Alpha Particle Closest Approach to Lead Nucleus

This problem requires us to calculate the distance of closest approach for an alpha particle when it approaches a lead nucleus head-on. This concept is fundamental to understanding Rutherford scattering and the structure of the atomic nucleus.

Energy Conservation Principle

In a head-on collision between a charged particle (like an alpha particle) and a stationary nucleus, the initial kinetic energy of the incoming particle is entirely converted into electrostatic potential energy at the point of closest approach. At this specific distance, the alpha particle momentarily stops before being repelled and reversing its direction due to the electrostatic force from the nucleus.

Therefore, we can use the principle of conservation of energy:

Initial Kinetic Energy (KE) = Final Electrostatic Potential Energy (PE)

\[ \text{KE} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r_0} \] Where:

• \(\text{KE}\) is the initial kinetic energy of the alpha particle.
• \(r_0\) is the distance of closest approach.
• \(\frac{1}{4\pi\epsilon_0}\) is Coulomb's constant, \(k = 9 \times 10^9 \text{ N m}^2/\text{C}^2\).
• \(q_1\) is the charge of the alpha particle.
• \(q_2\) is the charge of the lead nucleus.

Given Parameters and Charges

Let's list the given information and relevant constants:

Formulating the Closest Approach Distance

From the energy conservation equation, we can express \(r_0\) as:

\[ r_0 = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{\text{KE}} \]

Substitute the charges \(q_1 = 2e\) and \(q_2 = Ze\):

\[ r_0 = \frac{k (2e)(Ze)}{\text{KE}} = \frac{2kZe^2}{\text{KE}} \]

Step-by-Step Calculation

First, convert the kinetic energy from MeV to Joules:

• \( \text{KE} = 3.20 \text{ MeV} \) \( \text{KE} = 3.20 \times 10^6 \text{ eV} \) Since \(1 \text{ eV} = 1.6 \times 10^{-19} \text{ J} \): \( \text{KE} = 3.20 \times 10^6 \times (1.6 \times 10^{-19}) \text{ J} \) \( \text{KE} = 5.12 \times 10^{-13} \text{ J} \)

Now, substitute all values into the formula for \(r_0\):

• \( r_0 = \frac{2 \times (9 \times 10^9 \text{ N m}^2/\text{C}^2) \times 82 \times (1.6 \times 10^{-19} \text{ C})^2}{5.12 \times 10^{-13} \text{ J}} \)
• \( r_0 = \frac{2 \times 9 \times 82 \times (1.6)^2 \times 10^{9-38}}{5.12 \times 10^{-13}} \)
• \( r_0 = \frac{18 \times 82 \times 2.56 \times 10^{-29}}{5.12 \times 10^{-13}} \)
• \( r_0 = \frac{3773.44 \times 10^{-29}}{5.12 \times 10^{-13}} \)
• \( r_0 = 737.0 \times 10^{-29 - (-13)} \)
• \( r_0 = 737.0 \times 10^{-16} \text{ m} \)

Finally, convert the distance from meters to femtometers (fm). Remember that \(1 \text{ fm} = 10^{-15} \text{ m}\):

• \( r_0 = 73.70 \times 10^{-15} \text{ m} \)
• \( r_0 \approx 73.7 \text{ fm} \)

The calculated value is approximately 73.7 fm, which closely matches 73.8 fm given in the options, the minor difference being due to rounding of constants. Therefore, the distance of closest approach is 73.8 fm.