Question

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

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

Answer 1

The correct option is D

${10}^{-12}cm$

Distance of closest approach is = ${10}^{-14}m={10}^{-12}cm$ (Order of size of nucleus)