Introduction
To determine the distance of closest approach of a 15.0 MeV proton to a gold nucleus (Z=79), we can use the concept of electrostatic potential energy and kinetic energy.
Key Concepts
- Coulomb's Law: The electrostatic force between two charged particles is given by:
\[ F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \]
where \( k \) is Coulomb's constant, \( q_1 \) and \( q_2 \) are the charges of the proton and the gold nucleus, and \( r \) is the distance between them.
- Kinetic Energy (KE): The kinetic energy of the proton can be expressed as:
\[ KE = \frac{1}{2}mv^2 = 15.0 \text{ MeV} \]
- Potential Energy (PE): At the closest approach, the kinetic energy of the proton will be converted into electrostatic potential energy:
\[ PE = \frac{k \cdot |q_1 \cdot q_2|}{r} \]
Calculating the Charge
- Proton Charge: \( q_1 = +e \) (approximately \( 1.6 \times 10^{-19} \) C)
- Gold Nucleus Charge:
\[ q_2 = Z \cdot e = 79 \cdot e \]
Distance of Closest Approach
Equating kinetic energy to potential energy at the closest approach:
\[ 15.0 \text{ MeV} = \frac{k \cdot (79e^2)}{r} \]
Converting 15.0 MeV to joules (1 MeV = \( 1.6 \times 10^{-13} \) J):
\[ 15.0 \text{ MeV} = 15.0 \cdot 1.6 \times 10^{-13} \text{ J} = 2.4 \times 10^{-12} \text{ J} \]
Now, substituting values and solving for \( r \):
\[ r = \frac{k \cdot (79e^2)}{2.4 \times 10^{-12}} \]
Using \( k \approx 8.99 \times 10^9 \text{ Nm}^2/\text{C}^2 \) and \( e^2 \):
- After calculations, you can find \( r \) to be approximately in the order of femtometers (fm).
Conclusion
The distance of closest approach when a 15.0 MeV proton approaches a gold nucleus is typically around 10-15 fm, indicating the strong electrostatic repulsion at short ranges.