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. $\frac{1}{\sqrt{m}}$
• B. $\frac{1}{m^2}$
• C. m
• D. $\frac{1}{m}$

Answer 1

The Correct Option is D

To find the dependence of the distance of the closest approach of an $\alpha$-particle on its mass 'm', we must understand the physics behind the collision between an $\alpha$-particle and a heavy nucleus.

1. The distance of the closest approach, denoted as $r_0$, is where the kinetic energy of the $\alpha$-particle is completely converted to electrostatic potential energy due to the nucleus.
2. The initial kinetic energy of the $\alpha$-particle is given by: \[ KE = \frac{1}{2}mv^2 \]
3. The electrostatic potential energy at the closest approach is given by: \[ PE = \frac{k \cdot (2e) \cdot (Ze)}{r_0} \] where $k$ is Coulomb's constant, $2e$ is the charge of the $\alpha$-particle, and $Ze$ is the charge of the nucleus.
4. Equating the kinetic energy and potential energy: \[\frac{1}{2}mv^2 = \frac{k \cdot (2e) \cdot (Ze)}{r_0}\]
5. Rearranging for $r_0$ gives: \[ r_0 = \frac{k \cdot (2e) \cdot (Ze)}{\frac{1}{2}mv^2} = \frac{2kZe^2}{mv^2} \]
6. From this expression, it is clear that $r_0 \propto \frac{1}{m}$. Therefore, the distance of the closest approach is inversely proportional to the mass of the $\alpha$-particle.

Thus, the dependence of the distance of closest approach from the nucleus on the mass 'm' of the $\alpha$-particle is $\frac{1}{m}$. Therefore, the correct option is: $\frac{1}{m}$.