Advanced Nuclear Physics

Advanced Nuclear Physics
MOHAMMAD IMRAN AZIZ Assistant Professor PHYSICS DEPARTMENT SHIBLI NATIONAL COLLEGE, AZAMGARH (India).

Nuclides 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. .

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 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. 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.

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
where Proton: g = / Neutron: g = / 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

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 = the quoted magnetic moment is m = nuclear magnetons. Nuclide Nuclear spin I Magnetic moment m in mN n 1/2 p 2H(D) 1 17O 5/2 57Fe 57Co 7/2 +4.733 93Nb 9/2

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
- depends weakly on A (number of nucleons in the nucleus) Geiger, Marsden, Rutherford,1910 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 (<310-14m for gold). -particles: bare He nuclei 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). 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. R – the fitting parameter

Nuclear Mass number of nucleons in the nucleus
(mass number of the nucleus) chemical symbol for the element 6 protons + 6 neutrons number of protons in the nucleus (atomic number of the element) The neutron number: Atomic Mass (the mass of a neutral atom): - attraction between nucleons in the nucleus and between electrons and the nucleus - we can neglect Uel.-nucl. and introduce a convenient mass unit: Mass Unit

Nuclear Density ! 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 quarks and gluons ~ m EM = electromagnetic, and it refers to the electromagnetic force.

Structure of Matter g u d u
El.-mag. interaction determines the size of atoms d quarks and gluons u ~ m ~ m Protons & Neutrons (nucleons) are almost an empty space (the quark size is <10-18 m) Atom is almost an empty space (the nuclear volume is ~10-15 of the atomic volume) Strong interaction determines the size of nuclei and nucleons ~ m

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”). n n p p add EB n n p We can compute the binding energy if we know masses of a system and its constituents: mass deficit

Binding Energy curve Because 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 decrease of the binding energy with increasing Z is caused by the long-range Coulomb repulsion of protons: Binding energy per nucleon (EB/A), MeV Mass number, A The binding energy ~ 10MeV/nucleon is ~1% of the nucleon’s rest energy: we can consider the nucleus as a system of individual nucleons

Liquid-Drop Model A “semi-classical” model of the nucleus: describes reasonably well the dependence EB(A):  - charge density

Liquid-Drop Model (cont’d)

Limitations of Liquid-Drop Model
Maria Goeppert-Mayer, J.H.D.Jensen

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 oxygen 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?
Both 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). energy r The potential experienced by nucleons is a 3D potential well. The ground-state configuration of the carbon-16 nucleus : If a nucleus is allowed to decrease its energy by transforming “excessive” protons (neutrons) into neutrons (protons), it will do it! protons neutrons 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:

Why N  Z for light nuclei
protons neutrons energy protons neutrons energy 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. energy energy 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. protons neutrons protons neutrons

Why N > Z for heavy nuclei
energy energy 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 protons neutrons 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 + DE Mass difference between X and Y+Z is DE/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. Detector Velocity selector Ion Source

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

Isotope Shifts

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 mm~ 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=-an + 4pR2T B=an-bn2/3 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 73Li and 74Be. Mass difference due to n/p mass and Coulomb energy.

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, a,b=free constants E=-an + 4pR2T  B=an-bn2/3 surface area ~ 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 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) … and remember R=R0A-1/3 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)

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 np). Eg 73Li 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 to find that Now lets measure mirror nuclei masse, assume that the model holds and derive DECoulomb 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 right

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

More charge symmetry Energy Levels of two mirror nuclei for a number of excited states Corrected for n/p mass difference and Coulomb Energy DEcorrected

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.

Charge Independence 2210Ne 2212Mg 2211Na
DEcorrected 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 2212Mg 2211Na 2210Ne

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

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 D ?

Asymmetry Term Assume:
p and n form two independent, non-interacting gases occupying their own square Fermi wells kT << D 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 here Nparticle could be the number of protons or neutrons These are all standard stat. mech. results!

Asymmetry Term Compute total energy of all protons by Z*<E>
call this K 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 (e) Binomial expansion keep lowest term in y/A note! linear terms cancel This terms is only proportional to volume (A). It has already been captured by the Volume term of the liquid drop model comes from a fit of the SEMF to measurements analytical ≈ 24 MeV

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:

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 Note: this only holds for nn and pp, not for np.  We don’t have a preference for even A

Pairing Term Measure that the Pairing effect smaller for larger A
Phenomenological*) fit to A dependence gives A-1/2 d e-e +ive e-o o-o -ive 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 *) For an even more insightful explanation of the A dependence read the book by Jelley

Semi Empirical Mass Formula
Put everything together: Volume Term Surface Term Asymmetry Term Coulomb Term Pairing Term 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
Semi Empirical Mass Formula Binding Energy vs. A for beta-stable odd-A nuclei 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) Iron Not smooth because Z not smooth function of A

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

Angular part If V(r) then can separate variables y(r,q,f) = R(r)Y( q,f) 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,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/ *** 1P3/ 1P1/ *** 1D5/ 2S1/ 1D3/ *** 1F7/ *** 2P3/ 1F5/ 2P1/ 1G9/ *** *** “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 He-4 6 C C-12 8 O O-16 20 Ca 28 Ni Cr-52(24,28) 50 Sn Ni-78 82 Pb 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

Nuclear Cross Sections
Definition of Cross Section Why its useful. Breit-Wigner Resonances Rutherford Scattering Outline Experimental definition of cross section Theory of how to calculate cross sections: QM  BW line shape. Fermi Golden rule  recipe to calculate cross section BW cross section formula QM Rutherford Scattering Cross section calculation

Cross-Sections Why concept is important Experimental Definition
Learn about dynamics of interaction and/or constituents (cf Feynman’s watches). Needed for practical calculations. Experimental Definition How to calculate s Fermi Golden Rule Breit-Wigner Resonances QM calculation of Rutherford Scattering

Definition of s a+bx Effective area for reaction to occur is s
Na(0) particles type a/unit time hit target b Nb atoms b/unit volume Number /unit area= Nb dx Probability interaction = s Nbdx dNa=-Na Nb dx s Na(x)=Na(0) exp(-x/l) ; l=1/(Nb s) Beam a Na dx

Reaction Rates Na beam particles/unit volume, speed v Flux F= Na v
Rate/target b atom R=Fs Thin target x<<l: R=(NbT) F sTotal This is total cross section. Can also define differential cross sections, as a function of reaction product, energy, transverse momentum, angle etc. dR(a+bc+d)/dE=(NbT) F ds(a+bc+d) /dE

Breit-Wigner Line Shape
Start with NR Schrödinger equation: X by f*n and integrate Start in state m  exponential decay

Breit-Wigner Line Shape - 2
For

Breit-Wigner Line Shape -3
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. , G=FWHM;

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 s 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

Cross Section Breit Wigner cross section. Definition of s and flux F:

Breit-Wigner Cross Section
Combine rate, flux & density states 

Breit-Wigner Cross Section
n + 16O 17O

Low Energy Resonances n + Cd total cross section.
Cross section scales s ~ 1/E1/2 at low E. B-W: 1/k2 and G~n(E)~k

Rutherford Scattering 1

Fermi Golden Rule:

pf pi Compare with experimental data at low energy Q: what changes at high energy ?

Low Energy Experiment Scattering of a on Au & Ag  agree with calculation assuming point nucleus dN/dcosq Sin4(q/2)

Higher Energy Electron - Gold Deviation from Rutherford scattering at higher energy  determine charge distribution in the nucleus. Form factors is F.T. of charge distribution.

Induced Fission (required energy)
Neutron DEsep≈6MeV per nucleon for heavy nuclei Nucleus Potential Energy during fission [MeV] Very slow n A= 238 DEf=Energy needed to penetrate fission barrier immediately ≈6-8MeV Neutrons

Induced Fission (required energy)
Spontaneous fission rates low due to high coulomb barrier (6-8 A≈240) Slow neutron releases DEsep as excitation into nucleus Excited nucleus has enough energy for immediate fission if Ef - DEsep >0 We call this “thermal fission” (slow, thermal neutron needed) But due to pairing term … even N nuclei have low DEsep for additional n odd N nuclei have high DEsep for additional n  Fission yield in n -absorption varies dramatically between odd and even N

Induced Fission (fissile nuclei)
DEsep(n,23892U) = 4.78 MeV only  Fission of 238U needs additional kinetic energy from neutron En,kin>Ef-DEsep≈1.4 MeV We call this “fast fission” (fast neutrons needed) Thermally fissile nuclei, 1160K 23392U, 23592U, 23994Pu, 24194Pu Fast fissile nuclei En,kin=O(MeV) 23290Th, 23892U, 24094Pu, 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)
<# prompt n> nprompt=2.5 t≈10-14 s t>10-10 s <n-delay> td=few s <# delayed n> nd=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)  3-12 from X+nY+g E(prompt g) Subtotal: (good for power production) Delayed (10-10<t<): Ekin(e from b-decays) E(g following b-decay) Subtotal: (bad, spent fuel heats up) Neutrinos: (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= 23592U and even-even= 23892U unresolved, narrow resonances unresolved, narrow resonances 235U 238U n -Energy

Induced Fission ((n,f) and (n,g) probabilities in natural Uranium)
23592U(n,f) 23592U(n,g) 23892U(n,g) 23892U(n,f) energy range of fission neutrons “good 235 ” “bad-238” neutron absorbtion probabilit per 1 mm “good 238 ” “bad-235” fast thermal

Induced Fission (a simple bomb)
Uranium mix 235U:238U =c:(1-c) rnucl(U)=4.8*1028 nuclei m-3 average n crossection: mean free path for fission n: mean time between collisions =1.5*10-9 s @ Ekin(n)=2MeV Simplify to c=1 (the bomb mixture) prob(235U(nprompt 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 l=3cm  lmp=√6*3cm≈7cm in tmp=10-8 s

After 10-8 s 1n is replaced with n=2.5 n, 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 (nq-1) new such n in tp=10-8 s (ignoring delayed n as bombs don’t last for seconds!) if nq>1  exponential growths of neutron number For 235U, n=2.5  if q>0.4 you get a bomb

If object dimensions << lmp=7 cm  most n escape through surface  nq << 1 If Rsphere(235U)≥8.7cm  M(235U)≥52 kg  nq = 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 s(23892U(n,g)) increases and becomes resonant s(23892U(n,f)) decreases rapidly and vanishes below 1.4 MeV only remaining chance for fission is s(23592U(n,f)) which is much smaller then s(23892U(n,g)) Conclusion: piling up natural U won’t make a reactor because n get “eaten” by (n,g) resonances. I said it is not SO simple 23592U(n,f) 23592U(n,g) 23892U(n,g) 23892U(n,f)

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  23992U + g  23993Np + e- + ne  23994Pu + e¯ + ne 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 n(239Pu)=2.96 Fewer delayed n & higher n-absorbtion, more later

Fission Reactors (Reactor control)
For bomb we found: “boom” if: nq > 1 where n 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 snc 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 products all n -rich  all b- active Some b- decays have excited states as daughters These can directly emit n (see table of nuclides, green at bottom of curve) Energy several sources of delayed n typical lifetimes t≈O(1 sec) Fraction nd ≈ 0.6% off syllabus

Since fuel rods “hopefully” remain in reactor longer then 10-2 s  must include delayed n fraction nd into our calculations New control problem: keep (n+nd)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 sn-abs reduce En chemistry other coolant H2O liquid small best reactive cheap good D2O none 2nd best rare C solid mild medium CO2press. gas passive ok He 3rd best very passi. leaks Na very react. difficult excellent off syllabus

Fission Reactors (Thermal Stability)
Want dq/dT < 0 Many mechanical influences via thermal expansion Change in n-energy spectrum Doppler broadening of 238U(n,g) 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)
Isotope Half-lifea Bare critical mass Spontaneous fission neutrons Decay heat years kg, Alpha-phase (gm-sec)-1 watts kg-1 Pu-238 87.7 10 2.6x103 560 Pu-239 24,100 22x10-3 1.9 Pu-240 6,560 40 0.91x103 6.8 Pu-241 14.4 49x10-3 4.2 Pu-242 376,000 100 1.7x103 0.1 Am-241 430 1.2 114 a. By Alpha-decay, except Pu-241, which is by Beta-decay to Am-241. ideal bomb fuel = pure 239Pu

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 a. Pu-241 plus Am-241. 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). 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 g-rays  shielding problem 240Pu : lots of n from spontaneous fission 238Pu : a-decays quickly (t1/2 = 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.

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 g sources from 241Am outside a reactor large n fluxes from 240Pu outside reactors very penetrating  easy to spot over long range Plutonium isotope composition as a function of fuel exposure in a pressurized-water reactor, upon discharge.

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

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

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)
 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 (t½(16N)=7s, main contaminant) lower temperature   lower efficiency if steam fraction too large (norm. 18%)   Boiling crisis = loss of cooling

Fission Reactors (“cool” reactors)

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 T>56C, P=1bar centrifugal separation high separation power per centrifugal step low volume capacity per centrifuge total 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

1-2 m 15-20 cm O(70,000) rpm  Vmax≈1,800 km/h = supersonic! & gmax=106g  difficult to build!

Fission Reactors (enrichment)

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  8Be 8Be unstable ( ~ 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: The energy of the decay 8Be  two alpha particles: 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 - 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. C burning T ~ 6*108 K  ~ 2*105 gcm-3 A massive star near the end of its lifetime has “onion ring” structure. 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

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. The carbon-nitrogen cycle: Sirius A carbon-nitrogen cycle Sun luminosity red dwarf proton cycle a “catalyst” 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). T, K 105 107 109

Explosive Nucleosynthesis (Elements Heavier than Iron)
Explosive nucleosynthesis  Endothermic fusion Elements heavier than iron are created (mostly) by neutron capture. e- s-process (slow neutron capture): n 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. n e- r-process (a succession of rapid neutron captures on iron seed nuclei): n n 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. These processes require energy, occur only at high densities & temperatures (e.g., r-processes occur in core-collapse supernovae).

Both occur during quiescent and explosive stages
Summary Abundance relative to Silicon (=106) 10-1 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

a 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 Ea,pot=34 MeV if we believe: Question: How does the a escape from the Th nucleus? Answer: by QM tunnelling which we really should!

a Decay Theory I II III r=t r=R r see also Williams, p.85 to 89
Exponential decay of y radial wave function in alpha decay in 3 regions oscillatory y I II III total energy of a r nucleus inside barrier (negative KE) small flux of real α r=t potential energy of a r=R see also Williams, p.85 to 89

QM Tunnelling through a square well (the easy bit)
Etot Potential :V r=0 r=t V=V0 I II III r V=0 Wave vector Ansatz: in regions I and III in region II Stationary Wavefunction Ansatz: unit incoming oscillatory wave reflected wave of amplitude A two exponential decaying waves of amplitude B and C transmitted oscillatory wave of amplitude D 4 unknowns ! Boundary condition for Y and dY/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]

a-decay DEsep≈6MeV per nucleon for heavy nuclei
DEbind(42a)=28.3 MeV > 4*6MeV Neutrons Protons Alphas

Tunnelling in a-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 Dr 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

Tunnelling in a-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 inserted into: and Z1=2 gives

Tunnelling in a-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:

a-decay Rates How can we turn the tunnelling probability into a decay rate? We need to estimate the “number of hits” that an a 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 Dt=2R/v0  it will hit the surface with a rate of w0=v0/2R Decay rate w is then “rate of hits” x tunnelling probability Note: w0 is a very rough plausibility estimate! Williams tells you how to do it better but he can’t do it either!

a-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 (DJ) and parity (DP) change from parent to daughter DJ=La DP=(-1)L

a-decay experimental tests
We expect: ln(decay rate)

Fermi b Decay Theory Consider simplest case: of b-decay, i.e. n decay
At quark level: du+W followed by decay of virtual W to electron + anti-neutrino this section is close to Cottingham & Greenwood p ff but also check that you understand Williams p ff W- e- ( ) ne d u n p

Fermi Theory u d 4 point interaction e- ne
( ) ne d u 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 Gb also assume that p is heavy and does not recoil (it is often bound into an even heavier nucleus for other b-decays) We ignore parity non-conservation

Fermi Theory as we neglect nuclear recoil energy
electron energy distribution is determined by density of states but pe and pn or Ee and En are correlated to conserve energy  we can not leave them both variable

Fermi Theory  Kurie Plot
FGR to get a decay rate and insert previous results: 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) Relative Intensity 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  Ye 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 (I(p)/p2K(Z,p))1/2 Ekin,e (keV) Kurie-Plot

Selection Rules Fermi Transitions: Gamow-Teller transitions:
en couple to give spin Sen=0 “Allowed transitions” Len=0  DJnp=0. Gamow-Teller transitions: en couple to give spin Sen=1 “Allowed transitions” Len=0  DJnp=0 or ±1 “Forbidden” transitions See arguments on slide 15 Higher order terms correspond to non-zero DL. Therefore suppressed depending on (q.r)2L Usual QM rules give: DJnp=Len+Sen

Electron Capture capture atomic electron Can compete with b+ decay.
Use FGR again and first look at matrix element For “allowed” transitions we consider Ye and Yn const. Only le=0 has non vanishing Ye(r=0) and for ne=1 this is largest.

Electron Capture Density of states easier now
only a 2-body final state (n,n) n is assumed approximately stationary  only n matters  final state energy = En 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 b-decay because all wave functions assume to be plane waves Fermi’s Golden Rule (only positron moves in final state!)

Anti-neutrino Discovery
Phase space factor: Neglect neutron recoil: Combine with FGR

The Cowan & Reines Experiment
for inverse En ~ 1MeV  s ~10-47 cm2 Pauli’s prediction verified by Cowan and Reines. Liquid Scint. PMT 1 GW Nuclear Reactor n -beam H20+CdCl2 Liquid Scint. PMT original proposal wanted to use a bomb instead! all this well under ground to reduce cosmic rays! Shielding

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!

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 hg= -1) can be defined wrt. vacuum For other particles we can define relative parity. e.g. arbitrarily define hp=+1, hn=+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

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 Y be Eigenfunctions of P and H with intrinsic parity hp insert Unity as POp=-OpP since [P, Op]=0 use E.V. of Y under parity <Op> = - <Op> = 0 QED it is often useful to think of parity violation as a non vanishing expectation value of a pseudo scalar operator

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 5+ 0+ 4+ 2+ b - allowed Gamov Teller decay DJ=1 2.51 MeV 1.33 MeV 0 MeV Excitation Energy 60Ni 60Co ~100%

The Wu Experiment g’s from late cascade decays of Ni* measure
degree of polarisation of Ni* and thus of Co gamma det. signals summed over both B orientations! electron signal shows asymmetry of the electron distribution scintillator signal sample warms up  asymmetry disappears see also Burcham & Jobes, P.370

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 violated

Improved Wu-Experiment
Polar diagram of angular dependence of electron intensity q 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

g 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  g-decays g-decays important in cascade decays following a and b decays. Practical consequences Fission. Significant energy released in g decays (see later lectures) Radiotherapy: g 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 g-rays don’t need to “eat” the detector as g ’s penetrate the body call this substance a tracer MeV interesting meta stable state both b decay leaves Tc in excited state. MeV

Introduction Particle Ranges a) b) c)
If smooth energy loss via many steps (i.e. ionisation from light ions)  sharply defined range, useful for rough energy measurement 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 c) Sometimes several types of processes happen (i.e. high energy electrons)  mixed curves, extrapolated maximum range

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 + n

Charged particles in matter (non radiating interactions, what to collide with)
What could a charged particle collide with Atomic electrons (“free”)  large energy loss DE≈q2/2me (small me, q=momentum transfer)  small scattering angle Nuclei  small energy loss (DE=q2/2mnucleus)  large scattering angle Unresolved atoms (predominant at low energies)  medium energy loss DE<q2/2meeff 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 a’s) scatters of nucleus Charge(X)=Ze Charge(nucleus)=Z’e Mnucl >> MX  no nuclear-recoil first order perturbation theory (Z*Z’*aem<<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*aem<<1) point  point scattering  no form-factors commonalities differences 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 P,V = momentum and relative velocity of electron wrt. nucleus Z = charge of nucleus q = scattering angle of the electron wrt. incoming electron direction W= stereo angle 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

Charged particles in matter (normal Rutherford Scattering: e- on nucleus, change of variables)
Change variables from W to q2 (q = momentum transfer to electron) to get to frame independent form Pelectron,in P’electron,out q

Charged particles in matter (normal Rutherford Scattering: e- on nucleus, change of variables)

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 =megV q2 is frame independent non-relativistic this is obvious (do it at home) Energy transfer to the electron is defined via: DE=n=|q2|/2me  dn/dq2=1/2me relativistic need to define q as 4-momentum transfer, but we assume non relativistic for Rutherford anyway.

Charged particles in matter (From inverse Rutherford scattering to the Bethe-Bloch Formula)
|q2|=2nme Above is crossection for a non relativistic heavy particle of charge z to loose energy between n and n+dn 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 number of collissions with electrons in length dx per unit crossection area crossection weighted avg. energy lost per collision

Charged particles in matter (Ionisation and the Bethe-Bloch Formula, simple integral)
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) Rutherford term Mott term If we perform the same transformations (Wq2n) with this crossection and then perform the integral: we get …

A list of limits for nmax follows:
Charged particles in matter (Ionisation and the Bethe-Bloch Formula, Mott integral) Mott term Valid for all charged particles (not limited to heavy particles) nmax can be computed via kinematics of “free” electron since Ebind << Ekin (see Williams problem 11.1 on p.246) A list of limits for nmax follows:

Charged particles in matter (Ionisation and the Bethe-Bloch Formula, nmin)
But what about nmin ? can not assume that e is free for small energy transfers n≠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 n need 2-D integral dn 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]

Charged particles in matter (Ionisation and the Bethe-Bloch Formula, Bethe-Bloch features)
d=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 bg BBF as a Function of bg is nearly independent of M of projectile except for nmax and very weak log dependence in d  if you know p and measure b  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) (off syllabus)

Charged particles in matter (Ionisation and the Bethe-Bloch Formula, variation with bg)
m+ can capture e- Emc = critical energy defined via: dE/dxion.=dE/dxBrem. Broad bg≈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 minimum

Charged particles in matter (Ionisation and the Bethe-Bloch Formula, variation with particle type)
in drift chamber gas P=mgv=mgbc 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 b 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 b.  ct/n ct O P A particle trajectory 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.

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- g 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-g 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 g) 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) g + atom  g + atom dominant at lg>size of atoms Compton scattering Incoherent scattering on electron from atom g + e-bound  g + e-free possible at all Eg > min(Ebind) to properly call it Compton requires Eg>>Ebind(e-) to approximate free e- Photoelectric effect absorption of photon and ejection of single atomic electron g + atom  g + e-free + ion possible for Eg < max(Ebind) + dE(Eatomic-recoil, line width) (just above k-edge)

Photons in matter (Overview-II)
Pair production absorption of g in atom and emission of e+e- pair Two varieties: a) dominant: g + nucleus  e+ + e- + nucleusrecoil b) weak: g + Z*atomic e-  e+ + e- + Z *atomic e-recoil Both variants need: Eg>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)
Lead Carbon R  Rayleigh PE  Photoeffect C  Compton PP  Pair Production on nucleus PPE  Pair Production on atomic electrons PN  Giant Photo-Nuclear dipole resonance 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 2*mec2

Photons in matter (Comparison of Bremsstrahlung and Pair Production)
Ze e- e-* g Bremsstrahlung Ze e-* e- g Pair production Typical Lenth = Pair Production Length L0 Typical Lenth = Radiation Length X0 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 g travels before prob. for non interaction reduced to 1/e P(g)=1/L0*exp(-x/L0) L0=mean free path Very similar Feynman Diagram Just two arms swapped L0=9/7 X0

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