Solution The key idea here is that throughout the scattering process, the total mechanical energy of the system consisting of an $\alpha$ -particle and a gold nucleus is conserved. The system's initial mechanical energy is $E_{i}$ before the particle and nucleus interact, and it is equal to its mechanical energy $E_{f}$ when the $\alpha$ -particle momentarily stops. The initial energy $E_{i}$ is just the kinetic energy $K$ of the incoming

$\alpha-$ particle. The final energy $E_{f}$ is just the electric potential energy $U$ of the system.

Let $d$ be the centre-to-centre distance between the $\alpha$ -particle and the gold nucleus when the $\alpha$ -particle is at its stopping point. Then we can write the conservation of energy $E_{i}=E_{f}$ as $K=\frac{1}{4 \pi \varepsilon_{0}} \frac{(2 e)(Z e)}{d}=\frac{2 Z e^{2}}{4 \pi \varepsilon_{0} d}$

Thus the distance of closest approach $d$ is given by

$d=\frac{2 Z e^{2}}{4 \pi \varepsilon_{0} K}$

The maximum kinetic energy found in $\alpha$ -particles of natural origin is

$7.7\, MeV$ or $1.2 \times 10^{-12} \,J .$

since $1 / 4 \pi \varepsilon_{0}=9.0 \times 10^{9}\, Nm ^{2} / C ^{2} .$ Therefore

with $e=1.6 \times 10^{-19} \,C ,$ we have,

$d=\frac{(2)\left(9.0 \times 10^{9} \,Nm ^{2} / C^{2}\right)\left(1.6 \times 10^{-19} \,C \right)^{2} Z }{1.2 \times 10^{-12} \,J }$

$=3.84 \times 10^{-16}\, Zm$

The atomic number of foil material gold is $Z=79,$ so that $d( Au )=3.0 \times 10^{-14} \,m =30 fm .\left(1 \,fm \text { (i.e. fermi) }=10^{-15}\, m .\right)$

The radius of gold nucleus is, therefore, less than $3.0 \times 10^{-14} \,m .$ This is not in very good agreement with the observed result as the actual radius of gold nucleus is $6\, fm$. The cause of discrepancy is that the distance of closest approach is considerably larger than the sum of the radii of the gold nucleus and the $\alpha$ -particle. Thus, the $\alpha$ -particle reverses its motion without ever actually touching the gold nucleus.