ could be relativistic electron is often bound
Charged particles in matter(Comparison between Rutherford Scattering and EM-scattering of free electrons)
• Will initially ignore the spin and relativistic effects when deriving first parts of Bethe Bloch formula
• Start with Rutherford like scattering using electron as projectile
• Later introduce more realistic scattering crossection (Mott) to get full Bethe Bloch formula
• Add effects for bound electrons at the end
Charged particles in matter(From Rutherford Scattering to the Bethe-Bloch Formula)
• Differential Rutherford-scattering crossection for electrons as projectiles
If we want to turn this process around to describe energy loss of a particle X scattering of electrons in a solid we need to initially assume: X scatters of free electrons i.e. Ekin,projectile >> Ebin,electron or Vprojectile>>Vbound-e (deal with
bound electrons later) MX>>me so that reduced Mreduced(X) ≈ Mrest(X) will need recoil corrections to apply
results to dE/dx of electrons passing through matter
22 24
2 2 sin4 2
Rutherfordz cd
d P V
h
P,V = momentum and relative velocity of electron wrt. nucleusZ = charge of nucleus = scattering angle of the electron wrt. incoming electron direction= stereo angle
2 22
2 2 2
12 sin
2 sind d d d dq d d p d
pd d d d d dq dq dq
2 2 2 2 cosq P P PP
Charged particles in matter(normal Rutherford Scattering: e- on nucleus, change of variables)
• Change variables from to q2 (q = momentum transfer to electron) to get to frame independent form
Pelectron,in
P’ electron,out
q
2 2 2 2
in elastic scattering of heavy nucleus:
2 (1 cos ) 4 sin 2
P P p
q p p
222 sin
dqp
d
1if no dependence: 2 sin
2 sind
d dd
2 1sin 1 cos
2 2
2 2
22 2
2 2 4
since:
4
d ddq p d
z cddq V q
h
Charged particles in matter(normal Rutherford Scattering: e- on nucleus, change of variables)
22 24
2 2 sin4 2
z cdd pV
h
22
2
22 2 2
2 4
since: sin2 4
4
qp
z cd pd V q
h
Charged particles in matter(Rutherford Scattering, change of frame to nucleus on e)
• Change frame to:– electron stationary (in matter), nucleus moving with V
towards electron– p in formula is still momentum of electron moving with
relative V p =meV
– q2 is frame independent• non-relativistic this is obvious (do it at home)• Energy transfer to the electron is defined via: • E==|q2|/2me d/dq2=1/2me
• relativistic need to define q as 4-momentum transfer, but we assume non relativistic for Rutherford [email protected]
• Above is crossection for a non relativistic heavy particle of charge z to loose energy between and +d in collision with a spin-less electron it approaches with velocity V
• We want as a useful quantity: – kinetic energy lost by projectile = -dT – per path length dx – in material of atomic number density n – with Z’ electrons per atom
Charged particles in matter(From inverse Rutherford scattering to the Bethe-Bloch Formula)
22 2
2 2
2 1
e
z cdd d
d mV
h
max
minv
ddT nZdx d
dnumber of collissions withelectrons in length dx per unit crossection area
crossection weightedavg. energy lost per collision
22 2
2 2 2 4
412 e
z cd d d ddq dq d m d V q
h
|q2|=2me
Charged particles in matter(Ionisation and the Bethe-Bloch Formula, simple integral)
max
min
2 2 2 2 2 2max
2 2 2min
1 2 ( ) 2 ( )ln
e e
dT Z c Z cnZ d nZ
dx mV mVh h
• Two of our assumptions justifying the use of Rutherford scattering were:– Electrons in matter have no spin– Projectile travels at non relativistic speed
• None of these are met in practise• We have to do all of the last 5 slides again starting from a
relativistic crossection for spin ½ electrons.
Charged particles in matter(Ionisation and the Bethe-Bloch Formula, Mott)
• Differential Mott-scattering crossection for relativistic spin ½ electrons scattering off a finite mass nucleus (finite mass e- could be target)
22 2 24 2
2 2 2sin 1 sin4 2 2Mott
z cd Vd P V c
h
If we perform the same transformations (q2) with this crossection and then perform the integral:
Rutherford term Mott term
max
min
Mott
v
ddT nZdx d
d we get …
Charged particles in matter(Ionisation and the Bethe-Bloch Formula, Mott integral)
Valid for all charged particles (not limited to heavy particles)
max can be computed via kinematics of “free” electron since Ebind << Ekin (see Williams problem 11.1 on p.246)
2 2 2
max 2
2
1 2
with: and
as properties of incoming
particle
e
e e
m c
m mM M
E PM E
max 2
if incoming particle is not relaticistic ( 1)
and
4( )
e
e
E T M T M
mMTM m
=
=
2 2
max 2
2
Note: c=1 from here downwards!
2( )
2ee
E MM
m Em
2
max
if and and then
(ultra relativistic incoming particle)e eE M E M m E m
E
? ? ?
A list of limits for max follows:
2 2 2 2max max min
2 2 2min
2 ( )ln 1
2e e
dT Z c VnZ
dx mV c m c
h
Mott term
Charged particles in matter(Ionisation and the Bethe-Bloch Formula, min)
• But what about min ?– can not assume that e is free for small energy transfers– ≠q2/2me because electron bound to atom– can get excited atoms in final state (not just ions) our integral was wrong for the lower limit! (can’t get from first to
second line on slide 15 any more)– For small need 2-D integral d dq depending on detailed atomic
structure– We need to find some average description of the atomic structure
depending only on Z and A if we want to find a universal formula– This gives sizable fraction of integral but is very hard to do– The result is the Bethe-Bloch Formula
Charged particles in matter(Ionisation and the Bethe-Bloch Formula = BBF)
• Stopping power = mean energy lost by ionisation upon perpendicularly traversing a layer of unit mass per area.
• Units: Mev g-1 cm2, Range: 4.1 in H to 1.1 in U• I=mean excitation energy; depends on atom type, I≈11*Z [eV]
2 2 2max
2 2 22 2 22
2
in the infinitely heavy projectile ("no recoil", 2 )
approximation =2 and this becomes:
21 4ln
2
e
e
eA
e
M m
m c
m cN ZdT Zdx A m I
?
h
21with , , mass density, density correction
1-
atomic number, Avogadro's number, mean excitation potential A
Vc
A N I
1 is called Stopping Power.
dTdx
2 2 22 2 22max
2 2
21 4 1ln
2 2eA
e
m cN ZdT Zdx A m I
h
Charged particles in matter(Ionisation and the Bethe-Bloch Formula, Bethe-Bloch features)
• =density correction: dielectric properties of medium shield growing range of Lorenz-compacted E-field that would reach more atoms laterally. Without this the stopping power would logarithmically diverge at large projectile velocities. Only relevant at very large
• BBF as a Function of is nearly independent of M of projectile except for max and very weak log dependence in if you know p and measure get M (particle ID via dE/dx): See slide 23
• Nearly independent of medium. Dominant dependence is Z’/A ≈½ for most elements.
• Limitations:– totally wrong for very low V (ln goes negative particle gains Energy = stupid)– correct but not useful for very large V (particle starts radiating, see next chapter)
2 2 22 2 22max
2 2
21 4 1ln
2 2eA
e
m cN ZdT Zdx A m I
h
1ln ln and 28.816 (
2 2p
p Z AI
(off syllabus)
Charged particles in matter(Ionisation and the Bethe-Bloch Formula, variation with )
+ can capture e-
Ec = critical energydefined via:dE/dxion.=dE/dxBrem.
BetheBloch
• Broad minimum @ ≈3.0(3.5) for Z=100(7) • At minimum, stopping power is nearly
independent of particle type and material
Stopping Power at minimum varies from 1.1 to 1.8 MeV g-1 cm2)
Particle is called minimum ionising (MIP) when at [email protected]
in drift chambergas
Charged particles in matter(Ionisation and the Bethe-Bloch Formula, variation with particle type)
• P=mv=mc • variation in dE/dx is useful
for particle ID• variation is most
pronounced in low energy falling part of curve
• if you measured P and dE/dx you can determine the particle mass and thus its “name”
e
Charged particles in matter(Radiating Interactions)
• Emission of scintillation light is secondary process occurring later in time. – Has no phase coherence with the incident charge and is
isotropic and thus SCINTILLATION NOT A RADIATING INTERACTION in this sense.
• Primary radiation processes which are coherent and not isotropic are:– Cherenkov radiation is emitted by the medium due to the
passing charged particle. – Bremsstrahlung and Synchrotron Radiation are emitted
by charged particle itself as result of its environment.
Charged particles in matter(Cherenkov Radiation)
• Source of E-field (Q) passing through medium at a v > vphase(light in medium) creates conical shock wave. Like sonic boom or bow wave of a planing speed boat.
• Not possible in vacuum since v<c. Possible in a medium when v>c/n. – The Cerencov threshold at = 1/n can be used to measure and thus do particle ID if
you can measure the momentum as well. • Huygens secondary wavelet construction gives angle of shockwave as cos =
1/n, This can be used to measure particle direction and .
In time that the particle goes from O to P, light goes from O to A.
Cherenkov radiation first used in discovery of antiproton (1954).
Now often used in large water-filled neutrino detectors and for other particle physics detectors (see Biller).
Total energy emitted as Cherenkov Radiation is ~0.1% of other dE/dx.
ct/n
ct
O P
A
particle trajectory
Charged particles in matter(Cherenkov Radiation)
• Picture of Cherenkov light emitted by beta decay electrons in a working water cooled nuclear reactor.
Charged particles in matter(Bremsstrahlung = BS = Brake-ing Radiation)
• Due to acceleration of incident charged particle in nuclear Coulomb field
• Radiative correction to Rutherford Scattering. • Continuum part of x-ray emission spectra. • Electrons “Brem” most of all particles because
– radiation ~ (acceleration)2 ~ mass-2. • Lorentz transformation of dipole radiation from incident particle
frame to laboratory frame gives “narrow” (not sharp) cone of blue-shifted radiation centred around cone angle of =1/.
• Radiation spectrum falls like 1/E (E=photon Energy) because particles loose many low-E photons and few high-E photons. I.e. It is rare to hit nuclei with small impact parameter because most of matter is “empty”
• Photon energy limits:– low energy (large impact parameter) limited through shielding of nuclear
charge by atomic electrons. – high energy limited by maximum incident particle energy.
Ze
e- e-
e-*
Charged particles in matter(Bremsstrahlung EM-showers, Radiation length)
• dT/dx|Brem~T (see Williams p.247, similar to our deriv. of BBF and plot on slide 22) dominates over dT/dx|ionise ~ln(T) at high T.
• Ecrit = Energy at which BR-losses exceed ionisation losses (see slide 22)• For electrons Bremsstrahlung dominates in nearly all materials above few 10
MeV. Ecrit(e-) ≈ 600 MeV/Z• If dT/dx|Brem~T T(x)=T0 exp(-x/X0) • Radiation Length X0 of a medium is defined as:
– distance over which electron energy reduced to 1/e via many small BS-losses – X0 ~Z 2 approximately as it is the charge that particles interact with
• Bremsstrahlung photon can produce e+e--pair (see later) and start an em-shower (also called cascade, next slide)
The development of em-showers, whether started by primary e or is measured in X0.
Charged particles in matter(simple EM-shower model)
• Simple shower model assumes:– e≈2– E0 >> Ecrit
– only single Brem- or pair production per X0
• The model predicts:– after 1 X0, ½ of E0 lost by primary
via Bremsstrahlung– after next X0 both primary and
photon loose ½ E again – until E of generation drops below
Ecrit
– At this stage remaining Energy lost via ionisation (for e+-) or compton scattering, photo-effect (for ) etc.
Abrupt end of shower happens at t=tmax = ln(E0/Ecrit)/ln2 Indeed observe logarithmic dependence of shower depth on E0
Charged particles in matter(Synchroton Radiation)
• Appears mainly in circular accelerators (mainly to electrons) and limits max. energy achievable.
• Similar to Bremsstrahlung• Replace microscopic force from E-field in Bremsstrahlung with
macroscopic force from vxB to keep electron on circular orbit• Electrons radiate only to the outside of circle because they
are accelerated inward• Angle of maximum intensity of synchrotron radiation with
tangent of ring =1/• Synchrotron radiation = very bright source of broad range of
photon energies up to few 10 keV used in many areas of science
• Many astrophysical objects emit synchrotron radiation from relativistic electrons in strong magnetic fields
Photons in matter(Overview-I)
• Rayleigh scattering – Coherent, elastic scattering on the entire atom (the blue sky)– + atom + atom– dominant at >size of atoms
• Compton scattering– Incoherent scattering on electron from atom – + e-
bound + e-free
– possible at all E > min(Ebind)– to properly call it Compton requires E>>Ebind(e-) to approximate free e-
• Photoelectric effect– absorption of photon and ejection of single atomic electron– + atom + e-
free + ion– possible for E < max(Ebind) + E(Eatomic-recoil, line width) (just above k-edge)
Photons in matter(Overview-II)
Pair production absorption of in atom and emission of e+e- pair Two varieties:
a) dominant: + nucleus e+ + e- + nucleusrecoil b) weak: + Z*atomic e- e+ + e- + Z *atomic e-
recoil Both variants need: E>2mec2 + Erecoil
bigger Mrecoil gives lower threshold because Erecoil = Precoil2/2Mrecoil
type a) has lower threshold then type b) because Mnucl>>Meeff
Nucleus/atom has to recoil to conserve momentum coupling to nucleus/atom needed strongly charge-dependent crossection (i.e. growing with Z) type a) has aproximately Z times larger coupling dominant
Photons in matter (Crossections)
R Rayleigh PE Photoeffect C Compton PP Pair Production on nucleusPPE Pair Production on atomic electrons PN Giant Photo-Nuclear dipole resonance
Carbon
Lead
As Z increases PE extends to higher E due to stronger atomic e- binding PP & PPE extend to lower E due to stronger coupling of projectile to target Threshold for PPE decreases as nucleus contributes more to recoil via stronger atomic electron-
nucleus bond As A increases Erecoil (nucleus) decreases and threshold for PP gets closer to minimum of
Photons in matter(Comparison of Bremsstrahlung and Pair Production)
Very similar Feynman Diagram Just two arms swapped
Typical Lenth =Radiation LengthX0
Typical Lenth =Pair Production Length L0
L0=9/7 X0
Ze
e-
e-*
Bremsstrahlung
e-
Ze
e-*
e-
Pair production
e-
X0 :distance high E e-travels before it reduces its energy by 1/e or
E(e-)=E0*exp(-x/X0)X0=attenuation length
L0 :distance high E
travels before prob. for non interaction reduced to 1/e
P()=1/L0*exp(-x/L0)L0=mean free path