Question

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) $ \dfrac{1}{m} $
B) $ \dfrac{1}{{\sqrt m }} $
C) $ \dfrac{1}{{{m^2}}} $
D) $ m $

Answer 1

Hint : In this solution, we will use the law of Coulomb force between two charges to determine the distance of the closest approach. At the closet point of approach, the kinetic energy of the $ \alpha $ -particle will be converted into the potential energy of the two charges.

Complete step by step answer
We’ve been told that an $ \alpha $ -particle of mass $ m $ moving with velocity $ v $ bombs on a heavy nucleus of charge $ Ze $ and we want to find the relation of the distance of closest approach on the mass of the nucleus.
At the point of closest approach, the kinetic energy of the alpha particle will be completely converted into the potential energy of the system. So, we can write
 $\Rightarrow \dfrac{1}{2}m{v^2} = \dfrac{{kqQ}}{r} $
Substituting the value of $ q = + 2e $ for an alpha particle and $ Q = Ze $ for the nucleus and multiplying both sides by 2, we get
 $\Rightarrow m{v^2} = \dfrac{{4k{e^2}}}{r} $
Hence the closest distance of approach $ r $ can be calculated as
 $\Rightarrow r = \dfrac{{4kZ{e^2}}}{{m{v^2}}} $
So, the dependence on mass $ m $ will be
$\Rightarrow r \propto \dfrac{1}{m} $ which corresponds to option (A).

Additional Information
This phenomenon of scattering of alpha particles by nuclear charges was used by Rutherford in discovering the existence of a positively charged nucleus in atoms.

Note
Even if we are not aware of the charges of the alpha particle and the nucleus, we can still determine the relation of the closest distance of approach with the mass of the alpha particle. Here we have assumed that the nucleus is comparably heavy with respect to the alpha particles and stays stationary in its position without moving itself so only the alpha particle will be deflected.