Question

An $\alpha -$particle of energy $5MeV$ is scattered through ${180}^{\circ }$ by gold nucleus. The distance of closest approach is of the order of

• A. ${10}^{-12}cm$
• B. ${10}^{-16}cm$
• C. ${10}^{-10}cm$
• D. ${10}^{-14}cm$

Answer 1

The correct option is A ${10}^{-12}cm$

Assume $\alpha -$ particle is thrwon (head- on) from very far
$\Rightarrow$ Applying conservation of mechanical energy we can write
$K_{\alpha }+V_{\alpha }=k_f+V_f$
$\Rightarrow 5MeV+0=0+\frac{9\times {10}^9\times 79e\times 2e}{r\times 1.6\times {10}^{-19}\times {10}^6}MeV$
$\Rightarrow r=\frac{9\times {10}^9\times 19\times 2\times 1.6\times {10}^{-19}}{{10}^6\times 5}m$
$=4.55\times {10}^{-12}cm$