# Mirror nuclei: accounting for the difference in mass between nuclei

I was wondering if anyone here could guide me in the right direction with respect to the following problem:

Two nuclei are considered mirror nuclei if interchanging the neutrons and protons turns one nucleus into the other. An example is $$^{11}B$$, which consists of five protons and six neutrons, and $$^{11}C$$, which consists of six protons and five neutrons. Determine the difference in mass of these two nuclei assuming that the nuclear binding energy is the same for both and the difference in mass of these nuclei is due to electrostatic energy differences as well as the difference in mass of the nucleon constituents. Assume both nuclei are uniformly charged spheres of the same radius.

Initially, I thought that the difference in the energy required to assemble a charged sphere with $$Z=5$$ and $$Z=6$$ would account for the difference in the energy between the two nuclei and these could be related to the mass by converting to electron-volts and fixing the corresponding units. Is this right?

• Yes this is is correct, but as the question suggests, you need to account also for the difference of mass of neutrons and protons, because remember that neutrons are heavier than protons. So when changing in this case, you will have an extra neutron excess mass to account here. Feel free to solve it and answer your own question, which will earn you a badge :) Aug 18, 2015 at 15:01

Just to put it in writing:

$$M(^{11}_6 C)-M(^{11}_5 B)=m_p-m_n+\frac{\Delta E_C}{c^2}$$

Where $$\Delta E_C$$ is the Coulomb energy difference between two mirror nuclei,

$$\Delta E_C=\frac 3 5 \frac 1 {4\pi\epsilon_0} \frac {e^2} {r_0} A^{2/3}=\frac {3}{5} \frac 1 {4\pi\epsilon_0} \frac {e^2} {r_0}11^{2/3}$$