22 101 Applied Nuclear Physics Fall 2006 Lecture 10 10 -Books Download

22 101 Applied Nuclear Physics Fall 2006 Lecture 10 10
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Nuclei on the other hand have no center of attraction the nucleons are held together by. their mutual interactions which are much more complicated than Coulomb interactions. All atomic electrons are alike whereas there are two kinds of nucleons This. allows a richer variety of structures Notice that there are 100 types of atoms but more. than 1000 different nuclides Neither atomic nor nuclear structures can be understood. without quantum mechanics,Experimental Basis, There exists considerable experimental evidence pointing to the shell like. structure of nuclei each nucleus being an assembly of nucleons Each shell can be filled. with a given number of nucleons of each kind These numbers are called magic numbers. they are 2 8 20 28 50 82 and 126 For the as yet undiscovered superheavy nuclei the. magic numbers are expected to be N 184 196 272 318 and Z 114 126 164. Marmier and Sheldon p 1262 Nuclei with magic number of neutrons or protons or. both are found to be particularly stable as can be seen from the following data. i Fig 9 1 shows the abundance of stable isotones same N is particularly large. for nuclei with magic neutron numbers,20 28 50 82,Number of Stable Isotones. 0 20 40 60 80 100,Neutron Number N,Figure by MIT OCW Adapted from Meyerhof. Fig 9 1 Histogram of stable isotones showing nuclides with neutron numbers 20 28. 50 and 82 are more abundant by 5 to 7 times than those with non magic neutron numbers. from Meyerhof, ii Fig 9 2 shows that the neutron separation energy Sn is particularly low for. nuclei with one more neutron than the magic numbers where. S n M A 1 Z M n M A Z c 2 9 1, This means that nuclei with magic neutron numbers are more tightly bound.
Neutron Separation Energy Sn Mev,2 6 10 14 18 22 26 30. Neutron Number N of Final Nucleus,Figure by MIT OCW Adapted from Meyerhof. Fig 9 2 Variation of neutron separation energy with neutron number of the final nucleus. M A Z from Meyerhof, iii The first excited states of even even nuclei have higher than usual energies at. the magic numbers indicating that the magic nuclei are more tightly bound. see Fig 9 3,Excitation Energy Mev,0 20 50 82 126,Neutron Number N. Figure by MIT OCW Adapted from Meyerhof, Fig 9 3 First excited state energies of even even nuclei from Meyerhof.
iv The neutron capture cross sections for magic nuclei are small indicating a. wider spacing of the energy levels just beyond a closed shell as shown in Fig. Fast neutron cross section mb,10 30 50 70 90 110 130. Neutron number N,Figure by MIT OCW Adapted from Meyerhof. Fig 9 4 Cross sections for capture at 1 Mev from Meyerhof. Simple Shell Model, The basic assumption of the shell model is that the effects of internuclear. interactions can be represented by a single particle potential One might think that with. very high density and strong forces the nucleons would be colliding all the time and. therefore cannot maintain a single particle orbit But because of Pauli exclusion the. nucleons are restricted to only a limited number of allowed orbits A typical shell model. potential is,1 exp r R a, where typical values for the parameters are Vo 57 Mev R 1 25A1 3 F a 0 65 F In. addition one can consider corrections to the well depth arising from i symmetry energy. from an unequal number of neutrons and protons with a neutron being able to interact. with a proton in more ways than n n or p p therefore n p force is stronger than n n and. p p and ii Coulomb repulsion For a given spherically symmetric potential V r one. can examine the bound state energy levels that can be calculated from radial wave. equation for a particular orbital angular momentum l. h d 2 u l l l 1 h 2,V r u l r Eu l r 9 2, Fig 9 5 shows the energy levels of the nucleons for an infinite spherical well and a.
harmonic oscillator potential V r m 2 r 2 2 While no simple formulas can be given. for the former for the latter one has the expression. E h 3 2 h n x n y n z 3 2 9 3, where 0 1 2 and nx ny nz 0 1 2 are quantum numbers One should. notice the degeneracy in the oscillator energy levels The quantum number can be. divided into radial quantum number n 1 2 and orbital quantum numbers l 0 1. as shown in Fig 9 5 One can see from these results that a central force potential is. able to account for the first three magic numbers 2 8 20 but not the remaining four 28. 50 82 126 This situation does not change when more rounded potential forms are used. The implication is that something very fundamental about the single particle interaction. picture is missing in the description,1i2g3d4s 168 6. 138 1h2f 3p 112 5,2f 1g2d3s 70 4,Energy Mev,30 1h 1f 2p 40 3. 1g 58 1d2s 20 2,1f 34 1p 8 1,Figure by MIT OCW Adapted from Meyerhof 6. Fig 9 5 Energy levels of nucleons in a infinite spherical well range R 8F and b a. parabolic potential well In the spectroscopic notation n l n refers to the number of. times the orbital angular momentum state l has appeared Also shown at certain levels. are the cumulative number of nucleons that can be put into all the levels up to the. indicated level from Meyerhof,Shell Model with Spin Orbit Coupling.
It remains for M G Mayer and independently Haxel Jensen and Suess to show. 1949 that an essential missing piece is an attractive interaction between the orbital. angular momentum and the intrinsic spin angular momentum of the nucleon To take. into account this interaction we add a term to the Hamiltonian H. H V r Vso r s L 9 4, where Vso is another central potential known to be attractive This modification means. that the interaction is no longer spherically symmetric the Hamiltonian now depends on. the relative orientation of the spin and orbital angular momenta It is beyond the scope of. this class to go into the bound state calculations for this Hamiltonian In order to. understand the meaning of the results of such calculations eigenvalues and. eigenfunctions we need to digress somewhat to discuss the addition of two angular. momentum operators, The presence of the spin orbit coupling term in 9 4 means that we will have a. different set of eigenfunctions and eigenvalues for the new description What are these. new quantities relative to the eigenfunctions and eigenvalues we had for the problem. without the spin orbit coupling interaction We first observe that in labeling the energy. levels in Fig 9 5 we had already taken into account the fact that the nucleon has an. orbital angular momentum it is in a state with a specified l and that it has an intrinsic. spin of in unit of h For this reason the number of nucleons that we can put into each. level has been counted correctly For example in the 1s ground state one can put two. nucleons for zero orbital angular momentum and two spin orientations up and down. The student can verify that for a state of given l the number of nucleons that can go into. that state is 2 2 l 1 This comes about because the eigenfunctions we are using to. describe the system is a representation that diagonalizes the square of the orbital angular. momentum operator L2 its z component Lz the square of the intrinsic spin angular. momentum operator S2 and its z component Sz Let us use the following notation to. label these eigenfunctions or representation,l ml s m s Ylml sms 9 5. where Ylml is the spherical harmonic we first encountered in Lec4 and we know it is the. eigenfunction of the square of the orbital angular momentum operator L2 it is also the. eigenfunction of Lz The function sms is the spin eigenfunction with the expected. properties,S 2 sms s s 1 h 2 sms s 1 2 9 6,S z sms ms h sms s ms s 9 7. The properties of sms with respect to operations by S2 and Sz completely mirror the. properties of Ylml with respect to L2 and Lz Going back to our representation 9 5 we. see that the eigenfunction is a ket with indices which are the good quantum numbers. for the problem namely the orbital angular momentum and its projection sometimes. called the magnetic quantum number m but here we use a subscript to denote that it goes. with the orbital angular momentum the spin which has the fixed value of and its. projection which can be 1 2 or 1 2, The representation given in 9 5 is no longer a good representation when the.
spin orbit coupling term is added to the Hamiltonian It turns out that the good. representation is just a linear combination of the old representation It is sufficient for. our purpose to just know this without going into the details of how to construct the linear. combination To understand the properties of the new representation we now discuss. angular momentum addition, The two angular momenta we want to add are obviously the orbital angular. momentum operator L and the intrinsic spin angular momentum operator S since they. are the only angular momentum operators in our problem Why do we want to add them. The reason lies in 9 4 Notice that if we define the total angular momentum as. we can then write,S L j 2 S 2 L2 2 9 9, so the problem of diagonalizing 9 4 is the same as diagonalizing j2 S2 and L2 This is. then the basis for choosing our new representation In analogy to 9 5 we will denote the. new eigenfunctions by jm j ls which has the properties. j 2 jm j ls j j 1 h 2 jm j ls l s j l s 9 10,j z jm j ls m j h jm j ls j mj j 9 11. L2 jm j ls l l 1 h 2 jm j ls l 0 1 2 9 12,S 2 jm j ls s s 1 h 2 jm j ls s 9 13. In 9 10 we indicate the values that j can take for given l and s 1 2 in our discussion. the lower upper limit corresponds to when S and L are antiparallel parallel as shown. in the sketch, Returning now to the energy levels of the nucleons in the shell model with spin orbit.
coupling we can understand the conventional spectroscopic notation where the value of j. is shown as a subscript, This is then the notation in which the shell model energy levels are displayed in Fig 9 6. r 10 12 cm,3p 126 3p3 2,2f 1i 13 2,v 5 1h 9 2,Energy above well bottom Mev. 30 3s 82 3s1 2,20 3 28 2p 1 2,10 1p 8 1d 5 2,1s 2 1p 3 2. Figure by MIT OCW Adapted from Meyerhof, Fig 9 6 Energy levels of nucleons in a smoothly varying potential well with a strong. spin orbit coupling term from Meyerhof, For a given n l j level the nucleon occupation number is 2j 1 It would appear that.
having 2j 1 identical nucleons occupying the same level would violate the Pauli. exclusion principle But this is not the case since each nucleon would have a distinct. value of mj this is why there are 2j 1 values of mj for a given j. We see in Fig 9 6 the shell model with spin orbit coupling gives a set of energy. levels having breaks at the seven magic numbers This is considered a major triumph of. the model for which Mayer and Jensen were awarded the Noble prize in physics For. our purpose we will use the results of the shell model to predict the ground state spin and. parity of nuclei Before going into this discussion we leave the student with the. following comments, 1 The shell model is most useful when applied to closed shell or near closed shell. 2 Away from closed shell nuclei collective models taking into account the rotation. and vibration of the nucleus are more appropriate, 3 Simple versions of the shell model do not take into account pairing forces the. effects of which are to make two like nucleons combine to give zero orbital. angula momentum, 4 Shell model does not treat distortion effects deformed nuclei due to the. attraction between one or more outer nucleons and the closed shell core When. the nuclear core is not spherical it can exhibit rotational spectrum. Prediction of Ground State Spin and Parity, There are three general rules for using the shell model to predict the total angular. momentum spin and parity of a nucleus in the ground state These do not always work. especially away from the major shell breaks, 1 Angular momentum of odd A nuclei is determined by the angular momentum of.
the last nucleon in the species neutron or proton that is odd. 2 Even even nuclei have zero ground state spin because the net angular momentum. associated with even N and even Z is zero and even parity. 3 In odd odd nuclei the last neutron couples to the last proton with their intrinsic. spins in parallel orientation, To illustrate how these rules work we consider an example for each case Consider. the odd A nuclide Be9 which has 4 protons and 5 neutrons With the last nucleon being. the fifth neutron we see in Fig 9 6 that this nucleon goes into the state 1p3 2 l 1. j 3 2 Thus we would predict the spin and parity of this nuclide to be 3 2 For an even. even nuclide we can take A36 with 18 protons and neutrons or Ca40 with 20 protons and. neutrons For both cases we would predict spin and parity of 0 For an odd odd nuclide. we take Cl38 which has 17 protons and 21 neutrons In Fig 9 6 we see that the 17th. proton goes into the state 1d 3 2 l 2 j 3 2 while the 21st neutron goes into the state. 1 f 7 2 l 3 j 7 2 From the l and j values we know that for the last proton the orbital. and spin angular momenta are pointing in opposite direction because j is equal to l 1 2. For the last neutron the two momenta are pointing in the same direction j l 1 2. Now the rule tells us that the two spin momenta are parallel therefore the orbital angular. momentum of the odd proton is pointing in the opposite direction from the orbital angular. momentum of the odd neutron with the latter in the same direction as the two spins. Adding up the four angular momenta we have 3 1 2 1 2 2 2 Thus the total angular. momentum nuclear spin is 2 What about the parity The parity of the nuclide is the. product of the two parities one for the last proton and the other for the last neutron. Recall that the parity of a state is determined by the orbital angular momentum quantum. number l 1 l So with the proton in a state with l 2 its parity is even while. the neutron in a state with l 3 has odd parity The parity of the nucleus is therefore. odd Our prediction for Cl38 is then 2 The student can verify using for example the. Nuclide Chart the foregoing predictions are in agreement with experiment. Potential Wells for Neutrons and Protons, We summarize the qualitative features of the potential wells for neutrons and. protons If we exclude the Coulomb interaction for the moment then the well for a. proton is known to be deeper than that for a neutron The reason is that in a given nucleus. usually there are more neutrons than protons especially for the heavy nuclei and the n p. interactions can occur in more ways than either the n n or p p interactions on account of. the Pauli exclusion principle The difference in well depth Vs is called the symmetry. energy it is approximately given by,Vs 27 Mev 9 14. where the and signs are for protons and neutrons respectively If we now consider. the Coulomb repulsion between protons its effect is to raise the potential for a proton In. other words the Coulomb effect is a positive contribution to the nuclear potential which. is larger at the center than at the surface, Combining the symmetry and the Coulomb effects we have a sketch of the. potential for a neutron and a proton as indicated in Fig 9 7 One can also estimate the. E Net Proton Potential,Net Neutron,Fermi Level EF,Symmetry Coulomb Effect.
Figure by MIT OCW Adapted from Marmier and Sheldon. Fig 9 7 Schematic showing the effects of symmetry and Coulomb interactions on the. potential for a neutron and a proton from Marmier and Sheldon. well depth in each case using the Fermi Gas model One assumes the nucleons of a fixed. kind behave like a fully degenerate gas of fermions degeneracy here means that the. states are filled continuously starting from the lowest energy state and there are no. unoccupied states below the occupied ones so that the number of states occupied is. equal to the number of nucleons in the particular nucleus This calculation is carried out. separately for neutrons and protons The highest energy state that is occupied is called. the Fermi level and the magnitude of the difference between this state and the ground. state is called the Fermi energy EF It turns out that EF is proportional to n2 3 where n is. the number of nucleons of a given kind therefore EF neutron EF proton The sum of. EF and the separation energy of the last nucleon provides an estimate of the well depth. The separation energy for a neutron or proton is about 8 Mev for many nuclei Based. on these considerations one obtains the results shown in Fig 9 8. 8 MeV 8 MeV 8 MeV 8 MeV,Fermi Fermi,level level, Figure by MIT OCW Adapted from Marmier and Sheldon. Fig 9 8 Nuclear potential wells for neutrons and protons according to the Fermi gas. model assuming the mean binding energy per nucleon to be 8 Mev the mean relative. nucleon admixture to be N A 1 1 8m Z A 1 2 2 and a range of 1 4 F a and 1 1 F b. from Marmier and Sheldon, We have so far considered only a spherically symmetric nuclear potential well. We know there is in addition a centrifugal contribution of the form l l 1 h 2 2mr 2 and a. spin orbit contribution As a result of the former the well becomes narrower and. shallower for the higher orbital angular momentum states Since the spin orbit coupling. is attractive its effect depends on whether S is parallel or anti parallel to L The effects. are illustrated in Figs 9 9 and 9 10 Notice that for l 0 both are absent. We conclude this chapter with the remark that in addition to the bound states in. the nuclear potential well there exist also virtual states levels which are positive energy. states in which the wave function is large within the potential well This can happen if. the deBroglie wavelength is such that approximately standing waves are formed within. the well Correspondingly the reflection coefficient at the edge of the potential is large. A virtual level is therefore not a bound state on the other hand there is a non negligible. probability that inside the nucleus a nucleon can be found in such a state See Fig 9 11. 1 x 2h2 2 x 3h2 3 x 4h2,2M r2 2M r2 2M r2,V0 E3s E2d. u1s r u1p r u1d r u1f r,u2s r u2p r u2d r u2f r,u3s r u3p r u3d r u3f r. l 0 s l 1 p l 2 d l 3 f,Figure by MIT OCW Adapted from Cohen.
Fig 9 9 Energy levels and wave functions for a square well for l 0 1 2 and 3 from.

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