Atoms - Result Question 1
1. When an $\alpha$-particle of mass ’ $m$ ’ moving with velocity ’ $v$ ’ bombards on a heavy nucleus of charge ‘Ze’, its distance of closest approach from the nucleus depends on $m$ as :
[2016]
(a) $\frac{1}{m}$
(b) $\frac{1}{\sqrt{m}}$
(c) $\frac{1}{m^{2}}$
(d) $m$
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Solution:
- (a) At closest distance of approach, the kinetic energy of the particle will convert completely into electrostatic potential energy.
Kinetic energy K.E. $=\frac{1}{2} mv^{2}$
Potential energy P.E. $=\frac{KQq}{r}$
$\frac{1}{2} mv^{2}=\frac{KQq}{r} \Rightarrow r \propto \frac{1}{m}$
