A Model for Nuclear Fission — Question 1

Question 1

Conserving drop volume forces the surface area to grow on splitting

A spherical liquid drop of radius R splits into two identical spherical drops without changing total volume. Surface energy is proportional to area, and the initial surface energy is 600 MeV.

1. Derive the daughter radius and the ratio of final to initial total area.

2. Calculate the increase in surface energy and explain its resistance to fission.

Question 1 - Solution

Conserving drop volume forces the surface area to grow on splitting

(a) Method and calculation

$$\frac43\pi R^3=2\frac43\pi r^3\Rightarrow\boxed{r=2^{-1/3}R}.$$$$\boxed{\frac{S_f}{S_i}=\frac{2(4\pi r^2)}{4\pi R^2}=2^{1/3}=1.25992}.$$

(b) Interpretation and further calculation

$$\boxed{\Delta U_s=600(2^{1/3}-1)=155.953\,\mathrm{MeV}}.$$$$\boxed{\Delta U_s>0}\quad\text{so creating additional surface costs energy}.$$

Physical check

The model preserves nuclear density. Surface energy is positive potential energy and therefore contributes negatively to binding.