Question

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

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

Answer 1

Answer: (C)

$30\,fm$

Answer 2

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 $K = \frac{1}{4\pi\varepsilon_{0}}\times\frac{\left(2e\right)\left(Ze\right)}{d}$ $\therefore d = \frac{1}{4\pi\varepsilon_{0}} \frac{2Ze^{2}}{K} ...\left(i\right)$ Here, $\frac{1}{4\pi\varepsilon_{0}} = 9 \times10^{9} N m^{2} C^{-2}$ $Z = 79, e = 1.6 \times 10^{-19} C$. $K = 7.7 MeV$ $= 7.7 \times 10^{6}\times 1.6\times 10^{-9} J$ $= 1.2 \times 10^{-12}J$ Substituting these values in $\left(i\right)$ $d = \frac{2\times 9 \times 10^{9} \times \left(1.6 \times 10^{-19}\right)^{2}\times 79}{1.2\times10^{-12}}$ $d = 3 \times 10^{-14} m$ $m = 30\, fm \left(\because 1\, fm = 10^{-15} \,m\right)$