Alpha Particle Head-on Collision Analysis

This problem involves the concept of conservation of energy during a head-on collision between an alpha particle and a uranium nucleus. In such a scenario, the initial kinetic energy of the alpha particle is converted entirely into electrostatic potential energy at the point of closest approach, where the alpha particle momentarily comes to rest before reversing its direction.

Energy Conservation Principle

The fundamental principle governing this interaction is the conservation of energy. As the alpha particle approaches the positively charged uranium nucleus, the electrostatic repulsive force causes it to slow down. At the distance of closest approach ($r_{min}$), all of its initial kinetic energy ($K$) is transformed into electrostatic potential energy ($U$).

Mathematically, this can be expressed as:

$$K_{initial} = U_{final}$$

Where:

• $K_{initial}$ is the initial kinetic energy of the alpha particle.
• $U_{final}$ is the electrostatic potential energy at the distance of closest approach.

Given Parameters and Constants

Let's list the given values and necessary physical constants:

• Kinetic energy of the alpha particle, $K = 4.50 \text{ MeV}$
• Atomic number of uranium, $Z = 92$
• Charge of an alpha particle, $q_{\alpha} = +2e$ (since it consists of two protons)
• Charge of a uranium nucleus, $q_U = +Ze = +92e$
• Elementary charge, $e = 1.6 \times 10^{-19} \text{ C}$
• Coulomb's constant, $k = \frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \text{ N m}^2/\text{C}^2$
• Conversion factor for energy: $1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$

Step-by-step Calculation

1. Convert Kinetic Energy to Joules:

First, we convert the kinetic energy of the alpha particle from Mega-electron Volts (MeV) to Joules (J):

$$K = 4.50 \text{ MeV}$$

$$K = 4.50 \times 10^6 \text{ eV}$$

$$K = 4.50 \times 10^6 \times (1.6 \times 10^{-19} \text{ J})$$

$$K = 7.2 \times 10^{-13} \text{ J}$$

2. Express Charges in Coulombs:

Next, we write the charges of the alpha particle and uranium nucleus in Coulombs:

• Charge of alpha particle ($q_1$):
  $$q_1 = 2e = 2 \times (1.6 \times 10^{-19} \text{ C}) = 3.2 \times 10^{-19} \text{ C}$$
• Charge of uranium nucleus ($q_2$):
  $$q_2 = 92e = 92 \times (1.6 \times 10^{-19} \text{ C}) = 1.472 \times 10^{-17} \text{ C}$$

3. Set up the Energy Conservation Equation:

The electrostatic potential energy ($U$) between two point charges $q_1$ and $q_2$ separated by a distance $r$ is given by:

$$U = k \frac{q_1 q_2}{r}$$

At the distance of closest approach ($r_{min}$), $K = U$. So, we have:

$$K = k \frac{q_1 q_2}{r_{min}}$$

Rearranging the equation to solve for $r_{min}$:

$$r_{min} = k \frac{q_1 q_2}{K}$$

4. Substitute Values and Calculate $r_{min}$:

Now, we substitute the calculated values into the equation for $r_{min}$:

$$r_{min} = (9 \times 10^9 \text{ N m}^2/\text{C}^2) \times \frac{(3.2 \times 10^{-19} \text{ C}) \times (1.472 \times 10^{-17} \text{ C})}{7.2 \times 10^{-13} \text{ J}}$$

$$r_{min} = \frac{9 \times 10^9 \times 3.2 \times 10^{-19} \times 1.472 \times 10^{-17}}{7.2 \times 10^{-13}} \text{ m}$$

$$r_{min} = \frac{4.23936 \times 10^{-26}}{7.2 \times 10^{-13}} \text{ m}$$

$$r_{min} = 5.888 \times 10^{-14} \text{ m}$$

5. Convert $r_{min}$ to Femtometers:

Since nuclear distances are typically expressed in femtometers (fm), where $1 \text{ fm} = 10^{-15} \text{ m}$, we convert our result:

$$r_{min} = 5.888 \times 10^{-14} \text{ m} = 58.88 \times 10^{-15} \text{ m}$$

$$r_{min} \approx 59 \text{ fm}$$

Summary Table of Calculations

The calculated distance of closest approach is approximately 59 fm, which aligns with one of the provided options.