Question

In a Geiger-Marsden experiment. Find the distance of closest approach to the nucleus of a $7.7MeV$ $\alpha$-particle before it comes momentarily to rest and reverses its direction. ($Z$ for gold nucleus $=79$)

• A. $10fm$
• B. $20fm$
• C. $30fm$
• D. $40fm$

Answer 1

Answer: (C)

Let $d$ be the distance of closest approach then by the conservation of energy.
Initial kinetic energy of incoming $\alpha$-particle $K$
$=$ Final electric potential energy $U$ of the system
As $K=\frac{1}{4\pi {\epsilon }_0}\times \frac{\left(2e\right)\left(Ze\right)}{d}$
$\therefore d=\frac{1}{4\pi {\epsilon }_0}\frac{2Z{e}^2}{K}...\left(i\right)$
Here,
$\frac{1}{4\pi {\epsilon }_0}=9\times 10^{9}N{m}^2{C}^{-2}$
$Z=79,e=1.6\times 10^{-19}C$.
$K=7.7MeV$
$=7.7\times 10^6\times 1.6\times 10^{-9}J$
$=1.2\times 10^{-12}J$
Substituting these values in $\left(i\right)$
$d=\frac{2\times 9\times 10^9\times {\left(1.6\times 10^{-19}\right)}^2\times 79}{1.2\times 10^{-12}}$
$d=3\times 10^{-14}m$
$m=30fm\left(\because 1fm=10^{-15}m\right)$