Nuclear Physics PHY303

8 Course Review

8.1 Content in pictures

This summary uses some of the pictures which illustrated the course to point back to the various sections. For the purposes of revision you should refer to the recommended text, the course handouts and your own notes for more detail.

Click on the picture itself to return to the point in the notes where it first appeared.

We start with the distribution of nuclei in the Z/N plane.



A great deal in terms of the properties of nuclei can be deduced from this figure. Another key figure in the general properties of nuclei is the plot of binding energy per nucleon B/A as a function of A. This is shown below in a form which indicates the main contributions.



The volume, surface and coulomb terms are based on the model of the nucleus as a drop of charged fluid while the symmetry term has its origins in a model with energy levels occupied by individual nucleons. The broad maximum in the B/A values around A ~ 60, divides nuclei into those which will release energy when undergoing fusion and the heavier nuclei for which fission is exoergic.

The third picture illustrating general nuclear properties is the plot of nuclear density as a function of radius.



Three conclusions can be drawn The first two are consistent with the liquid drop model of the nucleus while the third points to the short range of the nuclear force. The study of the nuclear force was illustrated by considering the deuteron, the simplest multi-particle nucleus.



The deuteron has only one bound state and that has a small binding energy and relatively large mean separation between the proton and the neutron. The figure above shows these in the context of the quantum mechanical picture. The wave function has its maximum only just inside the nuclear potential well and the consequent long exponential tail of the function stretches the rms separation of the proton and neutron to a distance significantly outside the well. A second illustration related to the nuclear force is the sketch of the Yukawa potential shown below.



The potential is associated with a force which is transmitted by the exchange of a particle with non-zero rest mass. The exponential factor restricts the force to a range which is inversely proportional to the mass of the exchanged particle - as indicated.

In discussing the Shell Model several examples of supporting experimental evidence were quoted, two of which are recalled here.





Both the elemental abundances and the total neutron cross-sections exhibit effects associated with so-called magic numbers of protons or neutrons.

The Shell Model energy levels obtain by using various 'nuclear-like' potentials in the Schrödinger equation were illustrated by the following figure.



Here the vital role of spin-orbit coupling in matching theory to the observed magic numbers is apparent. As a final part of the discussion of nuclear models, distorted nuclei were considered and in particular the associated observation of vibrational and rotational energy levels. Characteristic examples are shown below.

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In the study of the decay of nuclei, the barrier penetration model of particle decay was illustrated with the next figure.



In particular this sketch shows how a positive energy emerges from a nucleus in which all the nucleons are in negative energy states.

In decay the following figures were used to show how the decay process moved between isobars.



For fixed A the atomic rest mass energies lie on parabolae - a single one for odd A and two for even A. This was used to explain the existence of more than one stable isobar in the latter case. The effect of the available phase space on the decay electron or positron momentum spectra was illustrated with the curves shown next.



These spectra are for a nuclide which can decay by either electron or positron emission and they show how the latter particles are repelled away from the positively charged nucleus to higher momentum while the electrons are slowed to lower momentum.

Evidence for the existence of a long lived intermediate state or compound nucleus in some nuclear reactions is shown in the following graph of neutron total cross-section.



The extremely sharp resonances indicate that the compound nucleus states are well defined with lifetimes several orders of magnitude longer than the time it takes a nucleon to traverse the nucleus.

Much of the discussion of nuclear fission was aided by the use of the liquid drop model. Competition between the terms representing the surface tension and the coulomb repulsion leads to a potential barrier as illustrated here.



This picture of nuclear breakup is also used to show the sequence of events in neutron induced fission below.



However the liquid drop model is not completely successful in describing the fission process. It predicts equal mass fragments while, as can be seen from the next figure, the observed fragment mass distribution is asymmetric.



These distributions are for fission at low energies of excitation. For induced fission at higher energies, fragments of equal mass are more likely to be seen.

That a chain reaction is not sustained by a lump of natural uranium was discussed in the context of the neutron cross-sections of the two isotopes which are illustrated next.



There are two possible routes for the conversion of four protons into a helium nucleus by fusion in stars, the pp chain and the CNO cycle. The figure below compares the cross-sections as a function of temperature for these two processes for the density conditions which prevail in the sun.



However attempts to develop a fusion reactor on earth have centred on the use of deuterium as fuel. The cross-sections for these reactions are shown in the next figure.



As can be seen, the most favoured process is the so-called D-T reaction. Finally a sketch of the basic elements of a Tokamak fusion reactor is given.




8.2 Solution of problems Most of the following problems can be found amongst those given for each Chapter for which brief solutions are also given. The solutions are presented here in a little more detail.

  1. In an experiment carried out with a beam of thermal neutrons it is found that on traversing a 2mm thick foil of 197Au, some 70% of the neutrons are removed. What is the total thermal neutron cross-section for this isotope of gold? Comment on the result of the cross-section measurement in the light of the fact that the radius of a gold nucleus is
    6.5 x 10-15 m. (Density of gold: 19300 kg m-3)

    Solution

    To solve this problem we note that absorption steadily decreases the intensity of the beam as it passes through the material and so we use the equation for exponential attenuation

    N(x) = N0exp(- nx)

    where N0 is the initial beam intensity, is the cross section, n is the number of absorbing or scattering nuclei per unit volume and x is the thickness of the absorber. You should be able to simply derive this equation. The value of n is obtained from

    n = NA/A

    where NA is Avogadro's number, is the density and A is the atomic mass number for gold. Putting the numbers in should give = 102 barns.

    Interpreting this as the cross-sectional area of the nucleus yields a radius of 5.7 x 10-14 m which is almost nine times larger than the actual radius given for the nucleus of the gold atom. This indicates that resonance capture is occuring and the wavefunction for the virtual state stretches far beyond the range of the nuclear potential.

  2. The isotope 14O8 is a positron emitter, decaying to an excited state of 14N7. The gamma rays from this latter have an energy of 2.313 MeV and the maximum energy of the positrons is 1.835 MeV. The mass of 14N7 is 14.003074 u and that of the electron is 0.000549 u. Write the equation for the decay of the oxygen isotope and sketch an energy level diagram for the process. Given that one unified mass unit (u) is equal to 931.502 MeV/c2 find the mass of 14O8.

    Solution

    A significant point to remember here is that when given the mass of an isotope it refers to the mass of the neutral atom. Hence in positron decay there is an atomic electron to be discarded by the daughter atom because its nucleus has lost one proton charge. The decay scheme should thus be written
    14O8 (14N7)* + e+ + e + e-

    where the latter is the excess atomic electron and the excited nitrogen isotope decays via gamma emission. Thus the energy level scheme looks like

    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.

  3. Given that the coefficients of the five terms which make up the binding energy in the mass formula have the approximate values (in MeV): volume, 15.5; surface, 16.8; coulomb, 0.72; asymmetry, 23; pairing, 34, show that the difference between the total binding energy of the uranium isotope 235U92 and the compound nucleus formed upon slow neutron absorption is 6.7 MeV while the corresponding difference for 238U92 is 5.2 MeV. What relevance does this have to the fission process in natural uranium?

    Solution

    The calculation can be carried out in full or directly in terms of the differences. Note that on neutron absorption 235U makes a transition from an odd A to an even-even nucleus while the reverse is true for 238U. Thus the pairing term appears in both of the expressions for the change in binding energy but with opposite sign and it is this which determines the different threshold behaviour for neutron induced fission. The values for the various terms are

    DifferenceVolume SurfaceCoulomb AssymetryPairing Total
    B(236) - B(235)15.5 -1.811.38 -8.960.56 6.7 MeV
    B(239) - B(238)15.5 -1.811.36 -9.31-0.56 5.2 MeV

    Natural uranium consists of 99.3% 238U and 0.7% 235U. The latter is FISSILE (fissions for neutrons of all energies) but the threshold for neutron induced fission of 238U is 1 MeV - a difference which arises, in the semi-empirical formula, from the contribution of the Pairing Term. This difference means that a neutron produced by fission in natural uranium is much more likely to suffer radiative capture in 238U than it is to cause fission in 235U. Consequently a chain reaction is not sustained.

  4. The observed magnetic moment of 3He2 is -2.128 µN. Explain any discrepency between this and the shell model prediction. (proton magnetic moment, 2.793 µN: neutron magnetic moment, -1.913 µN)

    Solution

    In the shell model description the two protons would occupy the 'spin up' and 'spin down' substates of the 1s energy level - their spins and consequently their magnetic moments cancelling out. This leaves the total spin and magnetic moment to be determined by the odd neutron. Thus the magnetic moment of 3He2 should be -1.913 µN. That it differs from this by about 10% can be partly explained by the mixing of states which was discussed in connection with the observed magnetic moment of the deuteron. However a more detailed study including comparison with the mirror nucleus 3H1 indicates this is not the whole story and that a further contribution is required - perhaps originating in the effective electrical currents resulting from the exchange of charged pions which bind the nucleons together.

  5. If the fusion reaction d + d 3He2 + n + 3.2 MeV occurs with the deuterons at rest what is the kinetic energy of the neutron?
    Based on the estimate that the deuterons have to come within 100 fm of each other for fusion to proceed, calculate the energy that must be supplied to overcome the electrostatic repulsion. Approximately what temperature is this equivalent to?

    Solution

    Since before the reaction there is no net momentum the momenta of the two particles in the final state must cancel out. Writing the mass of the neutron as m and approximating the mass of 3He2 as 3m leads to the following equalities
    3mV = mv or V2 = v2/9

    Now the total kinetic energy in the final state is
    3.2 MeV = 3mV2/2 + mv2/2
    or 3.2 MeV = (mv2/2)4/3
    thus mv2/2 = 2.4 MeV or 3.84 10-13 J


    In the next part the electrostatic energy is just e2/(40r) which is equal to 14.4 keV or 2.3 x 10-15 J, and putting this equal to kT gives a temperature of about 108 K.


8.3 Style of the examination - The written Examination is two hours long and three questions have to be answered - a COMPULSORY question and two others chosen from four.

The COMPULSORY question requires the student to briefly explain each of ten terms which have been used during the course - e.g. "exchange force".



The content and information presented here is for the academic session 2005-2006.
Back to Nuclear Physics Syllabus
In the case of difficulties with this course contact Dr. Chris Booth.

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© 1999 - FH Combley;   2005 - CN Booth