A $4.78$-MeV alpha particle from a $226$Ra decay makes a head-on collision with a uranium nucleus. A uranium nucleus has $92$ protons.
(a) What is the distance of closest approach of the alpha particle to the center of the nucleus? Assume that the uranium nucleus remains at rest and that the distance of closest approach is much greater than the radius of the uranium nucleus.
(b) What is the force on the alpha particle at the instant when it is at the distance of closest approach?

Verified step by step guidance

1. Step 1: Understand the problem. The alpha particle (charge = +2e) is repelled by the uranium nucleus (charge = +92e) due to the Coulomb force. At the distance of closest approach, the kinetic energy of the alpha particle is completely converted into electric potential energy. Use this energy conservation principle to find the distance of closest approach.
2. Step 2: Write the energy conservation equation. The initial kinetic energy of the alpha particle is given as 4.78 MeV. At the closest approach, this energy is equal to the electric potential energy between the alpha particle and the uranium nucleus: \( KE = U \), where \( U = \frac{k \cdot q_1 \cdot q_2}{r} \). Here, \( k \) is Coulomb's constant, \( q_1 = +2e \) (charge of the alpha particle), \( q_2 = +92e \) (charge of the uranium nucleus), and \( r \) is the distance of closest approach.
3. Step 3: Rearrange the equation to solve for \( r \), the distance of closest approach: \( r = \frac{k \cdot q_1 \cdot q_2}{KE} \). Substitute \( k = 8.99 \times 10^9 \ \text{N·m}^2/\text{C}^2 \), \( e = 1.6 \times 10^{-19} \ \text{C} \), \( q_1 = 2e \), \( q_2 = 92e \), and \( KE = 4.78 \ \text{MeV} \) (convert MeV to joules: \( 1 \ \text{MeV} = 1.6 \times 10^{-13} \ \text{J} \)).
4. Step 4: To find the force on the alpha particle at the distance of closest approach, use Coulomb's law: \( F = \frac{k \cdot q_1 \cdot q_2}{r^2} \). Substitute the values of \( k \), \( q_1 \), \( q_2 \), and the previously calculated \( r \) into this formula to compute the force.
5. Step 5: Verify the assumptions. Ensure that the distance of closest approach is much greater than the radius of the uranium nucleus, as stated in the problem. This validates the use of the point charge approximation for the uranium nucleus.

Key Concepts

Coulomb's Law

Coulomb's Law describes the electrostatic force between two charged particles. It states that the force is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. This principle is crucial for understanding the interactions between the positively charged alpha particle and the uranium nucleus, which contains protons.

Kinetic Energy and Potential Energy

In the context of particle collisions, kinetic energy is the energy of motion, while potential energy is the stored energy due to position in a force field. The alpha particle's kinetic energy at the start of the interaction will convert into potential energy as it approaches the uranium nucleus, allowing us to calculate the distance of closest approach using energy conservation principles.

Distance of Closest Approach

The distance of closest approach refers to the minimum distance between two charged particles during a collision, where the kinetic energy of the moving particle is fully converted into electrostatic potential energy. This concept is essential for determining how close the alpha particle can get to the uranium nucleus before being repelled due to the electrostatic force.