A Model for Nuclear Fission — Question 5

Question 5

A polynomial deformation path has a finite saddle energy

Consider the schematic potential $U(q)=12q^2-8q^3$ MeV for $q\ge0$, with dimensionless q. The polynomial models only the initial deformation path, not separated fragments.

1. Find the stationary points and identify the barrier above q=0.

2. Find the nonzero q where U returns to zero, and explain why a particle with energy 3 MeV cannot cross the barrier classically.

Question 5 - Solution

A polynomial deformation path has a finite saddle energy

(a) Method and calculation

$$U\prime=24q(1-q)=0\Rightarrow q=0,1;\quad U^{\prime\prime}=24-48q.$$$$\boxed{q=0:\text{ minimum};\quad q=1:\text{ maximum};\quad B=U(1)=4\,\mathrm{MeV}}.$$

(b) Interpretation and further calculation

$$U(q)=4q^2(3-2q)=0\Rightarrow\boxed{q=1.50\text{ (nonzero root)}}.$$$$\boxed{K(q)=E-U(q),\quad K(1)=3-4=-1\,\mathrm{MeV}<0}\quad\text{classically forbidden}.$$

Physical check

An exothermic region beyond the hill does not remove the need to cross or tunnel through the hill. The cubic potential is not a global nuclear potential.