advanced nuclear physics

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Information about advanced nuclear physics

Published on July 23, 2010

Author: MImranaziz


Advanced Nuclear Physics : Advanced Nuclear Physics MOHAMMAD IMRAN AZIZ Assistant Professor PHYSICS DEPARTMENT SHIBLI NATIONAL COLLEGE, AZAMGARH (India). Slide 2: 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 : 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 Slide 4: 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 : 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. Slide 6: 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 : 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 : 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. Slide 9: 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 : 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 where Proton: g = 5.5856912 +/- 0.0000022 Neutron: g = -3.8260837 +/- 0.0000018 Slide 11: 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 m = 2.7928 nuclear magnetons. Nuclei : 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 : Size of Nuclei and Rutherford Scattering 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). 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 (<3?10-14m 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 : Nuclear Mass 6 protons + 6 neutrons Atomic Mass (the mass of a neutral atom): - attraction between nucleons in the nucleus and between electrons and the nucleus Mass Unit 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: - we can neglect Uel.-nucl. and introduce a convenient mass unit: Nuclear Density : 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” : 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. Structure of Matter : 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 Binding energy: mass deficit 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 n p p n n n Binding Energy curve : 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 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: Binding energy per nucleon (EB/A), MeV Mass number, A Liquid-Drop Model : Liquid-Drop Model A “semi-classical” model of the nucleus: describes reasonably well the dependence EB(A): ? - charge density Liquid-Drop Model (cont’d) : Liquid-Drop Model (cont’d) Limitations of Liquid-Drop Model : Limitations of Liquid-Drop Model Maria Goeppert-Mayer, J.H.D.Jensen Stable Nuclei : 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? : 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 : protons neutrons energy 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). 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: 0 r Why N ? Z for light nuclei : 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 energy protons neutrons 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. Why N > Z for heavy nuclei : 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 energy . . . . . protons neutrons energy . . . . . The proton-neutron disbalance becomes more pronounced with increasing Z. Nuclear Masses and Sizes : 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 : 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 : 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 : 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 : 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 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 : 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 : 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 : 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 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 : 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) 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 : 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) “Charge symmetry” More charge symmetry : 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 : 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 : Charge Independence 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 Charge Independence : 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 : 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 : 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. 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 : 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 : 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 : Asymmetry Term Binomial expansion keep lowest term in y/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 (e) note! linear terms cancel Asymmetry term : 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 : 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. Pairing Term : Pairing Term Measure that the Pairing effect smaller for larger A Phenomenological*) fit to A dependence gives A-1/2 *) 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 : Semi Empirical Mass Formula Put everything together: 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 : Semi Empirical Mass Formula Binding Energy vs. A for beta-stable odd-A nuclei Iron Semi Empirical Mass Formula : 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 : 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 : 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 : 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 : 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 : 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 : 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 : 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 : 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 : Nuclear Cross Sections Definition of Cross Section Why its useful. Breit-Wigner Resonances Rutherford Scattering Cross-Sections : 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 s Fermi Golden Rule Breit-Wigner Resonances QM calculation of Rutherford Scattering Definition of s : Definition of s a+b?x Effective area for reaction to occur is s 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 = s Nbdx dNa=-Na Nb dx s Na(x)=Na(0) exp(-x/l) ; l=1/(Nb s) Reaction Rates : 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 : 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 : Breit-Wigner Line Shape - 2 For Breit-Wigner Line Shape -3 : 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 : Breit-Wigner Resonance Important in atomic, nuclear and particle physics. Uncertainty relationship Determine lifetimes of states from width. , G=FWHM; Fermi Golden Rule : 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 : Discrete ? Continuum Decays to a channel i (range of states n). Density of states ni(E). Assume narrow resonance Cross Section : Cross Section Breit Wigner cross section. Definition of s and flux F: Breit-Wigner Cross Section : Breit-Wigner Cross Section Combine rate, flux & density states ? Breit-Wigner Cross Section : Breit-Wigner Cross Section n + 16O? 17O Low Energy Resonances : 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 : Rutherford Scattering 1 Rutherford Scattering 2 : Rutherford Scattering 2 Rutherford Scattering 3 : Rutherford Scattering 3 Fermi Golden Rule: Rutherford Scattering 4 : Rutherford Scattering 4 Compare with experimental data at low energy Q: what changes at high energy ? pi pf Low Energy Experiment : Low Energy Experiment Scattering of a on Au & Ag ? agree with calculation assuming point nucleus Sin4(q/2) dN/dcosq Higher Energy : 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 Slide 83: Induced Fission(required energy) Neutrons A= 238 Nucleus Potential Energy during fission [MeV] Induced Fission(required energy) : Induced Fission(required energy) Spontaneous fission rates low due to high coulomb barrier (6-8 MeV @ 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) : 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, En,kinthermal=0.1eV @ 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) : 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> nprompt=2.5 <n-delay> td=few s <# delayed n> nd=0.006 Induced Fission (the fission process) : 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+n?Y+g E(prompt g) 6 Subtotal: 178 (good for power production) Delayed (10-10<t<?): Ekin(e from b-decays) 8 E(g following b-decay) 7 Subtotal: 15 (bad, spent fuel heats up) Neutrinos: 12 (invisible) Grand total: 205 Induced Fission(n -induced fission crossections (n,f) ) : 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/vA˜0.8 MeV between odd-even= 23592U and even-even= 23892U unresolved, narrow resonances unresolved, narrow resonances 238U 235U n -Energy Induced Fission((n,f) and (n,g) probabilities in natural Uranium) : Induced Fission((n,f) and (n,g) probabilities in natural Uranium) neutron absorbtion probabilit per 1 mm Induced Fission(a simple bomb) : Induced Fission(a simple bomb) mean free path for fission n: 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 l=3cm ? lmp=v6*3cm˜7cm in tmp=10-8 s Uranium mix 235U:238U =c:(1-c) rnucl(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) : Induced Fission(a simple bomb) 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 Induced Fission(a simple bomb) : Induced Fission(a simple 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) : 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 Fission Reactors(two ways out) : 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) : 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) : 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) : 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 Reactors (Reactor control) : Fission Reactors (Reactor control) 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) several sources of delayed n typical lifetimes t˜O(1 sec) Fraction nd ˜ 0.6% off syllabus Fission Reactors (Reactor control) : Fission Reactors (Reactor control) 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) : 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: off syllabus Fission Reactors (Thermal Stability) : 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) : Fission Bombs (fission fuel properties) ideal bomb fuel = pure 239Pu Fission Bombs (where to get Pu from? Sainsbury’s?) : Fission Bombs (where to get Pu from? Sainsbury’s?) 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) : 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) : 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 g 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) : 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 Slide 107: Slide 108: Slide 109: Fission Reactors (Thermal vs. Fast) : 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) : 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) : 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) : 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) : 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) : 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) 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) : 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 Slide 119: 15-20 cm 1-2 m O(70,000) rpm ? Vmax˜1,800 km/h = supersonic! & gmax=106g ? difficult to build! Fission Reactors (enrichment) : Fission Reactors (enrichment) Nuclear Fusion,as a source of stellar energy : 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 + 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). Slide 122: 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) : 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. Sun red dwarf Sirius A proton cycle carbon-nitrogen cycle T, K 107 109 105 luminosity 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). The carbon-nitrogen cycle: a “catalyst” Explosive Nucleosynthesis (Elements Heavier than Iron) : 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 e- n n n r-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. Slide 125: 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 Abundance relative to Silicon (=106) a Decay Theory : 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 : r nucleus inside barrier (negative KE) small flux of real a a Decay Theory I II III r=t r=R see also Williams, p.85 to 89 QM Tunnelling through a square well (the easy bit) : QM Tunnelling through a square well (the easy bit) 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] in regions I and III in region II 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 ! a-decay : a-decay Neutrons Protons Alphas Tunnelling in a-decay : 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 : 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 Tunnelling in a-decay : 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 : 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 : 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 : a-decay experimental tests We expect: ln(decay rate) Fermi b Decay Theory : 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.166 - ff but also check that you understand Williams p. 292 - ff Fermi Theory : 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 : 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 : Fermi Theory ? Kurie Plot FGR to get a decay rate and insert previous results: let’s plot that from real data Electron Spectrum : 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 Simple Spectrum Kurie Plot : 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 Selection Rules : Selection Rules Fermi 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 : 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 : 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 : 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 : Anti-neutrino Discovery Phase space factor: Neglect neutron recoil: Combine with FGR The Cowan & Reines Experiment : The Cowan & Reines Experiment for inverse b-decay @ En ~ 1MeV ? s ~10-47 cm2 Pauli’s prediction verified by Cowan and Reines. 1 GW Nuclear Reactor PMT H20+CdCl2 Liquid Scint. Shielding original proposal wanted to use a bomb instead! Liquid Scint. PMT all this well under ground to reduce cosmic rays! Parity Definitions : 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 : 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 Slide 150: 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 <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 Y under parity Q: Is Parity Conserved In Nature? : 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 : 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 ~100% The Wu Experiment : 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! scintillator signal electron signal shows asymmetry of the electron distribution see also Burcham & Jobes, P.370 sample warms up ? asymmetry disappears Interpreting the Wu Experiment : 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 : 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 : 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 : 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 both b decay leaves Tc in excited state. MeV MeV interesting meta stable state Introduction : Introduction Particle Ranges If smooth energy loss via many steps (i.e. ionisation from light ions) ? sharply defined range, useful for rough energy measurement Sometimes several types of processes happen (i.e. high energy electrons) ? mixed curves, extrapolated maximum range 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) : 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

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