Animated Solution for Physics - Atoms and Nuclei: An alpha particle of energy $5\text{ MeV}$ is scattered through $180^{\circ }$ by a fixed uranium nucleus. The distance of closest approach is of the order of

A Subatomic Game of Chicken

Imagine you are an observer in the subatomic realm, watching a tiny, positively charged $\alpha$-particle being fired directly at a massive, highly charged Uranium nucleus. It's a game of subatomic chicken!

As the $\alpha$-particle hurtles towards the nucleus, it encounters a rapidly growing wall of electrostatic repulsion. This invisible force acts like a powerful spring, continuously doing negative work on the $\alpha$-particle and draining its kinetic energy. Eventually, the particle's speed drops to absolute zero for a fleeting moment. This exact point of turnaround is what physicists call the distance of closest approach ($r_{\text{min}}$). After this momentary pause, the particle is violently repelled back along its original path, scattering through $180^{\circ }$.

Conservation of Energy

Because the electrostatic force is conservative and there are no dissipative forces like friction at play, the total mechanical energy of the system remains perfectly conserved.

This means that the initial kinetic energy ($K$) of the $\alpha$-particle when it was far away is entirely converted into electrostatic potential energy ($U$) at the exact moment it stops. We can express this profound physical truth with a beautifully simple equation:

We know the formula for the electrostatic potential energy between two point charges $q_1$ and $q_2$ separated by a distance $r$:

For our specific scenario, the $\alpha$-particle has a charge $q_1=+2e$, and the Uranium nucleus, with its 92 protons, has a charge $q_2=+92e$. Substituting these into our energy conservation equation gives us the master equation for this problem:

Unit Conversion

Before we rush into plugging in the numbers, we must navigate a classic physics trap: unit inconsistency. The kinetic energy is given as $5\text{ MeV}$ (Mega electron-volts), but our standard SI formulas require Joules.

Let's convert this step-by-step. First, we convert Mega to standard units by multiplying by $10^6$, giving us $5\times 10^6\text{ eV}$. Next, we convert electron-volts to Joules by multiplying by the elementary charge ($1.6\times 10^{-19}\text{ C}$):

Peering into the Nucleus

Now, we are ready for the final execution. Let's rearrange our master equation to isolate the distance of closest approach, $r_{\text{min}}$:

Substituting the known values, including Coulomb's constant ($\frac{1}{4\pi {\epsilon }_0}=9\times 10^9{\text{ N m}}^2/{\text{C}}^2$):

Carefully crunching these numbers yields:

Since our options are presented in centimeters and Angstroms, we convert our result to centimeters by multiplying by $10^2$:

Looking at our final result, we can confidently conclude that the distance of closest approach is of the order of $10^{-12}\text{ cm}$. This perfectly matches option (c). This incredibly tiny distance gives us a profound appreciation for the immense density and minuscule scale of the atomic nucleus!