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

Answer: (B)

$2\times {10}^{-14}m$

The distance of closest approach is given us in under :

$r_0=14\pi {ϵ}_0\times 2\times z\times e_{2}KE$

given $KE=6MeV$

and assuming the nucleus to be heat of copper, then $z=2q$ as $d.e=1.6\times {10}^{-19}C$.

Substituting the values in the above equation and calculating we have

$r_0=2\times {10}^{-14}m$.