गाइगर-मार्सडन प्रयोग में $7.7 \,MeV$ के किसी ऐल्फा कण की स्वर्ण-नाभिक से क्षण भर के लिए विरामावस्था में आने से पहले तथा दिशा प्रतिलोमन से पूर्व समीपतम दूरी क्या है?

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.