Question

In scattering experiment, find the distance of closest approach, if a $6MeV\alpha -particle$ is used.

• A. $3.2\times {10}^{-16}m$
• B. $2\times {10}^{-14}m$
• C. $4.6\times {10}^{-15}m$
• D. $3.2\times {10}^{-15}m$

Answer 1

The correct option is B $2\times {10}^{-14}m$

Distance of closet approach of an $\alpha$-particle is given by,

$r=\frac{2kZ{e}^2}{K_{\alpha }}$
ATQ, $Z=79$

$K_{\alpha }=6MeV=6\times 1.6\times {10}^{-13}J$

Substituting these values we get,

$r=\frac{2\times 9\times {10}^9\times 79\times {(1.6\times {10}^{-19})}^2}{6\times 1.6\times {10}^{-13}}=3.792\times {10}^{-16}m$