Forces in the Nucleus

The mass of the charged particles (protons, nuclei, and electrons) can be deter­mined by injecting them at a high speed into a magnetic field, where depending on their charge and mass, the path of particles deviates from a straight line. Neutrons, however, have no charges, so the mass of a neutron cannot be measured in this way; rather, its mass must be deduced. This can be achieved by the dissociation of the deuterium nucleus (one proton and one neutron) to a proton and neutron under the effect of gamma radiation.

The masses of free protons, neutrons, and electrons are listed in Table 2.1. When comparing the mass of the nucleus of an atom to the total mass of the free protons and neutrons, we can see that the sum of the mass of the free nucleons is always greater than the mass of the corresponding nucleus in the atom.

This difference will be equal to the binding energy of the nucleus (AE). [Note that Einstein’s formula for the equivalence of mass and energy (shown in Eq. (1.1)) can be used to calculate the binding energy.] When the binding energy of the nucleus is divided by the mass number, the binding energy per nucleon is obtained (A E/A):

AE (M — Z X mp — N X mn — Z X me )c2 ^ лл

“A A ( ‘ )

where M is the mass of the atom (not the nucleus!).

Mass number, A Figure 2.2 The binding energy

image012

The stability of a given nucleus can be characterized by the value of the binding energy per nucleon. The binding energy per nucleon as a function of atomic mass is shown in Figure 2.2.

The characteristic binding energy per nucleon for the most stable nuclei is in the range of 7—9 MeV. The absolute value of the binding energy per nucleon—the mass number function shows a maximum of about mass numbers 50—60. This mass represents the elements of the iron group; thus, these elements are the most stable ones in the periodic table. The smallest nucleus, where the term of the bind­ing energy per nucleon can be defined, is the deuterium, which has the smallest binding energy per nucleon (around 1.8 MeV).

The binding energy is usually expressed in millions of electron volts. One electron volt is the amount of energy gained by an electron (elementary charge, 1.6 X 10219C) when it is accelerated through an electric potential of 1 V. Transferring the electron volt to the SI unit of energy (joule), 1 eV = 1.6 X 10"19 C X 1V = 1.6 X 10"19 J. For every 1 mol of electrons [found by multiplying by the Avogadro’ number (6 X 1023 particles/mol)], about 105 J is obtained. The energy of an atomic mass unit (931 MeV), mentioned in Table 2.1, is в 1013 J. The binding energy per nucleon (7—9 MeV) is about 1011 J that is 108kJ. Therefore, the binding energy of the nuclei is about 108 kJ/mol.

Now let’s compare the binding energy of the nuclei to the energy of chemical bonds. The energy of primary (ionic, covalent) chemical bonds is a few hundred kJ/mol (an amount of electron volts). Thus, the difference is about six orders of magnitude: the binding energy of the nuclei is about a million times higher than the energy of the chemical reactions.

In 1935, Yukawa provided an interpretation of the nature of the forces in the atomic nuclei using quantum mechanics. He constructed a model similar to the one for electrostatic forces, where two charged particles interact through the electro­magnetic field. In Yukawa’s model, the so-called meson field should be substituted for the electromagnetic field. In the case of the meson field, the range of the interaction is very short (about 10_15m), while the electromagnetic field has a much bigger range. The potential between two particles in the nuclei, known as the

Yukawa potential (U), can be expressed as a function of the distance of the two particles (r):

Подпись: (2.5)U _ g2 exp(r/R)

image014 Подпись: (2.6)

In Eq. (2.5), the potential is negative, indicating that the force is attractive. The constant g is a real number; it is equal to the coupling constant between the meson field and the field of the protons and neutrons. R is the range of the nuclear forces, expressed as follows:

where h is Planck’s constant, c is the velocity of light in a vacuum, and mn is the rest mass of the meson. Assuming that the meson field range is about 10-15m, Yukawa suggested that there must be a particle with a rest mass of about 200 times that of an electron. In fact, this particle was observed in the cosmic ray in 1948. It is called п-meson, and its rest mass is 273 times higher than the rest mass of the electron. The meson is a kind of elementary particle (as discussed in Section 2.2).

The total nuclear binding energy (ДE) can be given approximately on the basis of nuclear forces, by the summation of the interaction energies of the nucleon pairs (Ur, kl) at the distance r:

ДЕ _-2XX Urk (2.7)

2 k I

where ДЕ is the total nuclear binding energy and k and I are the number of protons and neutrons, respectively. The protons and neutrons are considered to be identical. The factor 1/2 is in Eq. (2.7) because of the two summations for protons and neu­trons, so each nucleon appears twice. The total binding energy of the nucleus is proportional to the product of the number of protons and neutrons:

Z X N _ Z X (A — Z) (2.8)

image016 Подпись: (2.9)

The function of the total binding energy has a maximum of the atomic number expressed as follows:

From here

Подпись:(2.10)

In conclusion, those nuclei should be stable, such that the number of protons and neutrons are equal. This is indeed the case for light nuclei (e. g., 4He, 12C, 14N, 16O, 24Mg). However, for heavier nuclei, the number of protons increases, so the electrostatic repulsion of the positively charged protons increases. For this reason, extra neutrons are needed for stability. So Eq. (2.10) is modified as:

A $ 2Z (2.10a)

which means that the nuclei with high atomic numbers are stable at the ratio of neutron/proton = 1.4. The nuclei with medium atomic numbers have a ratio of neu — tron/proton between two values (1 — 1.4), i. e., the ideal neutron/proton ratio of the stable nuclei continuously changes in the periodic table (Figure 2.3).

image019 Подпись: (2.11)

The energy of the electrostatic repulsion can be calculated as follows:

where e is the elementary charge and Ra is the radius of the nucleus.

Подпись: Figure 2.3 The stability of nuclei: atomic number as a function of the number of the neutrons. Source: Conventional nuclear chart taken from Bes (1965) with permission from Elsevier.

The nuclei are classified as isotope, isobar, isoton, or isodiaphere based on the number of nucleons (Table 2.2).

Neutron number