Question

Determine the distance of the closest approach when an alpha particle of kinetic energy $$4.5 \, MeV$$ strikes a nucleus of $$Z = 80$$, stops, and reverses its direction.

Answer 1

Let r be the center to center distance between the alpha particle and the nucleus ($$Z = 80$$). When the alpha particle is at the stopping point, then
$$K=\dfrac{1}{4\pi\epsilon_0} \,\dfrac{(Ze)(2e)}{r}$$
or $$r=\dfrac{1}{4\pi\epsilon_0} \,\dfrac{(2Ze^2)}{K}$$
$$=\dfrac{9 \times 10^9\times2\times80\,e^2}{4.5\,MeV}=\dfrac{9 \times 10^9\times2\times80\,(1.6\times10^{-19})^2}{4.5 \times 10^{6}\times1.6\times10^{-19}\,J}$$
$$=\dfrac{9 \times 160 \times 1.6 }{4.5}\times10^{-16}=512 \times 10^{-16}m$$
$$=5.14 \times 10^{-14}\,m$$