3. Stability of nuclei

Chapter 3 from BASICS OF NUCLEAR PHYSICS AND OF RADIATION DETECTION AND MEASUREMENT – An open-access textbook for nuclear and radiochemistry students by Jukka Lehto

In the nature there are 92 elements and these have altogether 275 stable nuclides. Of these nuclides, 60% have both even number of protons and neutrons. They are called even-even nuclides. About 40% of nuclides have either even number of protons or neutrons. If the proton number is even and neutron number is odd the nuclide is called even-odd nuclide and in the opposite case and odd-even nuclide. Odd-odd nuclides where both proton and neutron numbers are odd are very rare. In fact, there are only four of them: ²H, ⁶Li, ¹⁰B and ¹⁴N.

Based on what was told above it is obvious that nuclei prefer nucleon parity and in the way that protons favor parity with other protons and neutrons with other neutrons. Single proton and single neutron do not form a pair with each other, which can be seen as the small number odd-odd nuclides. When looking at the nuclide chart, one can see that the elements with an even atomic number have considerably more stable isotopes than those having odd atomic number. For example ₃₂Ge has five stable isotopes, of which four have also an even number of neutrons, while ₃₃As has only one stable isotope and ₃₁Ga two.

Another important factor affecting stability of nuclei is the neutron to proton ratio (n/p). There are two forces affecting the stability of nucleus and they work in opposite directions. The nuclear force is binding the nucleons together, while the electric repulsion force due to protons' positive charges is pushing the protons away from each other. The stable isotopes of the lightest elements have the same, or close to same number of protons and neutrons. As the atomic number of the elements is increasing the repulsion force due to increasing number of protons increases. To keep the nucleus as one piece heavier elements have increasing number of neutrons compared to protons. In the heaviest stable element, bismuth, the n/p ratio is around 1.5, while the heaviest naturally occurring element, uranium, has 92 protons and 143 (²³⁵U) or 146 (²³⁸U) neutrons, the n/p ratio thus being 1.6. Figure III.1 (left side) shows all nuclides as a nuclide chart where the neutron number is on x-axis and the proton number of y-axis. The stable nuclides are seen as black boxes and one can see that they are not on the diagonal line (n/p=1), except in case of the lightest elements, but on a bended curve due to systematically increasing n/p ratio. The nuclides above the stable nuclides are proton-rich radionuclides and those below neutron-rich radionuclides. As seen from the right side of the Figure III.1 the n/p ratio remains close to unity up to atomic number 20 (Ca) where after clear and increasing excess of neutrons appear.

Figure III.1 Nuclide chart (left) (http://oscar6echo.blogspot.fi/2012/11/nuclide-chart.html). Neutron numbers of the most abundant isotopes of the stable elements as a function of atomic number (right).

The masses of atoms, nuclei and nucleons are not presented in grams due to their very small numbers but instead in relative atomic mass units (amu). Atomic mass unit is defined as a twelfth part of the mass of ¹²C isotope in which there are six protons and six neutrons. The mass of ¹²C is 12 amu and amu presented in grams

1 amu = (12 g/ 12) / N = 1g / N = 1.66·10^-24 g [III.I]

where N is the Avogadro number 6.023·1023 mol^-1. The mass of a proton is 1.00728 amu and that of neutron 1.00867 amu. The mass of an electron is only one 2000^th part of the masses of nucleons, 0.000548597 amu. The masses in amu units are presented with the capital M while those in grams are presented as lower case m.

At first sight, it would look logical that the mass of a nucleus (M_A) is the sum of masses of neutrons and protons.

M_A (calculated) = Z × M_p + N × M_n [III.II]

For example the mass of deuteron would thus be 1.007825 amu + 1.008665 amu = 2.016490 amu. The measured mass, however, is 2.014102 amu, being 0.002388 amu smaller than the calculated mass. This mass difference is called the mass defect ΔM and its general equation is:

ΔM_A = M_A (measured) - M_A (calculated) [III.III]

Mass defect is the mass a nucleus loses when it is “constructed” from individual nucleons.

Mass (m) and energy (E) are interrelated and converted to each other by the Einstein equation E = m×c² (c = speed of light). Thus, the mass defect can be converted to energy and one amu corresponds to 931.5 MeV of energy. In radioactive decay and nuclear reaction processes, the energies are presented as electron volts (eV) which is the energy needed to move an electron over 1 V potential difference. Electron volts are used as energy unit instead of joules (J) since the latter would yield very low numbers and the formers are thus more convenient to use. In joules one electron volt is as low as 1.602·10^-19 J. The mass defect of deuteron 0.002388 amu corresponds an energy of 2.224 MeV. This energy is released when one proton combines with one neutron to form a deuteron nucleus. Accordingly, at least this much of energy is needed to break deuteron nucleus to a single proton and a neutron. This mass defect, presented as energy, is thus the binding energy E_B) of the nucleus.

E_B (MeV) = -931.5 ΔM_A (amu) [III.IV]

Table III.I. gives the mass defects and binding energies to some elements. As seen, the total binding energy systematically increases with the mass of the nuclide, being already 1800 MeV in uranium In addition, binding energies are given per nucleon (E_B/A) in Table III.II. This value is also presented for a number on elements as a function of their mass numbers in Figure III.2.

Table III.II. Atomic masses (M_A), mass defects (ΔM_A), binding energies (E_B) and binding energies per nucleon (E_B/A) for some elements.

Figure III.2. Binding energy per nucleon E_B/A (MeV/A) of elements as a function of their mass number (http://www.daviddarling.info/encyclopedia/B/binding_energy.html).

The Figure III.2 shows that the binding energy per nucleon increases dramatically from the lightest elements to mass numbers between 50 and 60. As the E_B/A value for tritium is 1.11 MeV, it is at 8.8. MeV for iron (A=55). Thereafter E_B/A slowly decreases, being for example 8.02 MeV for thallium and 7.57 MeV for uranium. Thus, the intermediate-weight elements, such as iron, cobalt and nickel are the most stable. At certain mass numbers, or strictly speaking at certain proton and neutron numbers, the lighter nuclides have exceptionally higher binding energy than their neighbors do as seen in Figure III.2 and these nuclides are also exceptionally stable. For example, for ⁴He the E_B/A value is as high as 7.07 MeV whereas for the next heavier nuclide ⁶Li it is only 5.33 MeV. The high stability of ⁴He makes it understandable why in alpha decay they are emitted from heavy nuclides.

When the lighter nuclides (Z<25) are presented in a three-dimensional coordinate system where the axes are mass excess (mass defect with positive sign), atomic number (Z) and neutron number (N), we see a picture presented in Figure III.3. The figure shows a formation of the so-called energy valley where the stable nuclides are at the valley bottom while the proton-rich radionuclides are at the left edge and neutron-rich radionuclides at the right edge. As will be later described the radionuclides decay by beta decay on diagonal isobaric lines from the edges to the bottom.

Figure III.3. Energy valley, i.e. mass excess as a function of neutron and proton numbers of lighter elements (Z = 1-25) (http://slideplayer.com/slide/1589163/)

One can see from Figure III.2. that when two light nuclides (A<30) combine into a heavier nuclide energy is released. For example, combination of two ²⁰Ne nuclides into ⁴⁰Ca nuclide releases 24 MeV of binding energy. The E_B/A value for ²⁰Ne is 8.0 MeV and the total binding energy of two ²⁰Ne nuclides is 2×20×8.0 MeV = 320 MeV, whereas the total binding energy of ⁴⁰Ca is 24 MeV higher (40×8.6 MeV = 344 MeV). Combination of two light elements to give a heavier and energetically more stable element is called fusion. Fusion has been exploited in fusion bombs, also called hydrogen bombs, in which fusion of deuterium and tritium creates huge amounts of energy. For energy production, fusion reactors are still being developed where the biggest problem is how to obtain high enough temperature to induce and maintain the fusion process. In fusion reactors, this is accomplished by using plasma but it cannot be done in energetically or economically profitable way yet. In fusion bombs, the high temperature is obtained by exploding a U or Pu fission bomb, covering the fusionable material.

An opposite reaction to fusion is fission where heavy elements split into two lighter, usually intermediate-sized, elements. For example, in the following fission reaction:

²³⁶U → ¹⁴⁰Xe + ⁹³Sr + 3n [III.V]

191.5 MeV energy is released since the binding energy of ²³⁶U per nucleon is 7.6 MeV and the total binding energy 236×7.6 MeV = 1793.6 MeV. The binding energies per nucleon for ¹⁴⁰Xe and ⁹³Sr are 8.4 MeV and 8.7 MeV, respectively, and their total binding energies 1176 MeV and 809.1 MeV. Thus the energy released in this reaction is 1176 MeV + 809.1 MeV – 1793.6 MeV = 191.5 MeV. Of the naturally occurring elements uranium partly (0.005%) decays by spontaneous fission. A more common way to obtain fission is the induced fission in which the heavy elements are bombarded by neutrons. For example, the ²³⁶U in equation III.V is obtained by exposing ²³⁵U to a neutron flux. After neutron absorption the forming ²³⁶U nucleus is excited which results in fission reaction. There are two terms for fissioning nuclide, fissionable and fissile. Fissionable is a general term for any nuclide able to undergo a fission while fissile means that a nuclide undergos fission with thermal neutrons. Fissile nuclei, the most important of which are ²³⁵U and ²³⁹Pu, have even proton number but odd neutron number. When a thermal neutron enters the nucleus of a fissile element the nucleus goes to an excited energy state due to paring of nucleons. Fission, as fusion, was first utilized in bombs in 1940s but from 1950s on the main exploitation field has been nuclear energy production.

Based on the liquid droplet model of nucleus in which nucleons are taken as incompressible droplets that have a binding interaction with only their closest neighbors, an equation has been derived to calculate the binding energies of nuclei. This equation is semi-empirical since some of its parameters have been obtained from experimental data.

E_B (MeV) = a_v×A – a_a ×(N-Z)²/A - a_c×Z²/A^1/3 – a_s×A^2/3 ± a_δ/A^3/4 [III.VI]

where A is the mass number, Z the atomic number, N the neutron number and a_v, a_a, a_c, a_s and a_δ are experimentally obtained coefficients, the values of which are a_v = 15.5, a_a = 23, a_c = 0.72, a_s = 16.8 and a_δ = 34. The basic starting point in the equation is that the total binding energy is directly proportional to the number of nucleons which is taken into in the equation by the term a_v×A. This ”volume energy” decreases by factors that are taken into account by the three further terms in the equation. The last term either increases or decreases the energy or has no effect on it. The second term a_a×(N-Z)²/A takes into account the variance in neutron to proton ratio and is called asymmetry energy, the third term a_c×Z²/A^1/3 takes into account the coulombic repulsion between the protons and the fourth term a_s×A^2/3 takes into account the energies on the surface of a nucleus which differ from those in its bulk. The fifth term a_δ/A^3/4 takes into account the parity of nucleons: in case of an even-even nuclide the term has a positive value while for odd-odd it is negative. For even-odd and odd-even nuclides term has a value of zero.

When N in the equation is substituted by A-Z we get an equation where the binding energy is presented as a function of atomic number Z:

E_B = a Z² + b Z + c ± d/A^3/4 [III.VII]

where the coefficients a, b and c are dependent only on the mass number A. At a fixed mass number, i.e. isobaric line, the equation has a form of a parabel. Figure III.4 shows an example of such parabels which are important in understanding the beta decay modes decribed in chapter IV. In beta decay processes the decay takes place on isobaric lines, a neutron is transformed into a proton or vice versa and thus no change in the mass number is taken takes place. The curves in Figure III.4 present the nuclides in beta decay chains. The nuclide, or nuclides, on the bottom are stable while nuclides on the edges are radioactive stepwisely decaying towards the stable nuclide(s), neutron-rich radionuclides on left side by β- decay and proton-rich nuclides on the right side by β+/EC decay. Identical parabel is obtained when taking a cut in the energy valley (Figure III.3) along an isobaric line. The third term in equation III.VI, ± d/A^3/4, explains why there is only one parabel on the left side of the Figure III.4 and two overlying parabels on the right side. When the mass number in the isobar is odd the ± d/A^3/4 term has value of zero and all the transformations occur from odd-even to even-odd or vice versa. In the other case, when the mass number is even, the transformations are from odd-odd to even-even (or vice versa) and correspondingly the term ± d/A^3/4 changes its sign in each transformation which results in the formation of two parabels. The upper parabel is for the less stable nuclide, that is the odd-odd nuclide, whereas the lower parabel for the even-even nuclides.

Figure III.4. Parabels derived from semiempiral equation of nuclear mass for a fixed mass number. a) odd mass number, b) even mass number. Points on the parabels represent nuclides and the lines between them beta decay processes.

Droplet model, on which the semiempirical equation of nuclear mass is based, widely but not completely explains the nuclear mass and systematics of energy changes in nuclear transformations. However, as can also be seen in Figure III.2, there are some exceptions which cannot be described with the equation. For example, ⁴H, ¹⁶O, ⁴⁰Ca, ⁴⁸Ca and ²⁰⁸Pb have exceptionally high stabilities. From this it has been concluded that certain neutron and proton numbers create higher stabilities. These numbers, called magical numbers, are 2, 8, 20, 28, 50, 82 for both protons and neutrons and 126 for neutrons. These cannot be described with the droplet model but instead a nuclear shell model has been applied to explain the nature of magical numbers. This shell model is analogous to atomic electron shell model: nucleons are located on certain shells which have sites for only a certain number of nucleons on and those nuclides having full shells are more stable than the others. Droplet model and shell model are not exlusive, they rather complement each other.

With superheavy elements the magical neutron and proton number are no more identical. After element with atomic number of 82 the magical proton numbers are 114, 126, 164, 228 while the corresponding values for neutrons above N=126 are 184, 196, 228 and 272.