A Model for Nuclear Fission — Question 4

Question 4

Small deformations test local stability of a nuclear drop

For a dimensionless quadrupole deformation q, a supplied model gives $U(q)-U(0)=(2U_s-U_C)q^2/5+O(q^3)$. Compare drop A with $U_s=500$ MeV, $U_C=800$ MeV and drop B with $U_s=500$ MeV, $U_C=1100$ MeV.

1. Calculate the curvature at the spherical state and classify each drop locally.

2. Estimate the energy change at q=.100 for each and state the limit of this expansion.

Question 4 - Solution

Small deformations test local stability of a nuclear drop

(a) Method and calculation

$$\boxed{U^{\prime\prime}(0)=\frac25(2U_s-U_C)}.$$$$\boxed{U_A^{\prime\prime}=80\,\mathrm{MeV}>0,\quad U_B^{\prime\prime}=-40\,\mathrm{MeV}<0}.$$

(b) Interpretation and further calculation

$$\boxed{\Delta U_A\simeq+.400\,\mathrm{MeV},\quad\Delta U_B\simeq-.200\,\mathrm{MeV}}.$$$$|q|\ll1:\quad O(q^3)\ll O(q^2)\quad\text{away from vanishing quadratic coefficient}.$$

Physical check

Positive curvature implies a local minimum, not global stability. If the quadratic term vanishes, higher-order terms decide the behavior.