Spin Parity Assignment Of Mortgage

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

Сьюзан услышала глухой хлопок, когда уже спустилась на несколько пролетов. Звук показался очень далеким, едва различимым в шуме генераторов. Она никогда раньше не слышала выстрелов, разве что по телевизору, но не сомневалась в том, что это был за звук. Сьюзан словно пронзило током. В панике она сразу же представила себе самое худшее.


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