Nuclear Systematics

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Before we go to the detailed study of nuclei we shall look at the general features of the properties of nuclei. That is, how the basic properties like masses, sizes, shapes, spins, excitation spectra etc behave as a function of atomic number and charge of the nucleus ( or equivalently, how they behave as a function of neutron and proton number ). We shall find that although the properties do differ in detail as one goes from one nucleus to other, the average or gross properties vary smoothly as a function of atomic number and charge. This behavior will be discussed in this section.

Contents

[edit] Masses

The mass of a nuclide is close to $ times proton mass plus $N$ times neutron mass. There is small departure because the nucleons are bound in the nucleus and the mass is smaller by an amount given by the mass-equivalent of the total binding energy of all nucleons. Thus, we have

MA = ZMp + NMn − Bc2

Here Mp and Mn are the masses of proton and neutron respectively, $B$ is the binding energy of the nucleus ( i.e. total energy released when Z protons and N neutrons are brought together to form the nucleus and $c$ is the velocity of light. Here we have used the celebrated mass-energy relation of Einstein. In passing, we may note that nuclear binding energies provided the first experimental test of Einstein's mass-energy relation.

The measurement of nuclear masses is done by two kinds of methods. One is the direct method in which one studies the track of a nucleus in presence of electric and magnetic field. The other is the indirect method in which one studies the energetics of nuclear reactions to deduce the mass of one of the nucleus taking part in the rection in terms of the masses of the rest of the participants. We shall discuss these below.

[edit] Direct Measurements of Nuclear Masses

The direct measurements are based on the motion of charged particles in presence of electric and magnetic fields. Different mass spectrographs have been developed using this principle and now these techniques are being used to study the structure of large molecules. Here we shall consider the basic principle of mass measurement.

As an example, consider the motion of a particle of charge Q and unknown mass M in uniform magnetic field B. If the initial velocity of the particle is perpendicular to the direction of magnetic field, the trajectory of the particle is circular and the force, which is in the direction perpendicular to the velocity as well as magnetic field is \frac{Q}{Mc}\vec p \times \vec B. The magnitude of the force is \frac{Q}{Mc}pB. Here p=\sqrt{2M e} is the momentum of the particle and e is the kinetic energy. The force acting on a charged particle in presence of uniform electric field is Q \vec E where \vec E is the electric field.

If the uniform electric and magnetic fields are perpendicular to each other and the charged particle is incident in the direction perpendicular to both the fields, the forces due to electric and magnetic fields are parallel to each other. One can arrange the magnitude and directions of electric and magnetic fields in such a way that the two forces cancel and the particle continues in straight line motion. Then one gets

\frac{QB\sqrt{2M e}} {Mc} = QE

or

M = \frac{2 B^2 e}{c^2 E^2}.

So, one can adjust electric and magnetic fields so that the particle traverses the region of the fields in straight line. Then the mass of the particle is determined from known electric and magnetic fields and the energy of the particle.

In actual instruments, the deflection of ions due to electric and magnetic fields are measured and the mass is determined from the known value of particle energy. The earliest instrument used for mass determination was Aston's mass spectrograph.

[edit] Indirect Measurements of Nuclear Masses

The direct measurement of masses of nuclei is possible for a limited number of nuclei. If the nuclei don't occur naturally or when they are produced in reactions, indirect methods are required. The indirect methods use the principle of conservation of energy along with Einstein's mass-energy relation to determine the unknown masses. To be specific, consider a nuclear reaction

A + B \rightarrow C + D

Typically, one of the nucleus ( say the nucleus A ) will be a projectile hitting the second nucleus ( nucleus B ) stationary in the laboratory ( the target ). Then the total energy ( including the rest mass energy ) of the initial system is (MA + MB)c2 + TA. Here TA is the kinetic energy of nucleus A. The kinetic energy of B in the laboratory is zero. Here we are using non-relativistic kinematics. After the reaction, the reaction products ( C and D ) will have some kinetic energy which can be measured. The energy of the final system is (MC + MD)c2 + TC + TD. Conservation of energy implies that these two quantities should be equal. So, if we measure the kinetic energies of nuclei A, B and C ae can determine the mass of one of the nucleus in terms of the masses of other three nuclei. Masses of majority of nuclei are determined by such a procedure.


[edit] Binding Energy Systematics

Instead of nuclear masses, one considers binding energy per nucleon ( B / A for discussion. One reason for this is that the nuclear masses are close to their mass number times mass of proton or neutron. The departure from this number is an important quantity. Further, the binding energy per nucleon remains more or less constant over the periodic table. Or the binding energy is (almost ) linearly increasing function of A, the number of nucleons in a nucleus. The variation of B / A as a function of A is shown in an accompanying figure.

Plot of B/A vs A
Plot of B/A vs A

From the figure one can observe that for medium to heavy nuclei this number is of the order of 8 MeV. Compare this number with the energy-equivalent of nucleon mass, which is about 940 MeV. This means that when nucleons bound into a nucleus, their mass reduces by less than a percent or less than a percent of their mass is released as energy in some form. Although the energy released is a small fraction, it is enormous in comparison with the energy released in typical atomic and molecular reactions. For example, when hydrogen atom is formed by combining a proton and an electron 13.6 eV of energy is released where as the mass of hydrogen atom ( which is essentially the sum of mass of electron and proton ) is about 936 MeV. So, in this case, the fraction of mass released as energy is about one part in 108. In typical chemical reactions, the energy release per molecular process is of the order of electron volt and the masses involved are tens or hundreds of proton mass and therefore the fraction of mass released as energy is even smaller.

B/A has a broad peak around A=60 ( which corresponds to iron nucleus ). Below A=60, the curve has sharp peaks at certain values of A. The peaks at A=4, 12 and 16 are dominant. The peaks become less sharp as A increases and around A=40, smooth behavior B/A curve sets in. The value of B/A at the peak is about 8.5 MeV. As A increases beyond 60, B/A decreases slowly and at about A=260$ the curve terminates. This is because there are no naturally occuring nuclides beyond this value of A. Heavier nuclei have been prepared in laboratory and these continue to show the decreasing trend. In any case, the curve eventually terminates because the nuclei with larger A decay immediately.

Although the B/A curve appears to be quite smooth for A larger than 40, there are actually local variations. In particular, the peaks similar to the dominant peaks at small values of A persist but their heights are not very large and decrease with increasing A. This is a typical quantum or shell effect and we shall discuss the detailed reasons for these peaks later.

Stable nuclei as a function of neutron and proton number
Stable nuclei as a function of neutron and proton number

The near constancy of B/A implies that the nuclear interaction has a short range and nucleons in a nucleus interact with only few of the other nucleons. If a nucleon in a nucleus had interacted with all of the nucleons, its binding energy ( for large enough A ) would be proportional to A or the total binding energy would be proportional to A(A-1) which is close to A2 for large A. So B/A would be proportional to A for large enough A. On the other hand, for short-ranged interaction and for large enough A, most of the nucleons would interact with the nucleons within a certian distance and this number will be same for all the nucleons. Then the binding energy will be proportional to the number of nucleons or B/A will be a constant.

This type of behavior of binding energy is analogous to the binding of molecules forming a liquid. In fact one can stretch the analogy further. The nucleons at the surface of the nucleus will interact with fewer nucleons and therefore their contribution to the binding energy would be less. In liquids, this behavior gives rise to surface tension and we should expect the nuclei also to have surface tension. Put in another way, there would be a correction to the constancy of $B/A$ because of the smaller contribution of the surface nucleons. Now, as the system gets bigger, the surface area increases slower than the volume of the system. As a result, the importance of the surface energy ( the binding energy contribution of the surface nucleons ) decreases. We then understand why the B/A curve first increases before reaching the very broad peak at A \sim 60.

Nuclei contain protons which are positively charged. So they have Coulomb interaction which is repulsive would give a negative contribution to B/A. The Coulomb interaction is also long-ranged so its contribution is proportional to the square of the number of protons in a nucleus. We therefore expect that B/A would decrease as the number of protons increase. The slow fall in $B/A$ with the increase in $A$ is because of this behavior of the Coulomb energy contribution.

The above mentioned behavior of binding energy of the nucleus is analogous to the behavior of a charged liquid drop. This analogy has led to a nuclear model called liquid drop model. The liquid drop model is successful in explaining the systematics of binding energy of nuclei. We shall be studying the liquid drop model later but at this stage we shall note that the model gives rise to a nuclear binding energy formula ( also called nuclear mass formula ) called Bethe-Weissecker mass formula. The formula is

\frac{B}{A} = a_V - a_S A^{-1/3} - a_C \frac{Z^2}{A^{4/3}} + a_{asym} (N-Z)^2 + \delta_{pair} + \Delta

The terms in the Bethe-Weissecker mass formula are volume, surface, Coulomb, asymmetry and pairing terms. The last term is the correction due to shell effects which depicts the departure from the smooth behavior. The details of the formula will be discussed later.

Another application of the liquid drop model is the fission reaction of heavy nuclei. Energetically, splitting of heavy nuclei into two medium mass nuclei is possible. The process is, however, slowed down by the stability of nuclei under deformation. Using liquid drop model one can compute the energy of the nucleus as a function of nuclear deformation. Later we shall show that the deformation energy first increases and beyond a certain deformation starts decreasing. So, classically, a heavy nucleus is stable but quantum mechanical tunneling allows the nucleus to fission. Later we shall be discussing nuclear fission in details.

[edit] Nuclear Sizes

Nuclei being composite objects, they are not point particles but they have extended structure. They have different sizes and shapes. They are essentially bound states of protons and neutrons interacting through nuclear interaction. The sizes and shapes of nuclei are probed by using different probe particles. Ideally, probe particles should interact with the constituents of nuclei but they should not disturb them. That is, the interaction of probe particles should not lead to excitations on nuclei. In case of nuclear excitations one will not be probing static properties but dynamics of nuclear excitations and we are not concerned with that here. Also, if the interaction of the probe is simple, extraction of nuclear static properties is relatively easier.

The best probe for investigating nuclear sizes and shapes is the electron. That is because the interaction here is the Coulomb interaction which is not only simple but also weak ( in comparison with the nuclear interaction ). But electrons interact with charges in nuclei ( protons ) and their interaction with neutrons is weak ( through neutron magnetic moment ) so one gets information of charge distribution in nuclei from electron scattering. Muons also interact with nucleons via Coulomb interaction and they can also be used to probe nuclei. In particular, μ mesic atoms provide a tool for investigating the nuclear charge distribution. Muon are 207 times heavier than electrons. So the Bohr radii of muonic atoms are smaller by a factor of 207. As a result, there is a reasonably large overlap of muon wave function with nuclear charge distribution and therefore there is a shift of muonic atom energies from the point Coulomb energies.

[edit] Electron Scattering

Elastic scattering cross section for electrons from a point charge Q is given by Mott Scattering formula

\frac{d\sigma}{d\Omega}|_{Mott} = \frac{e Q}{4 k^4 \sin^4 \theta/2} (1 + \cos \theta)/2

In non-relativistic limit, this reduces to the Rutherford formula

\frac{d \sigma}{d\Omega}_{Ruth} = \frac{e Q}{4 k^4 \sin^4 \theta/2}

where θ is the scattering angle ( angle between the incident and final directions of electron ) and p is the momentum of the electron. Here we are assuming that the nucleus is heavy and we are neglecting the recoil of the nucleus. The Rutherford formula is obtained by considering the scattering of electrons from the Coulomb potential V_C(\vec r) = \frac{Q e}{r}. In Born approximation the scattering cross section is given by

\frac{d \sigma}{d \Omega} = | f(\vec q)|^2

where \vec q = \vec k - \vec k' is the momrntum transfer in the scattering and \vec k and \vec k' are the momenta of the electron before and after scattering and

f(\vec q) = \int d^3r e^{i \vec q \cdot \vec r} V_C(r).

Given the charge distribution \rho(\vec r) of a nucleus, the potential produced by this distribution is V_C(\vec r) = \int d^3r' \frac{\rho(\vec r')}{|\vec r - \vec r'|} so the scattering amplitude f(\vec q) is

f(\vec q) = \int d^3r d^3r' e^{i\vec q \cdot \vec r} \rho(\vec r') \frac{1}{|\vec r - \vec r'|}

Adding and subtracting i \vec q \cdot \vec r' in the exponential, and transforming the integration variable \vec r to \vec r'' = \vec r - \vec r', the integral separates into

f( \vec q) = \int d^3r'' e^{i\vec q \cdot \vec r''} \frac{Q}{r''} \times \frac{1}{Q}\int d^3r' \rho(\vec r').

Note that the first term is the Fourier transform of the Coulomb potential of a charge of strength Q, which is also the scattering amplitude for scattering from a point Coulomb source. The second term is the Fourier transform of the charge density ( normalized to unity ) of the nucleus.

From this result, we conclude that the scattering cross section of electrons from a nucleus can be written as a product of scattering cross section from a point Coulomb source times the square of the Fourier transform of nuclear charge density. Thus, if the ratio of electron-nucleus scattering cross section and equivalent point-Coulomb cross section is known for all values of q, one can compute the nuclear charge distribution in a model-independent way.

This, of course, is not possible because the cross section cannot be measured for all values of q. We have q^2 = (\vec k - \vec k')^2 = 2k^2 ( 1 - \cos \theta) where k is the momentum of incoming electron and θ is the scattering angle or the angle between the incident and final directions of the electron as measured in the center of mass frame. So the momentum transfer is the largest when cosθ = − 1 or when the scattering angle is π and it is limited by the momentum or energy of the incident electron.

A number of approximations have been used while deriving the result above. First, we have neglected relativistic effects. Those are not small in electron scattering because the electrons are relativistic when the electron energy is few MeV. We then need to solve Dirac equation for electrons in the presence of the electric potential produced by the nucleus to obtain electron scattering amplitude. Further, for heavy nuclei, the charge is large and perturbative result should not be used. In other words, one cannot determine the charge distribution of nuclei directly from the scattering cross section.

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