[NOTES/EM-03013] Electrostatic Energy of Nuclei

A pair of isotopes of two different elements are called {\tt mirror nuclei} having the number of protons and neutrons, \((Z_1, N_1)\) and \((Z_2,N_2)\) for the two nuclei are such that \(Z_1=N_2\) and \(N_1=Z_2\). An example of mirror nuclei is B\(^{11}\) and C\(^{11}\). These nuclei have the same mass number \(A=11\) and B\(^{11}\)  has five protons and six neutrons, and C\(^{11}\) has six protons and five neutrons.

Consider two nuclei \(X_1, X_2\) having \(Z_1, Z_2\) protons and mass number \(A\). The definition of binding energy gives the relations.

\begin{eqnarray} M_{X_1}c^2&=&Z_1m_pc^2+(A-Z_1)m_nc^2-(BE_1)\nonumber \\ M_{X_2}c^2&=&Z_2m_pc^2+(A-Z_2)m_nc^2-(BE_2) \\ \nonumber
\nonumber(M_{X_1}-M_{X_2})c^2&=&(Z_1-Z_2)(m_p-m_n)c^2+(BE_2-BE_1)\\
BE_2-BE_1&=&(M_{X_1}-M_{X_2})c^2+(Z_1-Z_2)(m_p-m_n)c^2 \label{eq38}
\end{eqnarray}

This difference in the binding energies can be computed using known masses of the two nuclei and the neutron proton mass difference.

If the nuclear forces are charge independent, the difference in the binding energies of these mirror nuclei should be accountable by the difference in electrostatic energies.

In order to test the assumption of charge independence of nuclear forces we estimate the electrostatic energy of nucleus under the assumption that the positive charge of protons is distributed uniformly over a sphere of  the size of nucleus.

If the number of protons in a nucleus is \(Z\) and the radius is \(R\), the electrostatic energy will be given  by

\begin{equation}
     U = \frac{3}{5}\frac{(Ze)^2}{4\pi\epsilon_0 R}.
\end{equation}


Of the mirror nuclei, B\(^{11}\) and C\(^{11}\), the mass difference is approximately 1.982 MeV. Substituting the numbers \(Z_1=5, Z_2=6\), \((m_p-m_n)c^2=1.29 MeV\), in \eqref{eq38}

\begin{eqnarray}
\text{BE(B)}^{11} - \text{BE(C)}^{11}= 1.982 + 0.784 = \text{2.7 66 MeV}.
\end{eqnarray}

Can we explain  this difference in terms of  difference in the electrostatic energy of the two nuclei? So, we now compute the difference in the electrostatic energies of the two nuclei.


Taking the sizes of the mirror nuclei to be equal, the difference in electrostatic energies works out to be
\begin{equation}
     U = \frac{3}{5}\frac{(Z_1^2-Z_2^2)e^2}{4\pi\epsilon_0 R}.
\end{equation}
Taking the nuclear radius of either  carbon, or boron we get the difference in the electrostatic  energies to be about 3.02 MeV. This is in agreement with the experimental value of the difference of 2.76 MeV in the binding energies  within 10\%.

Considering the simplified nature of the estimate, the agreement with the experiments is fairly good.