Question

In the different experiments, $\alpha -$particles, proton, deuteron and neutron are projected towards gold nucleus with the same kinetic energy. The distance of closest approach will be minimum for

• A. $\alpha -$ particle
• B. proton
• C. deuteron
• D. neutron

Answer 1

Answer: (A)

$\alpha -$ particle

We know, closest approach for $\alpha -rayscattering$.
$\frac{1}{2}m{v}^2=\frac{q_1{q}_2}{4\pi {ϵ}_{o}b}$

$KE=\frac{q_1{q}_2}{4\pi {ϵ}_{o}b}$

Here KE is same for all cases. For getting minimum value of $b$, $q$ should also be less thus, atomic number also will be less.