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 the closest approach is of the order of

Imagine an alpha particle, which is essentially a helium nucleus, hurtling through space with a kinetic energy of $5\text{ MeV}$. It is on a direct collision course with a massive, stationary Uranium nucleus. Because both the alpha particle and the Uranium nucleus are positively charged, they repel each other. As the alpha particle gets closer, this electrostatic force of repulsion grows stronger, acting like an invisible brake.

At a certain point, the alpha particle's speed drops to zero. It momentarily stops before being pushed back along its original path. This minimum distance between the alpha particle and the center of the Uranium nucleus is called the distance of closest approach, denoted by $r_0$.

According to the law of conservation of mechanical energy, the initial kinetic energy of the alpha particle is completely converted into electrostatic potential energy at this point. We can write this mathematically as:

We are given the initial kinetic energy $K_i=5\text{ MeV}$. The charge of the alpha particle is $q_1=2e$, and the charge of the Uranium nucleus is $q_2=92e$. Before substituting, we must convert the energy into Joules to maintain SI units:

Now, substituting the values into our energy equation:

Rearranging to solve for $r_0$:

Since the options are given in centimeters, we convert our result: This tells us that the distance of closest approach is of the order of $10^{-12}\text{ cm}$.