III. Topic : Nuclear Properties

The atomic nucleus is now known to be composed of protons and neutrons known as

nucleon. The number of protons and neutrons in the nucleus is its mass number and the

number of protons is its atomic number . A nucleus, of chemical symbol is uniquely designated by:

The atomic nuclei has some properties of interest:

Nuclear Size: In general atomic nuclei have spherical shape with radius roughly given by:

Charge: - The electric charge distribution within the nucleus is the same as thenuclear mass distribution Experimental results suggest that the ‘electrical radius of the nucleus’ and ‘nuclear matter radius’ are nearly the same.

Nuclear Spin: For each nucleon orbital angular momentum .. and spin combine to

the total angular momentum The total angular momentum of a nucleus is therefore the vector sum of the angular momenta of the nucleons

Angular momentum: The angular momentum has all of the usual properties of quantum mechanical angular momentum vectors:

The total angular momentum is usually referred to as nuclear spin and the

corresponding spin quantum number is used to describe nuclear states.

Nuclear stability is related to the number of nucleons constituting the nucleus. Stable nuclei only occur in a very narrow band in the Z-N plane. All other nuclei are unstable and decay spontaneously in various ways.

It is now well known that an atom

3.1. Basic Properties of the Atomic nucleus,

Charge and Mass of the Nucleus

The most important characteristics of a nucleus are its charge and its mass . The charge on the atomic nucleus is determined by the number of positive charges it contains. The carrier

of an elementary charge, , on the nucleus is proton. Since an atom as a whole is electrically neutral, the nuclear charge simultaneously determines the number of electrons around the nucleus. In other words, chemical elements are identified by their nuclear charge or, by their atomic numbers.

The mass of an atomic nucleus is practically the same as that of the entire atom because the

mass of the electrons in an atom is negligible. The mass of an electron is that of a proton. It is customary to measure the mass of an atom in atomic mass units, abbreviated

amu. The atomic mass unit is equal to one-twelfth of the mass of the neutral atom.

Spin And Magnetic Moment of The Nucleus:

In atomic physics module you have seen that the spin of an electron results in the fine structure of atomic spectrum. For atoms having one valence electron the relative orientation of the orbital and spin moments of the electron leads to the splitting of all energy levels (except the s-level) and as a result, to the splitting of spectral lines. With further improvement of spectroscopic instruments, investigators were able to investigate such lines. It was found that each of the two D-lines of sodium was in turn a doublet, that is , consisting of two very closely spaced spectral lines.

D1(5896A)

o

D1(5890

A)

o

Fig. D-lines of Na

Pauli suggested that the hyperfine structure might be due to an occurrence of angular momentum in the atomic nucleus. The total angular momentum, or nuclear spin, along with nuclear charge and nuclear mass, is the most important characteristic of the nucleus.

The nucleus is made up of protons and neutrons each of which has spin . The nuclear spin is the vector sum of the spin angular momenta of all the component particles. A ucleus

made up of an even number of nucleons has integral spin (in units of ) or zero spin. In addition to nuclear spin, the nucleus has a magnetic moment. Thus, all atomic particles (the nucleus and electrons) have a magnetic moment.

The magnetic moment of a nucleus is determined by those of its component particles. By analogy with the Bohr magneton, the magnetic moments of nuclei are expressed in terms of the so-called nuclear magneton defined as

where is the nuclear gyromagnetic ratio.

Nuclear constituents:

The nuclear model of the atom brought more questions than it answered when it was forwarded. What is the composition of the nucleus? How can a nuclear atom become stable? Answers to these questions could only be given after the discovery of various properties of the nucleus, notably nuclear charge Z, nuclear mass, and nuclear spin.

The nuclear charge was found to be defined by the sum of the positive charges it contains. Since an elementary positive charge is associated with the proton, the presence of protons in the nucleus appeared to be beyond any doubt from the outset Two more facts were also established, namely:

a. The masses of the isotopes (except ordinary hydrogen), expressed in proton mass units, were found to be numerically greater than their nuclear charges expressed in

elementary charge units, this difference growing with increases in . For the elements in the middle of the periodic Table the isotopic masses (in amu) are about twice as great as the nuclear charge. The ratio is still greater for the heavier nuclei. Hence one was forced to think that the protons were not the only particles that make up the nucleus.

b. The masses of the isotopic nuclei of all chemical elements suggested two possibilities, either the particles making up the nucleus had about the same mass, or the nucleus contained particles differing in mass to a point where the mass of some was negligible in comparison with that of the others, theta is, their mass did not contribute to the isotopic mass to any considerable degree.

The latter possibility appeared especially attractive because it fitted nicely with the proton-electron model of the nucleus. That the nucleus might contain electrons seemed to follow from the fact that natural beta-decay is accompanied by the emission of electrons. The proton-electron model also explained the fact why the isotopic atomic weights were nearly integers. According to this model, the mass of the nucleolus should be partially equal to the masses of the protons that make it up, because the electronic mass is about 1/2000th that of the proton. The number of electrons in the nucleus must be such that the total charge due to the positive protons and the negative electrons is the true positive charge of the nucleus.

For all its simplicity and logic, the proton-electron model was refuted by advances in nuclear physics. In fact, it ran counter to the most important properties of the nucleus.

If the nucleus contained electrons, the nuclear magnetic moment would be of the same order of magnitude as the electronic Bohr magneton Notice that the nuclear magnetic moment is defined by the nuclear magneton which is about 1/2000th the electronic magneton.

Data on nuclear spin also witnessed against the proton-electron model. For example,

according to this model the beryllium nucleus, , would contain nine protons and five electrons so that the total charge would be equal to four elementary positive charges. The proton and the electron have each a half-integral spin, h/2. The total spin of the nucleus made up of 14 particles (nine protons and five electrons) would have to be integral. Actually, the

beryllium nucleus, , has half-integral spin of magnitude 3h/2. Many more examples might be cited.

Last but not least, the proton-electron model conflicted with the Heisenberg uncertainty

principle. If the nucleus contained electrons, then the uncertainty in the electron position,

would be comparable with the linear dimensions of the nucleus, that is, or m. Let

us choose the greater value, From the Heisenberg uncertainty relation for the electron momentum we have

The momentum P is directly related to its uncertainty, that is Once the momentum of the electro is known, one can readily find its energy. Since in the above

example , one should use the relativistic relation for energy and momentum

Then we get

This figure is greatly in excess of that (7-8MeV)found for the total binding energy by experiment and is many times the energy of electrons emitted in beta-decay. If, on the other hand, the electrons in the nucleus were assumed to have the energy comparable with that associated with the particles emitted in beta-decay (usually a few MeV), then the region where the electrons must be localized, that is, the size of the nucleus as found from the uncertainty relations would be much greater than that found by observation.

A way out was found when in 1932 Chadwick discovered a new fundamental particle. From an analysis of the paths followed by the particles produced in some nuclear reactions and applying the law of conservation of energy and momentum, Chadwick concluded that these paths could only be followed by a particle with a mass slightly greater than that of the proton and with a charge of zero. Accordingly, the new particle was called the neutron.

According to the present views, a nucleus consists of nucleons: protons and neutrons. As the mass of a nucleon is about 2000 times the mass of an electron the nucleus carries practically all the mass of an atom

A nuclid is a specific combination of a number of protons and neutrons. The complete symbol for a nuclide is written as:

where is the chemical symbol of the element, is the atomic number, giving the number

of protons in the nucleus. is the totla number of nucleons in the nuclues. It is also known

as the mass number. is the number of neutrons.

In nucleus physics it is said that the proton and the neutron are two charge states of the same particle, the nucleon. The proton is the protonic state of the nucleon with a charge +e, and the neutron is its neutronic state with zero charge. According to the latest data, the rest mass of a proton and of a neutron respectively is

The proton and the neutron have the same mass number equal to unity. In the nucleus, the nucleons are in states substantially differing from their free states. This is because in all nuclei, except that of ordinary hydrogen, there are at least two nucleons between which a special nuclear interaction or coupling exists.

The proton-neutron model of the nucleus accounts for both the observed values of isotopic masses and, the magnetic moments of the nuclei. For, since the magnetic moments of the proton and the neutron are of the same order of magnitude as the nuclear magneton, it follows that a nucleus built up of nucleons should have a magnetic moment of the same order as the nuclear magneton. Therefore, with protons and neutrons as the building blocks of nuclei, the magnetic moment should be of the same order of magnitude. Observations have confirmed this.

is the typical length scale of nuclear physics.

Also with protons and neutrons as the constituents of nuclei, the uncertainty principle leads to reasonable value of energy for these particles in a nucleus, in full agreement with the observed energies per particle

Finally, with the assumption that nuclei are composed of neutrons and protons, the difficulty arising from nuclear spin has likewise been resolved. For if a nucleus contains an even

number of nucleons, it has integral spin (in units of ). With an odd number of nucleons, its

spin will be half-integral (in units of ).

3.2. Nuclear Binding Energy

Atomic nuclei containing positively charged protons and uncharged neutrons make up stable systems despite the fact that the protons experience Coulomb repulsion. The stability of nuclei is an indication that there must be some kind of binding force between the nucleons. The binding force can be investigated on the energy basis alone, without invoking any considerations concerning the nature and properties of nuclear forces.

An idea about the strength of a system can be gleaned from the effort required to break it up i.e. to do work against the binding. This approach leads to several important facts about the forces that hold the nucleons in a nucleus.

The energy required to remove any nucleon from the nucleus is called the binding (or separation) energy of that nucleon in the nucleus. It is equal to the work that must be done in order to remove the nucleon from the nucleus without imparting it any kinetic energy. The total binding energy of a nucleus is defined as the amount of work that must be done in order to break up the nucleus into its constituent nucleons. From the law of conservation of energy it follows that in forming a nucleus, the same amount of energy must be released as is put in to break it up.

The magnitude of the binding energy of nuclei may be estimated from the following considerations. The rest mass of any permanently stable nucleus has been found to be less than the sum of the rest masses of the nucleons that it contains. It appears as if in “packing up’’ to form a nucleus the protons and neutrons lose some of their masses.

An explanation of this phenomenon is given by the special theory of relativity. This fact is accounted for by the conversion of part of the mass energy of the particles into binding

energy. The rest energy of a body, , is related to its rest mass by:

o

where is the velocity of light in a vacuum. Designating the energy given upon the

formation of a nucleus as , then the mass equivalent of the total binding energy

o

is the decrease in the rest mass as the nucleons combine to make up the nucleus. The quantity

is also known as mass defect or mass decreament. If a nucleus of mass M is composed

of a number Z of protons with a mass and of a number A-Z of neutrons with a mass ,

the quantity is given by

The quantity gives a measure of the binding energy:,

In nuclear physics, energies are expressed in atomic energy units (aeu) corresponding to atomic mass units:

Thus, in order to find the binding energy in MeV, one should use the equation

Where the masses of the nucleons and the mass of the nucleus are expressed in atomic mass units. On the average, the binding energy per nucleon is about 8MeV, which is a fairly large amount.

Fig: A plot of the binding energy per nucleon as a function of mass number A

As is seen from the plot, the strength of binding varies with the mass number of the nuclei. The binding is at its strongest in the middle of the periodic Table, in the range 28<A<138,

that is, from In these nuclei, the binding energy is very close to 8.7 MeV. With further increases in the number of nucleons in the nucleus, the binding energy per nucleon

decreases. For the nuclei at the end of the periodic Table (for example, uranium), is about 7.6 MeV.

In the region of small mass numbers, the binding energy per nucleon shows characteristic maximua and minima. Minima in the binding energy per nucleon are shown by nuclei

containing an odd number of protons and neutrons, such as

Maxima in the binding energy per nucleon are associated with nuclei having an even number

of protons and neutrons, such as

The general course of the curve gives a clue to the mechanisms by which nuclear energy is released. We find that nuclear energy can be released either by the fission of heavy nuclei and the fusion of light nuclei from still lighter ones. It is clear from general considerations that energy will be released in nuclear reactions for which the binding energy per nucleon in the end products exceeds the binding energy per nucleon in the original nuclei.

3.2. Nuclear Stability

Not all nuclei are stable. Unstable nuclei undergo radioactive decay into different nuclei.

Stable nuclei have approximately equal numbers of neutrons and protons for small

and a small excess of neutrons for large as shown in the diagram.

The Pauli exclusion principle helps to understand the fact that nuclei with equal and are stable. Imagine filling a 1-deminsional box with protons and neutrons. We want the minimum

energy configuration for a given value of , say 5. Since both neutrons and protons have spin ½ they are fermions (like electrons) and so obey the Pauli exclusion principle. This principle restricts the number of protons and neutrons to 2 of each at each energy level.

Recall that the energy of the nth energy in a 1-dimensional box is given by , where

is the energy of the round level.

If all 5 nucleons were neutrons, the total energy of the nucleus would be

as shown in diagram . In contrast, if 3 were neutrons and 2

were protons (as shown in B), the energy would be which is far less.

This simple picture shows that it is more favourable energetically to have

If we include the Coulomb repulsion between the protons, the energy levels of the protons

become higher than the energy levels of the neutrons. As increases, it becomes more favourable to have a small excess of neutrons.

Some elements have more stable isotopes than others. The elements with the most number of

stable isotopes have values of 2, 8, 20, 28, 50, 82 and 126. These are called magic numbers, as the reason for stability was not understood at the time they were discovered. For

example, calcium has 6 stable isotopes whereas potassium and scandium

have only 2 stable isotopes each. Similarly, nuclei with N equal to a magic number

have a larger than average number of isotones (an isotone has the same value but a

different value).

Nuclei with are more tightly bound together and so they are at lower energy compared to the rest. (Binding energy is analogous to the energy required to lift a bucket of water from a well. A large binding energy means the water is low in the well, i.e. the water is at a low energy). If

two light nuclei with are brought together they create a new nuclei at lower rest

energy (this is called fusion). Also a heavy with can split into two nuclei of lower rest energy (this is called fission).

3.3. Mass and Isotopic Abundance

Properties of the atomic nucleus, discussed in the prevous sections, binding energies; decay rates, etc are the basic quantities determining the elemental and isotopic abundances in nature.

The relative abundance of an isotope in nature compared to other isotopes of the same element is relatively constant. The Chart of the Nuclides presents the relative abundance of the naturally occurring isotopes of an element in units of atom percent. Atom percent is the percentage of the atoms of an element that are of a particular isotope. Atom percent is

abbreviated as a/o. For example, if a cup of water contains atoms of oxygen, and

the isotopic abundance of oxygen-18 is 0.20%, then there are atoms of oxygen-18 in the cup.

The atomic weight for an element is defined as the average atomic weight of the isotopes of the element. The atomic weight for an element can be calculated by summing the products of the isotopic abundance of the isotope with the atomic mass of the isotope.

Example:

Calculate the atomic weight for the element lithium. Lithium-6 has an atom percent

abundance of 7.5% and an atomic mass of 6.015122 amu. Lithium-7 has an atomic abundance of 92.5% and an atomic mass of 7.016003 amu.

Solution:

The other common measurement of isotopic abundance is weight percent (w/o). Weight percent is the percent weight of an element that is a particular isotope. For example, if a sample of material contained 100 kg of uranium that was 28 w/o uranium-235, then 28 kg of uranium-235 was present in the sample.