Question

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

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

Answer 1

Answer: (C)

${10}^{-12}cm$

Distance of closet approx$:-$

$r_0=\frac{ze2e}{4\pi {\in }_{0}E}$

$KE=PE$

$\frac{1}{2}m{v}^2=\frac{1}{4\pi {\in }_0}\frac{q_1{q}_2}{r}$

$r=\frac{9\times {10}^9\times 2\times 92\times {\left(1.6\times {10}^{-19}\right)}^2}{5\times {10}^6\times 1.6\times {10}^{-19}}$

$r=5.3\times {10}^{-14}m\approx {10}^{-12}cm$

Hence,

option $\left(C\right)$ is correct answer.