Question

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

• A. ${10}^{-15}cm$
• B. ${10}^{-13}cm$
• C. ${10}^{-12}cm$
• D. ${10}^{-19}cm$

Answer 1

Answer: (C)

${10}^{-12}cm$

$1MeV=1.602\times {10}^{-13}Nm$

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

Charge of alpha particle and uranium nucleus is $2\times 1.602\times {10}^{-19}C$ and $92\times 1.602\times {10}^{-19}C$ respectively.

$\frac{1}{2}\times 1.602\times {10}^{-13}=\frac{9\times {10}^9\times (92\times 2)\times {(1.602\times {10}^{-19})}^2}{r}$

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