Q. 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) $1\ A{^\circ}$
B) $10^{- 10}\ cm$
C) $10^{- 12}cm$
D) $10^{- 15}cm$
Explanation
(c) $r_{0} = \frac{(Ze)(2e)}{4\pi\varepsilon_{0}(E)} = \frac{2 \times 92\left( 1.6 \times 10^{- 19} \right)^{2} \times 9 \times 10^{9}}{5 \times 1.6 \times 10^{- 13}}$
$= 0.53 \times 10^{- 14}m \approx 10^{- 12}cm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.