Nuclear Models — Question 1

Question 1

Short-range saturation motivates volume and surface binding terms

A liquid-drop model approximates binding energy as $B=a_vA-a_sA^{2/3}$ before Coulomb and other corrections. Use $a_v=15.5\,\mathrm{MeV}$ and $a_s=17.0\,\mathrm{MeV}$.

1. Derive binding energy per nucleon and its large-A limit.

2. Compare the surface penalty per nucleon for A=8 and A=64, and explain its geometric scaling.

Question 1 - Solution

Short-range saturation motivates volume and surface binding terms

(a) Method and calculation

$$\boxed{B/A=a_v-a_sA^{-1/3}}.$$$$A\to\infty:\quad\boxed{B/A\to a_v=15.5\,\mathrm{MeV}}\quad\text{within this truncated model}.$$

(b) Interpretation and further calculation

$$\boxed{(B_s/A)_8=17/2=8.50\,\mathrm{MeV},\quad(B_s/A)_{64}=17/4=4.25\,\mathrm{MeV}}.$$$$R\propto A^{1/3}\Rightarrow\boxed{\text{surface}/\text{volume}\propto R^{-1}\propto A^{-1/3}}.$$

Physical check

Surface nucleons have fewer nearby binding partners. Omitting Coulomb and asymmetry terms makes this truncated model unsuitable for predicting stability at arbitrarily large A.