Question

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

• A. ${10}^{-12}cm$
• B. ${10}^{-10}cm$
• C. $1A$
• D. ${10}^{-15}cm$

Answer 1

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

Given,

$E=5MeV=5\times {10}^6\times 1.6\times {10}^{-19}=8\times {10}^{-13}V$

$Q_{\alpha }=2\times 1.6\times {10}^{-19}=3.2\times {10}^{-19}C$

$Q_{ur}=92\times 1.6\times {10}^{-19}=147.2\times {10}^{-19}C$

At the closest distance approach, the kinetic energy will become zero.

$\frac{1}{2}\times 1.6\times {10}^{-19}=\frac{9\times {10}^9\times 147.2\times {10}^{-19}\times 3.2\times {10}^{-19}}{r}$

$r={10}^{-12}cm$

The correct option is A.