Question

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

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

Answer 1

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


$K.E.=\frac{1}{2\pi {ϵ}_0}\frac{Ze2e}{d}$
$5\times {10}^6\times 1.6\times {10}^{-19}=\frac{9\times {10}^9\times 235\times 2\times {1.6}^2\times ({10}^{-19}{)}^2}{d}$
$d=\frac{9\times {10}^9\times 470\times 1.6\times {10}^{-19}}{5\times {10}^6}m$
$d={10}^{-12}cm$