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Shell-Model Spin And Parity Assignments Due

In nuclear physics and nuclear chemistry, the nuclear shell model is a model of the atomic nucleus which uses the Pauli exclusion principle to describe the structure of the nucleus in terms of energy levels.[1] The first shell model was proposed by Dmitry Ivanenko (together with E. Gapon) in 1932. The model was developed in 1949 following independent work by several physicists, most notably Eugene Paul Wigner, Maria Goeppert Mayer and J. Hans D. Jensen, who shared the 1963 Nobel Prize in Physics for their contributions.

The shell model is partly analogous to the atomic shell model which describes the arrangement of electrons in an atom, in that a filled shell results in greater stability. When adding nucleons (protons or neutrons) to a nucleus, there are certain points where the binding energy of the next nucleon is significantly less than the last one. This observation, that there are certain magic numbers of nucleons: 2, 8, 20, 28, 50, 82, 126 which are more tightly bound than the next higher number, is the origin of the shell model.

The shells for protons and for neutrons are independent of each other. Therefore, one can have "magic nuclei" where one nucleon type or the other is at a magic number, and "doubly magic nuclei", where both are. Due to some variations in orbital filling, the upper magic numbers are 126 and, speculatively, 184 for neutrons but only 114 for protons, playing a role in the search for the so-called island of stability. Some semimagic numbers have been found, notably Z=40 giving nuclear shell filling for the various elements; 16 may also be a magic number.[2]

In order to get these numbers, the nuclear shell model starts from an average potential with a shape something between the square well and the harmonic oscillator. To this potential a spin orbit term is added. Even so, the total perturbation does not coincide with experiment, and an empirical spin orbit coupling must be added with at least two or three different values of its coupling constant, depending on the nuclei being studied.

Nevertheless, the magic numbers of nucleons, as well as other properties, can be arrived at by approximating the model with a three-dimensional harmonic oscillator plus a spin-orbit interaction. A more realistic but also complicated potential is known as Woods–Saxon potential.

Deformed harmonic oscillator approximated model[edit]

Consider a three-dimensional harmonic oscillator. This would give, for example, in the first two levels ("l" is angular momentum)

level nlmlms

We can imagine ourselves building a nucleus by adding protons and neutrons. These will always fill the lowest available level. Thus the first two protons fill level zero, the next six protons fill level one, and so on. As with electrons in the periodic table, protons in the outermost shell will be relatively loosely bound to the nucleus if there are only few protons in that shell, because they are farthest from the center of the nucleus. Therefore, nuclei which have a full outer proton shell will have a higher binding energy than other nuclei with a similar total number of protons. All this is true for neutrons as well.

This means that the magic numbers are expected to be those in which all occupied shells are full. We see that for the first two numbers we get 2 (level 0 full) and 8 (levels 0 and 1 full), in accord with experiment. However the full set of magic numbers does not turn out correctly. These can be computed as follows:

In a three-dimensional harmonic oscillator the total degeneracy at level n is .
Due to the spin, the degeneracy is doubled and is .
Thus the magic numbers would be
for all integer k. This gives the following magic numbers: 2,8,20,40,70,112..., which agree with experiment only in the first three entries. These numbers are twice the tetrahedral numbers (1,4,10,20,35,56...) from the Pascal Triangle.

In particular, the first six shells are:

  • level 0: 2 states (l = 0) = 2.
  • level 1: 6 states (l = 1) = 6.
  • level 2: 2 states (l = 0) + 10 states (l = 2) = 12.
  • level 3: 6 states (l = 1) + 14 states (l = 3) = 20.
  • level 4: 2 states (l = 0) + 10 states (l = 2) + 18 states (l = 4) = 30.
  • level 5: 6 states (l = 1) + 14 states (l = 3) + 22 states (l = 5) = 42.

where for every l there are 2l+1 different values of ml and 2 values of ms, giving a total of 4l+2 states for every specific level.

These numbers are twice the values of triangular numbers from the Pascal Triangle: 1,3,6,10,15,21....

Including a spin-orbit interaction[edit]

We next include a spin-orbit interaction. First we have to describe the system by the quantum numbersj, mj and parity instead of l, ml and ms, as in the hydrogen-like atom. Since every even level includes only even values of l, it includes only states of even (positive) parity. Similarly, every odd level includes only states of odd (negative) parity. Thus we can ignore parity in counting states. The first six shells, described by the new quantum numbers, are

  • level 0 (n = 0): 2 states (j = ​12). Even parity.
  • level 1 (n = 1): 2 states (j = ​12) + 4 states (j = ​32) = 6. Odd parity.
  • level 2 (n = 2): 2 states (j = ​12) + 4 states (j = ​32) + 6 states (j = ​52) = 12. Even parity.
  • level 3 (n = 3): 2 states (j = ​12) + 4 states (j = ​32) + 6 states (j = ​52) + 8 states (j = ​72) = 20. Odd parity.
  • level 4 (n = 4): 2 states (j = ​12) + 4 states (j = ​32) + 6 states (j = ​52) + 8 states (j = ​72) + 10 states (j = ​92) = 30. Even parity.
  • level 5 (n = 5): 2 states (j = ​12) + 4 states (j = ​32) + 6 states (j = ​52) + 8 states (j = ​72) + 10 states (j = ​92) + 12 states (j = ​112) = 42. Odd parity.

where for every j there are 2j+1 different states from different values of mj.

Due to the spin-orbit interaction the energies of states of the same level but with different j will no longer be identical. This is because in the original quantum numbers, when is parallel to , the interaction energy is positive; and in this case j = l + s = l + ​12. When is anti-parallel to (i.e. aligned oppositely), the interaction energy is negative, and in this case j=ls=l−​12. Furthermore, the strength of the interaction is roughly proportional to l.

For example, consider the states at level 4:

  • The 10 states with j = ​92 come from l = 4 and s parallel to l. Thus they have a positive spin-orbit interaction energy.
  • The 8 states with j = ​72 came from l = 4 and s anti-parallel to l. Thus they have a negative spin-orbit interaction energy.
  • The 6 states with j = ​52 came from l = 2 and s parallel to l. Thus they have a positive spin-orbit interaction energy. However its magnitude is half compared to the states with j = ​92.
  • The 4 states with j = ​32 came from l = 2 and s anti-parallel to l. Thus they have a negative spin-orbit interaction energy. However its magnitude is half compared to the states with j = ​72.
  • The 2 states with j = ​12 came from l = 0 and thus have zero spin-orbit interaction energy.

Deforming the potential[edit]

The harmonic oscillator potential grows infinitely as the distance from the center r goes to infinity. A more realistic potential, such as Woods–Saxon potential, would approach a constant at this limit. One main consequence is that the average radius of nucleons' orbits would be larger in a realistic potential; This leads to a reduced term in the Laplace operator of the Hamiltonian. Another main difference is that orbits with high average radii, such as those with high n or high l, will have a lower energy than in a harmonic oscillator potential. Both effects lead to a reduction in the energy levels of high l orbits..

Predicted magic numbers[edit]

Together with the spin-orbit interaction, and for appropriate magnitudes of both effects, one is led to the following qualitative picture: At all levels, the highest j states have their energies shifted downwards, especially for high n (where the highest j is high). This is both due to the negative spin-orbit interaction energy and to the reduction in energy resulting from deforming the potential to a more realistic one. The second-to-highest j states, on the contrary, have their energy shifted up by the first effect and down by the second effect, leading to a small overall shift. The shifts in the energy of the highest j states can thus bring the energy of states of one level to be closer to the energy of states of a lower level. The "shells" of the shell model are then no longer identical to the levels denoted by n, and the magic numbers are changed.

We may then suppose that the highest j states for n = 3 have an intermediate energy between the average energies of n = 2 and n = 3, and suppose that the highest j states for larger n (at least up to n = 7) have an energy closer to the average energy of n−1. Then we get the following shells (see the figure)

  • 1st shell:  2 states (n = 0, j = ​12).
  • 2nd shell:  6 states (n = 1, j = ​12 or ​32).
  • 3rd shell: 12 states (n = 2, j = ​12, ​32 or ​52).
  • 4th shell:  8 states (n = 3, j = ​72).
  • 5th shell: 22 states (n = 3, j = ​12, ​32 or ​52; n = 4, j = ​92).
  • 6th shell: 32 states (n = 4, j = ​12, ​32, ​52 or ​72; n = 5, j = ​112).
  • 7th shell: 44 states (n = 5, j = ​12, ​32, ​52, ​72 or ​92; n = 6, j = ​132).
  • 8th shell: 58 states (n = 6, j = ​12, ​32, ​52, ​72, ​92 or ​112; n = 7, j = ​152).

and so on.

Note that the numbers of states after the 4th shell are doubled triangular numbers PLUS TWO. Spin-orbit coupling causes so-called 'intruder levels' to drop down from the next higher shell into the structure of the previous shell. The sizes of the intruders are such that the resulting shell sizes are themselves increased to the very next higher doubled triangular numbers from those of the harmonic oscillator. For example, 1f2p has 20 nucleons, and spin-orbit coupling adds 1g9/2 (10 nucleons) leading to a new shell with 30 nucleons. 1g2d3s has 30 nucleons, and addition of intruder 1h11/2 (12 nucleons) yields a new shell size of 42, and so on.

The magic numbers are then

  •   2
  • 8=2+6
  • 20=2+6+12
  • 28=2+6+12+8
  • 50=2+6+12+8+22
  • 82=2+6+12+8+22+32
  • 126=2+6+12+8+22+32+44
  • 184=2+6+12+8+22+32+44+58

and so on. This gives all the observed magic numbers, and also predicts a new one (the so-called island of stability) at the value of 184 (for protons, the magic number 126 has not been observed yet, and more complicated theoretical considerations predict the magic number to be 114 instead).

Another way to predict magic (and semi-magic) numbers is by laying out the idealized filling order (with spin-orbit splitting but energy levels not overlapping). For consistency s is split into j = 1⁄2 and j = -1⁄2 components with 2 and 0 members respectively. Taking leftmost and rightmost total counts within sequences marked bounded by / here gives the magic and semi-magic numbers.

  • s(2,0)/p(4,2) > 2,2/6,8, so (semi)magic numbers 2,2/6,8
  • d(6,4):s(2,0)/f(8,6):p(4,2) > 14,18:20,20/28,34:38,40, so 14,20/28,40
  • g(10,8):d(6,4):s(2,0)/h(12,10):f(8,6):p(4,2) > 50,58,64,68,70,70/82,92,100,106,110,112, so 50,70/82,112
  • i(14,12):g(10,8):d(6,4):s(2,0)/j(16,14):h(12,10):f(8,6):p(4,2) > 126,138,148,156,162,166,168,168/184,198,210,220,228,234,238,240, so 126,168/184,240

The rightmost predicted magic numbers of each pair within the quartets bisected by / are double tetrahedral numbers from the Pascal Triangle: 2,8,20,40,70,112,168,240 are 2x 1,4,10,20,35,56,84,120..., and the leftmost members of the pairs differ from the rightmost by double triangular numbers: 2 − 2 = 0, 8 − 6 = 2, 20 − 14 = 6, 40 − 28 = 12, 70 − 50 = 20, 112 − 82 = 30, 168 − 126 = 42, 240 − 184 = 56, where 0,2,6,12,20,30,42,56... are 2 × 0,1,3,6,10,15,21,28....

Other properties of nuclei[edit]

This model also predicts or explains with some success other properties of nuclei, in particular spin and parity of nuclei ground states, and to some extent their excited states as well. Take 17
8O (oxygen-17) as an example: Its nucleus has eight protons filling the three first proton "shells", eight neutrons filling the three first neutron "shells", and one extra neutron. All protons in a complete proton shell have total angular momentum zero, since their angular momenta cancel each other. The same is true for neutrons. All protons in the same level (n) have the same parity (either +1 or −1), and since the parity of a pair of particles is the product of their parities, an even number of protons from the same level (n) will have +1 parity. Thus the total angular momentum of the eight protons and the first eight neutrons is zero, and their total parity is +1. This means that the spin (i.e. angular momentum) of the nucleus, as well as its parity, are fully determined by that of the ninth neutron. This one is in the first (i.e. lowest energy) state of the 4th shell, which is a d-shell (l = 2), and since , this gives the nucleus an overall parity of +1. This 4th d-shell has a j = ​52, thus the nucleus of 17
8O is expected to have positive parity and total angular momentum ​52, which indeed it has.

The rules for the ordering of the nucleus shells are similar to Hund's Rules of the atomic shells, however, unlike its use in atomic physics the completion of a shell is not signified by reaching the next n, as such the shell model cannot accurately predict the order of excited nuclei states, though it is very successful in predicting the ground states. The order of the first few terms are listed as follows: 1s, 1p​32, 1p​12, 1d​52, 2s, 1d​32... For further clarification on the notation refer to the article on the Russell-Saunders term symbol.

For nuclei farther from the magic numbers one must add the assumption that due to the relation between the strong nuclear force and angular momentum, protons or neutrons with the same n tend to form pairs of opposite angular momenta. Therefore, a nucleus with an even number of protons and an even number of neutrons has 0 spin and positive parity. A nucleus with an even number of protons and an odd number of neutrons (or vice versa) has the parity of the last neutron (or proton), and the spin equal to the total angular momentum of this neutron (or proton). By "last" we mean the properties coming from the highest energy level.

In the case of a nucleus with an odd number of protons and an odd number of neutrons, one must consider the total angular momentum and parity of both the last neutron and the last proton. The nucleus parity will be a product of theirs, while the nucleus spin will be one of the possible results of the sum of their angular momenta (with other possible results being excited states of the nucleus).

The ordering of angular momentum levels within each shell is according to the principles described above - due to spin-orbit interaction, with high angular momentum states having their energies shifted downwards due to the deformation of the potential (i.e. moving from a harmonic oscillator potential to a more realistic one). For nucleon pairs, however, it is often energetically favorable to be at high angular momentum, even if its energy level for a single nucleon would be higher. This is due to the relation between angular momentum and the strong nuclear force.

Nuclear magnetic moment is partly predicted by this simple version of the shell model. The magnetic moment is calculated through j, l and s of the "last" nucleon, but nuclei are not in states of well defined l and s. Furthermore, for odd-odd nuclei, one has to consider the two "last" nucleons, as in deuterium. Therefore, one gets several possible answers for the nuclear magnetic moment, one for each possible combined l and s state, and the real state of the nucleus is a superposition of them. Thus the real (measured) nuclear magnetic moment is somewhere in between the possible answers.

The electric dipole of a nucleus is always zero, because its ground state has a definite parity, so its matter density (, where is the wavefunction) is always invariant under parity. This is usually the situations with the atomic electric dipole as well.

Higher electric and magnetic multipole moments cannot be predicted by this simple version of the shell model, for the reasons similar to those in the case of deuterium.

Including residual interactions[edit]

For nuclei having two or more valence nucleons (i.e. nucleons outside a closed shell) a residual two-body interaction must be added. This residual term comes from the part of the inter-nucleon interaction not included in the approximative average potential. Through this inclusion different shell configurations are mixed and the energy degeneracy of states corresponding to the same configuration is broken.[3][4]

These residual interactions are incorporated through shell model calculations in a truncated model space (or valance space). This space is spanned by a basis of many-particle states where only single-particle states in the model space are active. The Schrödinger equation is solved in this basis, using an effective Hamiltonian specifically suited for the model space. This Hamiltonian is different from the one of free nucleons as it among other things has to compensate for excluded configurations.[4]

One can do away with the average potential approximation entirely by extending the model space to the previously inert core and treat all single-particle states up to the model space truncation as active. This forms the basis of the no-core shell model, which is an ab initio method. It is necessary to include a three-body interaction in such calculations to achieve agreement with experiments.[5]

Related models[edit]

Igal Talmi developed a method to obtain the information from experimental data and use it to calculate and predict energies which have not been measured. This method has been successfully used by many nuclear physicists and has led to deeper understanding of nuclear structure. The theory which gives a good description of these properties was developed. This description turned out to furnish the shell model basis of the elegant and successful interacting boson model.

A model derived from the nuclear shell model is the alpha particle model developed by Henry Margenau, Edward Teller, J. K. Pering, T. H. Skyrme, also sometimes called the Skyrme model.[6][7] Note, however, that the Skyrme model is usually taken to be a model of the nucleon itself, as a "cloud" of mesons (pions), rather than as a model of the nucleus as a "cloud" of alpha particles.

See also[edit]



Recent Lecture[edit]

Internet streaming broadcasting both on WM and QT at (at 64 kbit/s, 256 kbit/s, 1 Mbit/s ) and DVD ISO (NTSC and PAL) delivery are now available at RIKEN Nishina Center.

Low-lying energy levels in a single-particle shell model with an oscillator potential (with a small negative l2 term) without spin-orbit (left) and with spin-orbit (right) interaction. The number to the right of a level indicates its degeneracy, (2j+1). The boxed integers indicate the magic numbers.
Residual interactions among valance nucleons are included by diagonalising an effective Hamiltonian in a valance space outside an inert core. As indicated, only single-particle states lying in the valance space are active in the basis used.
  1. ^"Shell Model of Nucleus". HyperPhysics. 
  2. ^Ozawa, A.; Kobayashi, T.; Suzuki, T.; Yoshida, K.; Tanihata, I. (2000). "New Magic Number, N=16, near the Neutron Drip Line". Physical Review Letters. 84 (24): 5493–5. Bibcode:2000PhRvL..84.5493O. doi:10.1103/PhysRevLett.84.5493. PMID 10990977.  (this refers to the nuclear drip line)
  3. ^Caurier, E.; Martínez-Pinedo, G.; Nowacki, F.; Poves, A.; Zuker, A. P. (2005). "The shell model as a unified view of nuclear structure". Rev. Mod. Phys. 77 (2): 427–488. arXiv:nucl-th/0402046. Bibcode:2005RvMP...77..427C. doi:10.1103/RevModPhys.77.427. 
  4. ^ abCoraggio, L.; Covello, A.; Gargano, A.; Itaco, N.; Kuo, T.T.S. (2009). "Shell-model calculations and realistic effective interactions". Progress in Particle and Nuclear Physics. 62 (1): 135–182. arXiv:0809.2144. Bibcode:2009PrPNP..62..135C. doi:10.1016/j.ppnp.2008.06.001. 
  5. ^Barrett, B.R.; Navrátil, P.; Vary, J.P. (2013). "Ab initio no core shell model". Progress in Particle and Nuclear Physics. 69: 131–181. Bibcode:2013PrPNP..69..131B. doi:10.1016/j.ppnp.2012.10.003. 
  6. ^Skyrme, T. H. R. (7 February 1961). "A Non-Linear Field Theory". Proceedings of the Royal Society A: Mathematical, Physical and Engineering Sciences. 260 (1300): 127–138. Bibcode:1961RSPSA.260..127S. doi:10.1098/rspa.1961.0018. 
  7. ^Skyrme, T.H.R. (March 1962). "A unified field theory of mesons and baryons". Nuclear Physics. 31: 556–569. Bibcode:1962NucPh..31..556S. doi:10.1016/0029-5582(62)90775-7. 

Nuclear Physics PHY303

Solutions 2


Charge of the electron e = 1.6 10-19 C
Mass of the electron me = 9.11 10-31 kg = 511 keV/c2
Mass of the proton mp = 1.673 10-27 kg = 938.272 MeV/c2
Mass of the neutron mn = 1.675 10-27 kg = 939.566 MeV/c2
Planck constant h = 6.626 10-34 J s = 4.136 10-15 eV s
Boltzmann constant k = 1.38 10-23 J K-1 = 8.617 10-5 eV K-1
Speed of light in free space c = 3.00 108 m s-1
Permittivity of free space 0 = 8.85 10-12 F m-1
Permeability of free space µ0 = 4 10-7 H m-1
Avogadro constant NA = 6.02 1026 kg-mol-1
Rydberg constant R = 1.10 107 m-1
Bohr magneton µB = 9.27 10-24 J T-1
Nuclear magneton µN = 5.0508 10-27 J T-1 = 3.1525 10-14 MeV T-1
Fine structure constant = 1/137

Useful data

Unified atomic mass unit 1u = 1.66 10-27 kg = 931.502 MeV/c2
Energy conversion 1 eV = 1.6 10-19 J
Year in seconds 1 yr = 3.16 107 s
Atmospheric pressure 1 atmosphere = 1.01 105 N m-2
Acceleration due to gravity on Earth's surface g = 9.81 m s-2
1 gram molecule at STP occupies 22.4 litres

Solutions 3

Go back to the Problems or to the constants and useful data.
    This problem can be handled by focusing on the right hand side of the energy level figure given in the notes and repeated here

    Note this latter asignment follows the ordering given in the question which slightly differs at this point from that in the figure. This just underlines the fact that the ordering of the levels does exhibit some local variations and so the figure should be considered as no more than a reasonable approximation.

    Across these three isotopes the number of protons remains the same of course but the number of neutrons rises from 7 to 9. This means that 15O has one neutron missing in the 1p1/2 level; 16O has a complete 1p1/2 neutron level; 17O has a single neutron in the 1d5/2 level. From this and the given data we can deduce:
    The binding energy of a neutron in 1p1/2 level is 127.62 - 111.96 = 15.66 MeV
    The binding energy of a neutron in 1d5/2 level is 131.76 - 127.62 = 4.14 MeV
    Thus the (1p1/2 - 1d5/2) difference is 15.66 - 4.14 = 11.52 MeV

    The numbers of nucleons associated with each of the listed levels are:


    From this it can be seen that the 3rd proton is in the 1p3/2 level; the 11th proton is in the 1d5/2 level; the 17th neutron is in the 1d3/2 level; the 21st proton is in the 1f7/2 level. Hence the ground and excited state spin and parity assignments are:


    The nucleus 209Bi83 has an odd number of protons and an even (magic) number of neutrons. Looking at the energy level diagram given above it would be expected that the odd proton would be in the 1h9/2 level indicating a (9/2)- ground state, as is observed to be the case.

    In this simple model the magnetic moment comes from the single proton, being a combination of its orbital and intrinsic moments. For the ground state = 5 and j = 9/2 thus we need the expression for the j = - 1/2 Schmidt limit. This is:

    µ = [g j(j + 3/2)/(j + 1) - gsj/(2(j + 1))] µN

    For the proton g = 1 and gs = 5.5857. Inserting these and the value of j gives µ = 2.62 µN which is quite a bit smaller than the observed value of 4.1 µN. This may well reflect a mixing of states - the nearest other state is (7/2)- which has the same parity and so is a likely candidate. This is also a j = + 1/2 level and so a proton in it would produce a relatively large magnetic moment (5.79 µN). It should also be noted that the value of gs which has been used is that for a free proton. The appropriate value for a proton inside a nucleus is most likely to be different and an examination of the distribution of the measured moments relative to the Schmidt limits suggests a value of something like 0.6 gs. Both of these factors would enlarge the magnetic moment of 209Bi83 beyond the simple Shell Model prediction.

    The Electric Quadrupole moment can be simply thought of, in these units, as (minus) the area enclosed by the orbit of the proton. Estimating the radius of the orbit as approximately that of the nucleus ro A1/3 would give a quadrupole moment -0.5 x 10-28 m2. This is pretty close to the observed value considering the crudeness of the model. A quantum mechanical treatment reveals the dependence on j and gives -0.4 x 10-28 m2 for this case.

    As indicated above this nucleus has a magic number (126) of neutrons and so the small neutron capture cross-section is not surprising.

    The nuclide 42Sc21 has an odd number of both protons and neutrons. From the energy level diagram above the 21st nucleon should be in the 1f7/2 level and the combination of two (7/2)- states will give (7)+ if all the angular momenta are "parallel".

    In terms of magnetic moments the expression for the j = + 1/2 Schmidt limit is required. This is:

    µ = [g (j - 1/2) + gs/2] µN

    For the proton g = 1 and gs = 5.5857 and for the neutron g = 0 and gs = -3.8261. Thus the two contributions to the magnetic moment are 5.79 µN from the proton and -1.91 µN from the neutron - a total of 3.88 µN.

    This is a straightforward if somewhat tedious classical slog. For those who like that sort of thing the quadrupole moment is given by an integral of the form:

    Then making use of the axial symmetry:

    We can equate dV to the elemental ring volume illustrated.

    The elipsoidal shape is represented by

    where b is the deformation parameter and the spherical harmonic

    Writing p for cos() and integrating over r from 0 to R reduces the integral to

    Inserting the expression for R and substituting Ze 4(Rav)3/3 for the density produces the integral

    Assuming that b is small, the expansion is only taken up to b2. Also note

    which just leaves two terms in the integral

    Evaluation of this yields

Solutions 4

Go back to the Problems or to the constants and useful data.
    The decay scheme is
    14O8 (14N7)* + e+ + e + e-

    where the latter is the excess atomic electron and the excited nitrogen isotope decays via gamma emission. The rest mass energies before and after are related by
    M(14,8)c2 = M(14,7)c2 + 2me + Q

    The Q = 1.835 + 2.313 = 4.148 MeV and converting this into unified mass units gives the mass of the original atom as 14.008625 u.

    The missing piece of information is the binding energy of 231Th90. Using the given coefficients the following values are obtained for the terms
    3187.8-489.42-757.16-225.1901716.0 MeV

    Using this with the binding energies given for the other nuclei gives 4.6 MeV for the alpha particle energy and 0.3 MeV for the maximum beta energy.

    Allowing a possible error of about 1 keV and ensuring that each state can reach the ground state a possible scheme is

    By looking at the curve for the binding energy per nucleon B/A as a function of A

    it can be deduced that fission is generally not favoured in terms of energy release unless A is greater that approximately 60. In order to pin down at which value of A alpha particle decay becomes energetically favoured it is necessary to be a little more specific. Consider just those nuclides which are stable against beta decay. To find the value of Z at the bottom of the valley for a given A essentially requires the differentiation of the expression for the binding energy with respect to Z at fixed A.

    avA- asA2/3- acZ2/(A)1/3- aaA(1 - 2Z/A)2
    00- 2Zac/(A)1/3+ 4aa(1 - 2Z/A)

    Equating the differential to zero produces a relationship between Z and A for the nuclei which lie on the line of beta stability. That relationship can be expressed as

    The term in the brackets shows how the ratio Z/A increasingly differs from 0.5 as A gets larger. This expression means that B can be written as a fuction of A alone. My prefered method at this point is to use a spreadsheet program such as Excel to calculate binding energies for a range of A values, and hence determine where the increase in B for an decrease in A of 4 becomes less than the binding energy of the deuteron. This occurs for

    An alternative, approximate method can be used. From the above sketch for B/A it appears that for A greater than about 60 the curve can be approximated by a straight line. An example of how this could be done using the values of av etc given in question 2 and making the appropriate substitution for Z/A is illustrated below.

    av-asA-1/3-ac(Z/A)2A2/3-aa(1 - 2Z/A)2
    13.8-2.80-2.34-0.378.29 MeV
    13.8-2.22-3.18-0.797.61 MeV

    The straight line passing through these two points has the equation

    Finally, this approximation together with the given binding energy of 28.3 MeV for the alpha particle, enables the simple calculation of the energy released in the alpha decay of a nucleus with mass number A. Setting this energy to zero yields about 141 for the limiting value of A.
    The actual lightest alpha emitter of this type is 146Sm62.

    The deduction for the normal density of states factor is given in the notes and from consideration of coordinate and momentum space for both the electron and the neutrino an expression

    dn = (V2162/h6). (pe)2dpe(p)2dp

    is obtained, which through the substitutions

    p = (T0 - Te)/c : dp/dE = 1/c


    dn/dE = (V2162/(h6c3)). (T0 - Te)2(pe)2dpe

    In order to find the effect of a neutrino mass it is necessary to re-evaluate the above substitutions which assumed zero mass. The expression E2 = (pc)2 + (mc2)2 must now be used for the neutrino where E is (T0 - Te). The substitutions become

    p = [(T0 - Te)/c].[1 - m2c4/(T0 - Te)2]1/2 : dp/dE = [1/c].[1 - m2c4/(T0 - Te)2]-1/2

    giving the required overall multiplicative factor.
    The slope of the normal spectrum goes to zero as Te approaches its limiting value T0. This is due to the presence of the factor (T0 - Te) throughout. In the case where the neutrino has mass the limit of T0 - Te is mc2 rather than zero. There is now a term in the slope of the spectrum which is proportional to the reciprocal of the multiplicative factor and since this latter goes to zero at the limit, the slope there is infinite.

    The maximum energy available is 1.29 MeV. Neglecting the recoil of the proton and assigning this energy to the electron gives a momentum of [(1.29)2 - (0.51)2]1/2 = 1.185 MeV/c. The maximum momentum that the anti-neutrino can carry is less because the electron still has its rest mass energy. Thus for the anti-neutrino the momentum is 1.29 - 0.51 = 0.78 MeV/c.

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In the case of difficulties with this course contact Prof Neil Spooner.

If you find this material at all helpful please let me know!

© 1999 - FH Combley, 2000 - CN Booth, 2009 - NJC Spooner

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