© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Understand the Physical Situation
An alpha particle (charge = 2e) with kinetic energy
$ \frac{1}{2} m v^2 $
is directed toward a heavy nucleus of charge
$ Ze $. The alpha particle will slow down under the repulsive electrostatic force, and its distance of closest approach is reached when its kinetic energy is fully converted into electrostatic potential energy.
Step 2: Write the Energy Conservation Relation
At the closest distance of approach $ r $, all the kinetic energy of the alpha particle is converted into the electrostatic potential energy between the two charges:
$ \frac{1}{2} m v^2 = \frac{k \,(2e)\,(Ze)}{r}, $
where $ k $ is the Coulomb’s constant ($ k = \frac{1}{4\pi \varepsilon_0} $), and
$ (2e) $ and
$ (Ze) $
are the charges of the alpha particle and the heavy nucleus respectively.
Step 3: Solve for the Distance of Closest Approach
Rearrange the above equation to solve for $ r $:
$ r = \frac{k \,(2e)\,(Ze)}{\frac{1}{2} m v^2}
= \frac{2\,k \,Z e^2}{\frac{1}{2} m v^2}
= \frac{4\,k \,Z e^2}{m\,v^2}.
$
Step 4: Identify the Proportionalities
From the expression
$ r = \frac{4\, k \,Z e^2}{m\,v^2}, $
we see that
$ r \propto \frac{1}{v^2} $
and
$ r \propto \frac{1}{m}. $
The question specifically asks about the dependence on $v^2$, so the distance of closest approach is inversely proportional to
$ v^2 $.
Answer
Thus, the distance of closest approach is proportional to
$ \frac{1}{v^2}. $
