3.5: Stability of Nuclei

In Figure $\PageIndex{1}$ we have color coded the nuclei of a given mass $A=N+Z$ by their mass, red for those of lowest mass through to magenta for those of highest mass. We can see that typically the nuclei that are most stable for fixed $A$ have more neutrons than protons, more so for large $A$ increases than for low $A$. This is the “neutron excess”.

Figure $\PageIndex{1}$: The valley of stability

$\beta$ decay

If we look at the mass of nuclides with fixed nucleon number $A$ (i.e., roughly perpendicular cuts through the valley of stability in Figure $\PageIndex{1}$), we can see that the masses vary strongly,

Figure $\PageIndex{2}$: The negative of binding energy per nucleon for nuclides with fixed $A$: (left) $A= 56$ and (right) $A= 150$. The profile of binding energy across the valley of stability is roughly a parabola (e.g., Iron-56 is stable, while Vandium-56 is unstable to $β^−$ decay.

It is known that a free neutron is not a stable particle, it actually decays by emission of an electron and an antineutrino,

$$n \rightarrow p + e^-+\bar{\nu}_e. \nonumber$$

The reason that this reaction can take place is that it is endothermic,

$$m_n c^2 > m_p c^2 + m_e c^2. \nonumber$$

Here we assume that the neutrino has no mass.

The degree of allowance of such a reaction is usually expressed in a $Q$ value, the amount of energy released in such a reaction,

$$\begin{align} Q &=m_n c^2 - m_p c^2 - m_e c^2 \\[5pt] &=939.6-938.3-0.5=0.8\text{ MeV}. \end{align} \nonumber$$

Generically it is found that two reaction may take place, depending on the balance of masses. Either a neutron “$\beta$ decays” as sketched above, or we have the inverse reaction

$$p \rightarrow n + e^++{\nu}_e. \nonumber$$

For historical reason the electron or positron emitted in such a process is called a $\beta$ particle. Thus in $\beta^-$ decay of a nucleus, a nucleus of $Z$ protons and $N$ neutrons turns into one of $Z+1$ protons and $N-1$ neutrons (moving towards the right in Figure $\PageIndex{2A}$. In $\beta^+$ decay the nucleus moves to the left. Since in that figure I am using atomic masses, the $Q$ factor is

$$\begin{aligned} Q_{\beta^-}&= M(A,Z)c^2-M(A,Z+1)c^2,\nonumber\\ Q_{\beta^-}&= M(A,Z)c^2-M(A,Z-1)c^2-2m_e c^2.\end{aligned} \nonumber$$

The double electron mass contribution in this last equation because the atom looses one electron, as well as emits a positron with has the same mass as the electron.

In similar ways we can study the fact whether reactions where a single nucleon (neutron or proton) is emitted, as well as those where more complicated objects, such as Helium nuclei ($\alpha$ particles) are emitted. I shall return to such processed later, but let us note the $Q$ values,

$$\begin{aligned} \text{neutron emission} && Q&=[M(A,Z)-M(A-1,Z)-m_n]c^2, \nonumber\\ \text{proton emission} && Q&=[M(A,Z)-M(A-1,Z-1)-M(1,1)]c^2, \nonumber\\ \text{$\alpha$ emission} && Q&=[M(A,Z)-M(A-4,Z-2)-M(4,2)]c^2, \nonumber\\ \text{break up} && Q&=[M(A,Z)-M(A-A_1,Z-Z_1)-M(A_1,Z_1)]c^2.\end{aligned} \nonumber$$