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# APPL ATOM & NUCL PHYS I MANE 4410

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This 14 page Class Notes was uploaded by Antone Mann on Monday October 19, 2015. The Class Notes belongs to MANE 4410 at Rensselaer Polytechnic Institute taught by Staff in Fall. Since its upload, it has received 30 views. For similar materials see /class/224921/mane-4410-rensselaer-polytechnic-institute in Mechanical and Aerospace Engineering at Rensselaer Polytechnic Institute.

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Date Created: 10/19/15

MANE 44106930 Applied Atomic amp Nuclear Physics Fall 2006 Lecture 6 Chapter III The Nuclear Force A The Neutron Proton System Bound State of the Deuteron The investigation of this nuclear force has turned out to be a truly monumental task Perhaps more manhours of work have been devoted to it than any other scienti c question in the history of mankind 7 B L Cohen Concepts of Nuclear Physics p 32 A direct way to study nuclear forces is to consider the simplest possible nucleus which is a twonucleon system 7 the deuteron H2 nucleus composed of a neutron and a proton We will discuss two properties of this system using elementary quantum mechanics the bound state of the nucleus this lecture and the scattering of a neutron by a proton next lecture The former problem is an application of our study of bound states in threedimensions while the latter is a new application of solving the timeindependent Schrodinger wave equation From these two problems a number of fundamental features of the nuclear potential will emerge Bound State of the Deutertm The deuteron is the only stable bound system of two nucleons 7 neither the dineutron nor the diproton are stable Experimentally it is known that the deuteron exists in a bound state of energy 223 Mev This energy is the energy of a gamma ray given off in the reaction where a thermal neutron is absorbed by a hydrogen nucleus 1 2 nH gt H V 223 Mev The inverse reaction of using electrons of known energy to produce external bremsstrahlung for y 2 2 n reaction on H also has been studied Besides the ground state no stable excited states of H have been found however there is a virtual state at N 230 Mev Suppose we combine the information on the boundstate energy of the deuteron with the boundstate solutions to the wave equation to see what we can learn about the potential well for neutronproton interaction Again we will assume the interaction between the two nucleons is of the form of a spherical well V r V0 r lt r0 0 r gt r0 Page 1 of4 MANE441069307Fall 2006 We ask what is the energy level structure and what values should V0 and r0 take in order to be consistent with a bound state at energy EB 223 Mev There is good reason to believe that the deuteron ground state is primarily ls n 1 Z 0 First the lowest energy state in practically all the model potentials is an sstate Secondly the 2 magnetic moment of H is approximately the sum of the proton and the neutron moments indicating that gquot and 3p are parallel and no orbital motion of the proton relative to the neutron This is also consistent with the total angular momentum of the ground state being I 1 We therefore proceed by considering only the Z 0 radial wave equation Em Vrur Eur 6 1 m r d 2 where E EB and m is the neutron or proton mass In view of our previous discussions of boundstate calculations we can readily write down the interior and exterior solutions ur AsinKr K mV0 EB 2 m r lt r0 62 ur 32 K mEB h r gt r0 62 Applying the boundary condition at the interface we obtain the relation between the potential well parameters depth and width and the boundstate energy EB V E 1 Z KcotKr0 K or tanKr0 J 64 B To go further we can consider either numerical or graphical solutions as discussed before or approximations based on some special properties of the nuclear potential Let us consider the latter option Suppose we take the potential well to be deep that is V0 gtgt EB then the RHS right hand side of 64 is large and we get an approximate value for the argument of the tangent K row TE 2 65 Then KN mV0 172392r0 Page 2 of4 MANE441069307Fall 2006 2 2 er mtg h 1Msvrbzm m m We see that a knuwledge hf Ea alluws us te aetemhe the pheauet hf m2 and hut va aha h n r V we will and that re ZF puthhg thsmtu a a we get vs 5 Maxi whchjusn es uurtakmg vs te mg 1 he MN Fig 1 The nuclear petehtaal fur the heutaehepmteh system m the farm hf a spheneal wen hf depth vs aha mathh Eats the buundsstate margy quhe deutemn expat at the mtehfaee the quantity Io must he shghdy geezer than 711 z amehe accurate estamate gves 1mquot msteaa emquot xfwevmte K ZTIX7Zg then the effemzve wavelmth39 x is appmxlmately 4F Whth suggests that mueh ef the wave fuhetaems hut m the mtenurregun Therelzxauun eehstaht eh decay length m Lhamtenurregun ean he esumated as 43F x 7 ImEl Ths means that the tvm hueleehs m 112 spend alarge amun quhexr tame at h gt rm the hegmh hf hegatave kmetae margy thaws elasseany fehhtaaeh page 3 an MANEAMEIEEDA J 2cm On class Example In classical mechanics a particle incident upon a potential with range r0 at an impact parameter b would be scattered if b lt r0 but if b were greater than r0 then there would be no interaction Use this simple picture to show that in the scattering of a neutron at low energy by which we mean kr0 ltlt1 withE fisz 2m only the s wave interaction is important Take r0 15 F what is the range of neutron energy where this approximation is valid Semi classical representation of a collision impact parameter Let us View the n p collision from a classical point of View where we can Visualize a de nite impact parameter b A collision with an impact parameter b would have an angular momentum of the two particle system equals to Lbph where pmv 1 II I I7 b i I l proton Now in order to have any appreciable re ection the neutron has to approach within the range of interaction with the proton Let us say that the range of n p interaction is V0 then in order to have scattering one has to have b S r0 2 Therefore from 1 amp 2 rep 2 M or Z S r0 rok 3 for 1 ltlt1 we can have only 3 0 i9 only the spartial wave would contribute to the scattering Now take V0 15 F Z Z 3 E ltlt 2 6626gtlt10 3quot2 27r2 925Mev kr lt2ltml 2mr02 2 X 167 gtlt103927 X 15 gtlt1039152 X 16 gtlt103919 0 E ltlt 925Mev Page 4 of4 MANE441069307Fa11 2006 MANE 44106390 Applied Atomic amp Nuclear Physics Fall 2006 Lec 8 10506 References M A Preston Physics of the Nucleus AddisonWesley Reading 1962 E Segre Nuclei armParticles W A Benjamin New York 1965 Chap X C Neutron Proton Scattering We continue the study of the neutronproton system by taking up the wellknown problem of neutron scattering in hydrogen The scattering cross section has been carefully measured to be 204 barns over a wide energy range Our intent is to apply the method of phase shifts summarized in the preceding lecture to this problem We see very quickly that the swave approximation the condition of interaction at low energy is very well justified in the neutron energy range of l 1000 eV The scatteringstate solution with E gt 0 gives us the phase shift or equivalently the scattering length This calculation yields a cross section of 23 barns which is considerably different from the experimental value The reason for the discrepancy lies in the fact that we have not taken into account the spindependent nature of the np interaction The neutron and proton spins can form two distinct spin configurations the two spins being parallel triplet state or antiparallel singlet each giving rise to a scattering length When this is taken into account the new estimate is quite close to the experimental value The conclusion is therefore that np interaction is spindependent and that the anomalously large value of the hydrogen scattering cross section for neutrons is really due to this aspect of the nuclear force For the scattering problem our task is to solve the radial wave equation for swave for solutions with E gt 0 The interior and exterior solutions have the form u r B sinK 39r r lt r0 81 and u r C sinkr 50 r gt r0 82 where K 39 W h and k M h Applying the interface condition we obtain K cotK39 r0 k cotkr0 50 83 which is the relation that allows the phase shift to be determined in terms of the potential parameters and the incoming energy E We can simplify the task of estimating the phase shift by recalling that it is simply related to the scattering length by 50 ak cf B25 Assuming the scattering length a is larger than r0 we see the RHS of 83 is approximately k cotc30 For the LHS we will ignore E relative to V0 in K39 and at the same time ignore EB relative to V0 in K Then K N K and the LHS Page 1 oflO MANE44106930 Fall 2006 can be set equal to K by virtue of 64 Notice that this series of approximations has enabled us to make use of the dispersion relation in the boundstate problem 64 for the scattering calculation As a result 83 becomes k cotc o K 84 which is a relation between the phase shift and the binding energy Once the phase shift 50 is known the differential scattering cross section is then given by 1323 039 a 1 k2 sin2 50 85 A simple way to make use of 84 is to note the trigonometric relation sin2 x l l cot 2x or 1 l 86 lcot260 1K2k2 2 sin 60 2 1 W mEEB mEB Thus 66 87 The last step follows because we are mostly interested in estimating the scattering cross section in the energy range 1 100 eV Putting in the numerical values of the constants h 1055 x 10727erg sec m 167 x 10724 g and EB 223 x 106 x 16 x 10712 ergs we get 04739chzmEBNZB bams 88 This value is considerably lower than the experimental value of the scattering cross section of H1 204 barns as shown in Fig 81 mm mm in mm m u a n voo 39 E Lav Fig 81 Experimental neutron scattering cross section of hydrogen showing a constant value of 204 barns over a wide range of neutron energy The rise in the cross section at energies below N 01 Page 2 of 10 MANEV44106930 77 Fall 2006 eV can be explained in terms of chemical binding effects in the scattering sample The explanation of this wellknown discrepancy lies in the neglect of spindependent effects It was suggested by E P Wigner in 1933 that neutronproton scattering should depend on whether the neutron and proton spins are oriented in a parallel configuration the triplet state total spin angular momentum equal to h or in an antiparallel configuration singlet state total spin is zero In each case the interaction potential is different and therefore the phase shifts also would be different Following this idea one can write instead of 87 l l 3 06 k ZZsm2 505 Zs1n2 502 89 We have already mentioned that the ground state of the deuteron is a triplet state at E EB If the singlet state produces a virtual state of energy E E then 88 would become 2 039 z i i L 810 m E B E Taking a value of E N 70 keV we find from 810 a value of 204 barns thus bringing the theory into agreement with experiment In summary experimental measurements have given the following scattering lengths for the two types of np interactions triplet and singlet configurations and their corresponding potential range and well depth Interaction Scattering length a F ro F Vo MeV np triplet 54 2 36 np singlet 237 N 25 18 Notice that the scattering length for the triplet state is positive while that for the singlet state is negative This illustrates the point of Fig 73 As a final remark we note that experiments have shown that the total angular momentum nuclear spin of the deuteron ground state is I l where I L s with L being the orbital angular momentum and g the intrinsic spin s It is also known that the ground state is mostly ls Z 0 therefore for this state we have S l neutron and proton spins are parallel We have seen from Lec 6 that the deuteron ground state is barely bound at E3 223 Mev so all the higher energy states are not bound The ls state with S 0 neutron and proton spins antiparallel Page 3 of10 MANE44106930 Fall 2006 is a Virtual state it is unbound by N 60 Key An important implication is that nuclear interaction varies with S or nuclear forces are spin dependent E ects 0f Pauli Exclusion Principle One might ask why are the dineutron and the diproton unstable The answer lies in the indistinguishability of particles and the Exclusion Principle no two fermions can occupy the same state Consider the two electrons in a helium atom Their wave function may be written as lI12 111 1 Na 2 sin klr1 sinkzr2 811 1 A r2 where w1r is the wave function of electron 1 at r But since we cannot distinguish between electrons 1 and 2 we must get the same probability of finding these electrons if we exchange their positions or exchange the particles 2 w122w21 e w12iw21 For fermions electrons neutrons protons we must choose the sign because of FermiDirac statistics the wave function must be antisymmetric under exchange Thus we should modify 811 and write lIL2 111111 lI2 z2 112 LOU1amp2 E 11 lIL2 111111 lI2 2 112 1 lI1 2 E 11 If we now include the spin then an acceptable antisymmetric wave function is P12 w W T so that under an interchange of particles 1 lt gt 2 1p 12 71p 21 This corresponds to S 1 symmetric state in spin space But another acceptable antisymmetric wave function is T 12 w 2 W l mm which corresponds to S 0 antisymmetric state in spin For the symmetry of the wave function in configurational space we recall that we have 2r r 1l N PZ 00596W Page 4 of10 MANE44106930 Fall 2006 which is even odd if X is even odd Thus since I has to be antisymmetric one can have two possibilities even S 0 space symmetric spin antisymmetric odd S 1 space antisymmetric spin symmetric These are called T 1 states T is isobaric spin available to the np nn pp systems By contrast states which are symmetric T 0 are even S 1 Z odd S 0 These are available only to the np system for which there is no Pauli Exclusion Principle The ground state of the deuteron is therefore a T 0 state The lowest T 1 state is Z 0 S 0 As mentioned above this is known to be unbound E N 60 Kev We should therefore expect that the lowest T 1 state in nn and pp to be also unbound ie there is no stable dineutron or diproton Essential Features of Nuclear Forces In closing we summarize a number of important features of the nucleonnucleon interaction potential several of which are basic to the studies in this class Meyerhof Chap 6 1 There is a dominant shortrange part which is central and which provides the overall shellmodel potential 2 There is a part whose range is much smaller than the nuclear radius which tends to make the nucleus spherical and to pair up nucleons 3 There is a part whose range is of the order of the nuclear radius which tends to distort the nucleus 4 There is a spinorbit interaction 5 There is a spinspin interaction 6 The force is charge independent Coulomb interaction excluded 7 The force saturates Page 5 of10 MANE44106930 Fall 2006 1quot XL X Fm 82 Two spins of call combine in four possible ways three of which yield triplet parallel spin 3mm whereas the fourth possibility corresponds to the ainglet antiparallel spin state Page 6 0f 10 MANE 44106930 Fa11 2006 a barns o lllllllll I I l 102 103 104 105 105 107 Neutron kinetic energy EV Figure 83 The neutron proton scattering cross section at low energy Data taken from a review by R K Adair Rev Mod Phys 22 249 1950 with additional recent results from T L Houk Phys Rev C 3 1886 1970 F 0 Reading Materials Pauli exclusion principle From Wikipedia the free encyclopedia The Pauli exclusion principle is a quantum mechanical principle formulated by Wolfgang Pauli in 1925 This principle is signi cant for the fact that it explains why matter occupies space exclusively for itself and does not allow other material objects to pass through it at the same time allowing lights and radiations to pass It states that no two identical fermions may occupy the same quantum state simultaneously A more rigorous statement of this principle is that for two identical fermions the total wavefunction is antisymmetric It is one of the most important principles in physics primarily because the three types of particles from which ordinary matter is madeielectrons protons and neutronsiare all subject to it consequently all material particles exhibit spaceoccupying nature The Pauli exclusion principle underpins many of the characteristic properties of matter from the largescale stability of matter to the existence of the periodic table of the elements The Pauli exclusion principle follows mathematically from the de nition of wave function for a system of identical particles it can be either symmetric or antisymmetric depending on particles39 spin Particles with antisymmetric wave functions are called fermionsiand obey the Pauli Page 7 of 10 MANE441069307Fall 2X16 exclusion principle Apart from the familiar electron proton and neutron these include the neutrinos the quarks from which protons and neutrons are made as well as some atoms like helium3 All fermions possess quothalfinteger spinquot meaning that they possess an intrinsic angular momentum whose value is h h27239 Planck39s constant divided by 271 times a halfinteger 12 32 52 etc In the theory of quantum mechanics fermions are described by quotantisymmetric statesquot which are explained in greater detail in the article on identical particles Particles with integer spin have a symmetric wave function and are called bosons in contrast to fermions they may share the same quantum states Examples of bosons include the photon and the W and Z bosons Connection to quantum state symmetry The Pauli exclusion principle was originally formulated as an empirical principle It was invented by Pauli in 1924 to explain experimental results in the Zeeman effect in atomic spectroscopy ferromagnetism and how the periodic table is regulated by the electron structure of atoms well before the 1925 formulation of the modern theory of quantum mechanics by Werner Heisenberg and Erwin Schrodinger However this does not mean that the principle is in any way approximate or unreliable in fact it is one of the most welltested and commonlyaccepted results in physics The Pauli exclusion principle can be derived starting from the assumption that a system of particles occupy antisymmetric quantum states According to the spinstatistics theorem particles with integer spin occupy symmetric quantum states and particles with halfinteger spin occupy antisymmetric states furthermore only integer or halfinteger values of spin are allowed by the principles of quantum mechanics Consequences The Pauli exclusion principle helps explain a wide variety of physical phenomena One such phenomenon is the quotrigidityquot or quotstiffnessquot of ordinary matter fermions the principle states that identical fermions cannot be squeezed into each other Cf Young and bulk moduli of solids hence our everyday observations in the macroscopic world that material objects collide rather than passing straight through each other and that we are able to stand on the ground without sinking through it etc Another consequence of the principle is the elaborate electron shell structure of atoms and of the way atoms share electrons thus variety of chemical elements and of their combinations chemistry An electrically neutral atom contains bound electrons equal in number to the protons Page 8 oflO MANE44106930 Fall 2006 in the nucleus Since electrons are fermions the Pauli exclusion principle forbids them from occupying the same quantum state so electrons have to quotpile on top of each otherquot within an atom For example consider a neutral helium atom which has two bound electrons Both of these electrons can occupy the lowestenergy ls states by acquiring opposite spin This does not violate the Pauli principle because spin is part of the quantum state of the electron so the two electrons are occupying different quantum states However the spin can take only two different values or eigenvalues In a lithium atom which contains three bound electrons the third electron cannot fit into a ls state and has to occupy one of the higherenergy 2s states instead Similarly successive elements produce successively higherenergy shells The chemical properties of an element largely depend on the number of electrons in the outermost shell which gives rise to the periodic table of the elements Astronomy provides another spectacular demonstrations of this effect in the form of white dwarf stars and neutron stars For both such bodies their usual atomic structure is disrupted by large gravitational forces leaving the constituents supported by quotdegeneracy pressurequot alone This exotic form of matter is known as degenerate matter In white dwarfs the atoms are held apart by the degeneracy pressure of the electrons In neutron stars which exhibit even larger gravitational forces the electrons have merged with the protons to form neutrons which produce a larger degeneracy pressure Neutrons are the most quotrigidquot objects known their Young modulus or more accurately bulk modulus is 20 orders of magnitude larger than that of diamond Another physical phenomenon for which the Pauli principle is responsible is ferromagnetism in which the exclusion effect results in exchange energy that induces neighboring electron spins to align whereas classically they would antialign References 0 Griffiths David J 2004 Introduction to Quantum Mechanics 2nd ed Prentice Hall ISBN 013805326X O Liboff Richard L 2002 Introductory Quantum Mechanics AddisonWesley ISBN 0805387145 O Massimi Michela 2005 Pauli39s Exclusion Principle Cambridge University Press ISBN 0521839114 O Tipler Paul Llewellyn Ralph 2002 Modern Physics 4th ed W H Freeman ISBN 0716743450 Page 9 of 10 MANE44106930 Fall 2006 Exam le n scatterin 12 Calculate the neutron scattering cross section of C for thermal neutrons Assume a potential well 13 with depth V0 36 MeV and range r0 l4gtltA F A mass number 12 and consider only the swave contribution Compare your result with the experimental value 6 5 barns and discuss any signi cance Page 10 oflO MANE44106930 Fall 2006

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