# Advanced Nuclear Physics by imran aziz

Post on 22-Nov-2014

107 views

Embed Size (px)

DESCRIPTION

Advanced Nuclear PhysicsMOHAMMAD IMRAN AZIZ Assistant Professor PHYSICS DEPARTMENT SHIBLI NATIONAL COLLEGE, AZAMGARH (India).aziz_muhd33@yahoo.co.inNuclidesThe total number of protons in the nucleus of an atom is called the atomic number of the atom and is given the symbol Z. The number of electrons in an electrically-neutral atom is the same as the number of protons in the nucleus. The number of neutrons in a nucleus is known as the neutron number and is given the symbol N. The mass numbTRANSCRIPT

<p>Advanced Nuclear PhysicsMOHAMMAD IMRAN AZIZAssistant ProfessorPHYSICS DEPARTMENTSHIBLI NATIONAL COLLEGE, AZAMGARH (India).aziz_muhd33@yahoo.co.inaziz_muhd33@yahoo.co.inThe 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..NuclidesNuclear constituents and their propertiesaziz_muhd33@yahoo.co.inwhy electrons cannot exist inside a nucleus:This can be explained mathematically also using Heisenberg's uncertainty principle as followsaziz_muhd33@yahoo.co.inConfinement CalculationElecttron 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 SpinThenuclearspinsforindividualprotonsandneutrons parallelsthetreatmentofelectronspin,withspin1/2 andanassociatedmagnetic moment.Themagnetic momentismuchsmallerthanthatoftheelectron.For thecombinationneutronsandprotonsintonuclei,the situation is more complicated.aziz_muhd33@yahoo.co.inIt is common practice to represent the total angular momentum of a nucleus by the symbol I and to call it "nuclear spin". For electrons in atoms we make a clear distinction between electron spin and electron orbital angular momentum, and then combine them to give the total angular momentum. But nuclei often act as if they are a single entity with intrinsic angular momentum I. Associated with each nuclear spin is a nuclear magnetic moment which produces magnetic interactions with its environment. A characteristic of the collection of protons and neutrons (which are fermions) is that a nucleus of odd mass number A will have a half-integer spin and a nucleus of even A will have integer spin. The suggestion that the angular momenta of nucleons tend to form pairs is supported by the fact that all nuclei with even Z and even N have nuclear spin I=0. The half-integer spins of the odd-A nuclides suggests that this is the nuclear spin contributed by the odd neutronaziz_muhd33@yahoo.co.inNuclear 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 momentumaziz_muhd33@yahoo.co.inElectric Quadrupole Moments of Nucleiaziz_muhd33@yahoo.co.inThe 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.aziz_muhd33@yahoo.co.inGenerally, 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 = 2For free protons and neutrons with spin I =1/2, the magnetic moments are of the formTheprotong-factorisfarfromthegS =2fortheelectron,andeven theunchargedneutronhasasizablemagneticmoment!Forthe neutron,thissuggeststhatthereisinternalstructureinvolvingthe movementofchargedparticles,eventhoughthenetchargeofthe neutroniszero.Ifg=2wereanexpectedvaluefortheprotonand g=0wereexpectedfortheneutron,thenitwasnotedbyearly researchersthatthetheprotong-factoris3.6unitsaboveits expected value and the neutron value is 3.8 units below its expected value.Thisapproximatesymmetrywasusedintrialmodelsofthe magneticmoment,andinretrospectistakenasanindicationofthe internalstructureofquarks inthestandardmodeloftheproton and neutronaziz_muhd33@yahoo.co.inwhereProton: g = 5.5856912 +/- 0.0000022Neutron: g = -3.8260837 +/- 0.0000018aziz_muhd33@yahoo.co.inNuclide Nuclear spin IMagnetic momentin Nn 1/2 -1.9130418p 1/2 +2.79284562H(D) 1 +0.857437617O 5/2 -1.8927957Fe 1/2 +0.0906229357Co 7/2 +4.73393Nb 9/2 +6.1705Note that the maximum effective magnetic moment of a nucleus in nuclear magnetons will be the g-factor multiplied by the nuclear spin. For a proton with g = 5.5857 the quoted magnetic moment is = 2.7928 nuclear magnetons. Nuclei Parameters of nuclei Strong Interaction Binding Energy Stable and Unstable Nuclei Liquid-Drop Model Numerous Applications:nuclear power applicationsinmedicine,biologyand chemistry evolution of stars and the Universe nuclear weaponsaziz_muhd33@yahoo.co.inSize of Nuclei and Rutherford Scattering1/30R R A 1501.2 10 1.2 R m fm R the fitting parameterCalculations were strictly classical. However, because of theCoulombinteractionbetweenalpha-particlesand nucleus, the result miraculously coincides with the exact quantum-mechanical one (recall the success of the Bohr model for atoms).( )( )1/3238921.2 238 7.4 R U fm fm Geiger, Marsden, Rutherford,1910- depends weakly on A (number of nucleons in the nucleus) -particles: bare He nucleiScatteringpatternwasconsistentwiththat expectedforscatteringofparticlesby pointlikeobjects havingachargeof+79e(the chargeofthegoldnucleus).Thisallowed Rutherfordtoputanupperlimitonthesizeof the nucleus ( nDpD -> pD If had potential could solve Schrod. Eq. Dont 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 well skip) N-N interactions are limited (at high A) due to Pauli exclusion.p + n -> p + n only if state is availableInfinite Radial Well Radial part of Scrod Eq</p>
<p>Easy to solve if l=0For L>0, angular momentum term goes to infinity at r=0. Reduces effective wavelength, giving higher energyGo 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 + ++ </p>
<p>]]] 2 222222 214md ud rV rml lru E uu r r R r P r u( )( )( ) ( ) ( ) u k r knaEpmkmh nm a s i n( ) ( ) 22 2 2 82 2 22Angular part If V(r) then can separate variables (r,,) = R(r)Y( ,) have spherical harmonics for angular wave function Angular momentum then quantized like in Hydrogen (except that L>0 for n=1, etc) Energy doesnt 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,nL l l L ml m l l n r q u a n t u mZ2 210 1 2 + ( ), , #. .L,S,J Coupling: Atoms vs NucleiATOMS: If 2 or more electrons, Hunds 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=0leftover odd p (or n) will have two possible J = L + orJ = L 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 + JpAtom called LS coupling nuclei called jj Note that magnetic moments add differently as different g-factor for p,nSpin Coupling in Nuclei All nucleons in valence shell have same J Strong pairing causes Jz antiparallel(3 and -3)spin wavefunction = antisymmetricspace 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 formulaEnergy Levels in NucleiLevels in ascending order (both p,n) State nLdegeneracy(2j+1) sum 1S1/210 22*** 1P3/211 46 1P1/21 128*** 1D5/212 6 14 2S1/220 216 1D3/2 12 4 20*** 1F7/21 3828*** 2P3/22 1432 1F5/21 3638 2P1/22 1240 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 NumbersLarge energy gaps between some filled shells and next (unfilled) shell give larger dE/A and more made during nucleosnthesis in stars # protons#neutrons 2He2 He-4 6C6 C-12 8O8 O-1620 Ca 2028 Ni28 Cr-52(24,28)50 Sn50 Ni-78 82Pb82126136 Ni-78 (2005) doubly magic. Whileit 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 stableNuclear Magnetic MomentsProtons and neutrons are made from quarks and gluons. Their magnetic moment is due to their spin andorbital angular momentumThe g-factors are different than electrons. orbital, p=1 and n=0 as the neutron doesnt have chargespin, g for proton is 5.6 and forneutron 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 -1/3A neutron is made from 1 up and 2 down and has more negative charge/momentsNo theory which explains hadronic magnetic moments orbital and spin magnetic moments arent aligned, need to repeat the exercise in atoms (Zeeman effect) to get values for the z-component of the moment + + L SNl S Npg L g Sem( )2Nuclear Cross SectionsDefinition of Cross SectionWhy its useful.Breit-Wigner ResonancesRutherford Scatteringaziz_muhd33@yahoo.co.inCross-SectionsWhy concept is importantLearn about dynamics of interaction and/or constituents (cf Feynmans watches).Needed for practical calculations.Experimental DefinitionHow to calculate Fermi Golden RuleBreit-Wigner ResonancesQM calculation of Rutherford Scatteringaziz_muhd33@yahoo.co.inDefinition of a+bxEffective area for reaction to occur is Beam adxNaNa(0) particles type a/unit time hit target bNb atoms b/unit volumeNumber /unit area= Nb dxProbability interaction = NbdxdNa=-Na Nb dx Na(x)=Na(0) exp(-x/ ) ; =1/(Nb ) aziz_muhd33@yahoo.co.inReaction RatesNa beam particles/unit volume, speed vFlux F= Na vRate/target b atom R=FThin target x / taziz_muhd33@yahoo.co.inBreit-Wigner Line Shape -3) E E ( P H2) t ( a4 / ) E E (H) t ( an m2mn2n2 2n m2mn2n + 4) E E (12) E E ( P22n mn m+ Normalised Breit-Wigner line shapeQ: where have you seen this shape before?We will see this many times in NP and PP.aziz_muhd33@yahoo.co.inBreit-Wigner ResonanceImportant in atomic, nuclear and particle physics.Uncertainty relationshipDetermine lifetimes of states from width., =FWHM; ~ t E ~ t E ~ t E / ~aziz_muhd33@yahoo.co.inFermi Golden RuleWant to be able to calculate reaction rates in terms of matrix elements of H.Warning: We will use this many times to calculate but derivation not required for exams, given here for completeness.aziz_muhd33@yahoo.co.inDiscrete Continuum Decays to a channel i (range of states n). Density of states ni(E). Assume narrow resonancedE ) E E ( P ) E ( n H2P0 i20 i i) E ( n H2P0 i20 i iTotal i i TotaliiR P R ; R ; P ) E ( n H2R0 i20 iiiaziz_muhd33@yahoo.co.inCross SectionBreit Wigner cross section.Definition of and flux F:) r . k i exp( V2 / 1 F Rv V F1 23k 4) 2 ( V) k ( n vdkdE;m 2) k (E2 vk 4) 2 ( V) E ( n23aziz_muhd33@yahoo.co.inBreit-Wigner Cross SectionCombine rate, flux & density states ( )4 / ) E E (E21k 4 Vv ) 2 (vV2 20 1f 1 i23 + </p>
<p>,`</p>
<p>.|</p>
<p>,`</p>
<p>.|f2 20 1201f2o4 / ) E E (H) t ( a R + ) E ( n H 2 ) E (210 i ( )4 / ) E E (E) E ( n121R2 20 1f 1 i + aziz_muhd33@yahoo.co.inBreit-Wigner Cross Section4 / ) E E ( k2 20 1f i2n + 16O 17Oaziz_muhd33@yahoo.co.inLow Energy Resonancesn + Cd total cross section.Cross section scales ~ 1/E1/2 at low E.B-W: 1/k2 and ~n(E)~kaziz_muhd33@yahoo.co.inRutherford Scattering 11 c ;c 4 e;rZ Z) r ( V022 1 ) r . k i exp( V ; ) r . k i exp( Vf2 / 1f i2 / 1i r d ) r . k i exp(rZ Z) r . k i exp( V H3f2 1i1fi</p>
<p>,`</p>
<p>.|f ik k q r dr) r . q i exp(Z Z V H32 11fi cos d dr rr) cos iqr exp(2 Z Z V H22 11fiaziz_muhd33@yahoo.co.inRutherford Scattering 2dr riqr) iqr exp( ) iqr exp(2 Z Z V H222 11fi a ) a / r exp( ); r ( xVdr r iq a r iq aiqZ ZV H fi + ) / 1 exp( ) / 1 exp(22 11]]]</p>
<p> + iq a / 1 1iq a / 1 1iq2 Z ZV H2 11fiaziz_muhd33@yahoo.co.inRutherford Scattering 3 Fermi Golden Rule:f2fidEdnH2Rv / 1dEdp;dEdpdpdndEdn;4dhVp 4dpdn32 ) 2 2 ( ;) 2 () (32 h dvV pE nv V F ; F / R1 vV) 2 ( vV pVq4 Z Z 2dd32222 1 4 222 12) ( 4q vZ Z pdd aziz_muhd33@yahoo.co.inRutherford Scattering 4) 2 / ( sin p 4 ) cos 1 ( p 2 ) p p ( q2 2 2 2f i2 ) 2 / ( sin 4) (4 2 222 1 v pZ ZddCompare with experimental data at low energyQ: what changes at high energy ?pipfaziz_muhd33@yahoo.co.inLow Energy ExperimentScattering of on Au & Ag agree with calculation assuming point nucleusSin4( /2)dN/dcosaziz_muhd33@yahoo.co.inHigher Energy Deviation from Rutherford scattering at higher energy determine charge distribution in the nucleus. Form factors is F.T. of charge distribution.Electron - Goldaziz_muhd33@yahoo.co.in Induced Fission(required energy)Neutrons Ef=Energy needed to penetrate fission barrier immediately 6-8MeVA=238NeutronNucleus Potential Energy during fission [MeV] Esep6MeV per nucleon for heavy nucleiVery slow naziz_muhd33@yahoo.co.in Induced Fission(required energy) Spontaneous fission rates low due to high coulomb barrier (6-8 MeV @ A240)Slow neutron releases Esep as excitation into nucleus Excited nucleus has enough energy for immediate fission if Ef - Esep>0 We call this thermal fission (slow, thermal neutron needed) But due to pairing term even N nuclei have low Esepfor additional n</p>
<p>odd N nuclei have high Esep for additional n Fission yield in n -absorption varies dramatically between odd and even N aziz_muhd33@yahoo.co.in Induced Fission(fissile nuclei) Esep(n,23892U) = 4.78 MeV only Fission of 238U needs additional kinetic energy from neutron En,kin>Ef- Esep1.4 MeVWe call this fast fission (fast neutrons needed)Th...</p>

Recommended

View more >