A Model for Nuclear Fission — Question 3

Question 3

Competing surface and Coulomb changes can favor separated fragments

A two-drop model has parent surface energy 600 MeV and Coulomb self energy 900 MeV. For an equal split at infinite separation use $U_{s,f}=2^{1/3}U_{s,i}$ and $U_{C,f}=2^{-2/3}U_{C,i}$. Other contributions cancel.

1. Compute the potential-energy change and resulting Q value.

2. Find the minimum parent Coulomb-to-surface ratio that makes the separated final state energetically favorable.

Question 3 - Solution

Competing surface and Coulomb changes can favor separated fragments

(a) Method and calculation

$$\Delta U=600(2^{1/3}-1)+900(2^{-2/3}-1).$$$$\boxed{\Delta U=-177.083\,\mathrm{MeV},\quad Q=-\Delta U=177.083\,\mathrm{MeV}}.$$

(b) Interpretation and further calculation

$$\Delta U<0\Rightarrow\boxed{\frac{U_{C,i}}{U_{s,i}}>\frac{2^{1/3}-1}{1-2^{-2/3}}=.702414}.$$$$\boxed{900/600=1.50>.702414}\quad\text{so the separated state is lower}.$$

Physical check

This comparison of endpoints says nothing about the highest energy along the path; it is not a criterion for barrier-free fission.