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

Proton-Nucleus Collision: Closest Approach Distance

The "distance of closest approach" signifies the minimum separation between an incoming proton and a nucleus, driven by electrostatic repulsion. This point is reached when the proton's kinetic energy is fully converted into potential energy.

Energy conservation is applied:

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

The potential energy \( U \) at the closest distance \( r \) is determined by Coulomb's law:

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

Where:

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

Equating initial kinetic energy to final potential energy:

\[ K.E. = U \]

The proton's kinetic energy \( K.E. \) is:

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

Equating energies:

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

Solving for \( r \):

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

Substituting values:

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

The calculated distance of closest approach is:

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

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