Discovering the Nucleus — Question 1

Question 1

A head-on alpha trajectory bounds the size of the nucleus

An alpha particle of kinetic energy $E=6.00\,\mathrm{MeV}$ approaches a fixed nucleus of charge $Ze$ with $Z=50$. Treat both charges as points until closest approach. Use $e^2/(4\pi\varepsilon_0)=1.44\,\mathrm{MeV\,fm}$.

1. Derive and evaluate the head-on turning distance.

2. If purely Coulomb scattering persists at this energy, what size bound is suggested, and what energy would give a turning distance of 8.00 fm in this model?

Question 1 - Solution

A head-on alpha trajectory bounds the size of the nucleus

(a) Method and calculation

$$E=\frac{2Ze^2}{4\pi\varepsilon_0r_{\min}}\Rightarrow\boxed{r_{\min}=2Z(1.44)/E}.$$$$\boxed{r_{\min}=144/6.00=24.0\,\mathrm{fm}}.$$

(b) Interpretation and further calculation

$$\boxed{R\lesssim24.0\,\mathrm{fm}}\quad\text{within the point-projectile, Coulomb-only interpretation}.$$$$\boxed{E(8.00\,\mathrm{fm})=144/8.00=18.0\,\mathrm{MeV}}.$$

Physical check

This is an upper-bound argument, not a direct radius measurement. Finite projectile size, recoil and the nuclear force matter when the surfaces approach.