Animated Solution for Physics - Atoms and Nuclei: An alpha nucleus of energy $\frac{1}{2}m{v}^2$ bombards a heavy nuclear target of charge $Ze$. Then, the distance of closest approach for the alpha nucleus will be proportional to

The concept of the "distance of closest approach" is a beautiful intersection of kinematics and electrostatics. It takes us back to the historic Rutherford alpha-scattering experiment, where the dense, positively charged nucleus was first discovered.

Imagine you are observing the microscopic world. A heavy nuclear target, carrying a positive charge of $+Ze$, sits perfectly still. It is massive enough that we can consider it completely stationary—an immovable object.

Now, an alpha particle is fired directly at this nucleus. The alpha particle is essentially a helium nucleus, meaning it carries a positive charge of $+2e$ and has a mass $m$. It starts its journey with an initial velocity $v$, giving it an initial kinetic energy:

As the alpha particle hurtles towards the target nucleus, it enters a powerful electrostatic field. Because both the alpha particle and the target nucleus are positively charged, they repel each other. This repulsive Coulomb force acts like an invisible brake, continuously slowing the alpha particle down.

Eventually, the alpha particle's velocity drops to zero. It momentarily halts before the repulsive force pushes it back in the opposite direction. The distance between the center of the alpha particle and the center of the target nucleus at this exact moment of halting is called the distance of closest approach, denoted by $r$.

To find this distance, we rely on one of the most powerful tools in physics: the Principle of Conservation of Mechanical Energy.

Since the electrostatic force is conservative, the total mechanical energy of the system remains constant. As the alpha particle slows down, its kinetic energy decreases, but this energy isn't lost; it is entirely converted into electrostatic potential energy.

At the point of closest approach, the kinetic energy is zero, and the potential energy is at its maximum. We can equate the initial kinetic energy to the final potential energy:

Substituting the known expressions for kinetic and potential energy, we get:

Now, we simply rearrange this equation to solve for the distance of closest approach, $r$:

This final expression is a goldmine of information. It tells us exactly how $r$ depends on the various parameters of the system: - $r$ is directly proportional to the target's atomic number $Z$ ($r\propto Z$). - $r$ is inversely proportional to the mass of the alpha particle $m$ ($r\propto \frac{1}{m}$). - $r$ is inversely proportional to the square of the initial velocity $v$ ($r\propto \frac{1}{v^2}$).

Looking at our options, the correct relationship is that the distance of closest approach is proportional to $\frac{1}{m}$.

A Quick Trap Warning: The problem explicitly stated that the target nucleus is "heavy." This was our cue to assume it remains stationary. If the target were a light nucleus (like hydrogen or helium), the repulsive force would cause the target itself to accelerate away! In that scenario, we would have to use the conservation of linear momentum alongside energy conservation to find the closest approach. Always read the constraints carefully!