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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
A
$\frac{1}{\sqrt{m}}$
B
$\frac{1}{m^2}$
C
m
D
$\frac{1}{m}$
Explanation
$r_0 = \frac{4Ze^2}{4\pi\varepsilon_0mv^2}$.
Detailed Solution
By conservation of energy, the initial KE of the $\alpha$-particle equals the potential energy at closest approach:
$\frac{1}{2}mv^2 = \frac{2Ze^2}{4\pi\varepsilon_0r_0}$
So $r_0 \propto \frac{1}{m}$
$\frac{1}{2}mv^2 = \frac{2Ze^2}{4\pi\varepsilon_0r_0}$
So $r_0 \propto \frac{1}{m}$