Question

An $\alpha -$particle passes through a potential difference of $2\times {10}^{6}V$ and then it becomes incident on a silver foil. The charge number of silver is 47. The distance of closest approach of the particle to the nucleus will be :

• A. $6.4\times {10}^{-13}m$
• B. $4.3\times {10}^{-13}m$
• C. $2.1\times {10}^{-13}m$
• D. $3.76\times {10}^{-15}m$

Answer 1

The correct option is D $3.76\times {10}^{-15}m$

For $\alpha$-ray scattering
$\frac{1}{2}m{v}^2=\frac{1}{4\pi {ϵ}_o}\times \frac{q_1{q}_2}{\sigma }$

$v_{\alpha }V=\frac{1}{4\pi {ϵ}_o}\times \frac{q_2{q}_2}{\sigma }$

$V=\frac{v_s}{4\pi {ϵ}_o\sigma }$

$\sigma =\frac{v_s}{4\pi {ϵ}_{o}V}$

$=\frac{47\times 1.6\times {10}^{-19}\times {10}^9}{2\times 106}$

$=3.76\times {10}^{-15}m$