Question

An $\alpha -$ particle having energy $10MeV$ collides with a nucleus of atomic number $50$.Then distance of closet approach will be

• A. $14.4\times {10}^{-16}m$
• B. $1.7\times {10}^{-7}m$
• C. $1.5\times {10}^{-12}m$
• D. $14.4\times {10}^{-15}m$

Answer 1

The correct option is D $14.4\times {10}^{-15}m$

At the distance of closest approach the whole kinetic energy get converted into potential energy.

So $KE=\left(1/4\pi {ϵ}_0\right)2Z{e}^2/r$

Where r is distance of closest approach.

$r=\frac{2Z{e}^2}{KE\times 4\pi {ϵ}_0}$

putting $Z=50,\left(1/4\pi {ϵ}_0\right)=9\times {10}^9,e=1.6\times {10}^{-19},KE=10\times 1.6\times {10}^{-13}Joule$

we get $r=14.4\times {10}^{-15}m$

Option D is correct.