A Model for Nuclear Fission — Question 2

Question 2

Charge separation reduces the self energy of two distant drops

A uniformly charged drop has Coulomb self energy $U_C=3Q^2/(20\pi\varepsilon_0R)$. It divides equally in charge and volume. Treat the two daughter drops as infinitely separated; initial self energy is 900 MeV.

1. Find the sum of daughter self energies relative to the parent value.

2. Compute the energy decrease and identify the extra Coulomb term needed at a finite center separation d.

Question 2 - Solution

Charge separation reduces the self energy of two distant drops

(a) Method and calculation

$$Q_d=Q/2,\quad r=R2^{-1/3}\Rightarrow\frac{U_{C,f}}{U_{C,i}}=2\frac{(Q/2)^2}{Q^2}\frac Rr.$$$$\boxed{U_{C,f}/U_{C,i}=2^{-2/3}=.629961}.$$

(b) Interpretation and further calculation

$$\boxed{\Delta U_C=900(2^{-2/3}-1)=-333.036\,\mathrm{MeV}}.$$$$\boxed{U_{\mathrm{mutual}}(d)=\frac{(Q/2)^2}{4\pi\varepsilon_0d}>0}\quad(d\ge2r).$$

Physical check

The mutual term vanishes only at infinite separation. Adding it at contact prevents confusing self-energy relief with the full deformation path.