Question

Define the term, ‘distance of closest approach’. A proton of 3.95 MeV energy approaches a target nucleus \( Z = 79 \) in head-on position. Calculate its distance of closest approach.

Hint:

The concept of the "distance of closest approach" helps in understanding how the kinetic energy of the proton is transformed into electrostatic potential energy. Remember to use Coulomb’s law for potential energy in these types of problems.

Answer 1

Distance of Closest Approach for a Proton Colliding with a Nucleus 

The "distance of closest approach" refers to the minimum distance between the incident particle (in this case, a proton) and the nucleus due to electrostatic repulsion. This occurs when the kinetic energy of the proton is entirely converted into potential energy at the closest point.

We use the concept of energy conservation here:

\[ \text{Initial kinetic energy} = \text{Final potential energy} \]

The potential energy \( U \) at the closest distance \( r \) is given by the Coulomb force formula:

\[ U = \frac{1}{4 \pi \epsilon_0} \frac{Z e^2}{r} \]

Where:

• \( Z \) is the atomic number of the nucleus (in this case, \( Z = 79 \)),
• \( e \) is the charge of the proton (\( e = 1.6 \times 10^{-19} \, \text{C} \)),
• \( r \) is the distance of closest approach,
• \( \epsilon_0 \) is the permittivity of free space (\( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2 \)).

Now, equate the initial kinetic energy to the final potential energy:

\[ K.E. = U \]

The kinetic energy \( K.E. \) is given by:

\[ K.E. = \text{Energy of the proton} = 3.95 \, \text{MeV} = 3.95 \times 10^6 \times 1.6 \times 10^{-13} \, \text{J} \]

Equating the two energies:

\[ 3.95 \times 10^6 \times 1.6 \times 10^{-13} = \frac{1}{4 \pi \epsilon_0} \frac{Z e^2}{r} \]

Substituting the known values and solving for \( r \):

\[ r = \frac{1}{4 \pi \epsilon_0} \frac{Z e^2}{K.E.} \]

After substituting values:

\[ r = \frac{(9 \times 10^9) \times (79) \times (1.6 \times 10^{-19})^2}{3.95 \times 10^6 \times 1.6 \times 10^{-13}} \, \text{m} \]

Solving for \( r \), we get:

\[ r = 28.8 \times 10^{-15} \, \text{m} = 28.8 \, \text{fm} \]

Thus, the distance of closest approach is \( 28.8 \times 10^{-15} \, \text{m} \).