#### Solution By Steps
***Step 1: Calculate the Initial Potential Energy***
The initial potential energy of the $\alpha$-particle is given by the equation:
\[U_i = \frac{k_e \cdot |q_1 \cdot q_2|}{r_i}\]
where
\(k_e = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2\) (Coulomb's constant),
\(q_1 = 2e\) (charge of the $\alpha$-particle in coulombs),
\(q_2 = 79e\) (charge of the gold nucleus in coulombs),
\(r_i\) is the initial distance from the gold nucleus.
Substitute the given values to find \(U_i\).
***Step 2: Calculate the Final Potential Energy***
The final potential energy of the $\alpha$-particle when it is at a large distance from the gold nucleus is given by:
\[U_f = \frac{k_e \cdot |q_1 \cdot q_2|}{r_f}\]
where
\(r_f\) is the large distance from the gold nucleus.
Substitute the given values to find \(U_f\).
***Step 3: Equate Initial and Final Potential Energies***
Set \(U_i = U_f\) and solve for \(r_f\).
***Step 4: Calculate the Maximum Radius***
The maximum radius of the gold nucleus is given by the equation:
\[r_{\text{max}} = r_i - r_f\]
Substitute the calculated values to find \(r_{\text{max}}\).
#### Final Answer
The radius of the gold nucleus must be less than $4.7 \times 10^{-14} \, \text{m}$.
#### Key Concept
Electrostatic Potential Energy
#### Key Concept Explanation
Electrostatic potential energy is the energy associated with the interaction of charged particles. In this context, it is used to calculate the maximum radius of the gold nucleus based on the initial and final potential energies of the $\alpha$-particle. This concept is crucial in understanding the behavior of charged particles and their interactions in atomic and nuclear physics.
Follow-up Knowledge or Question
What is the relationship between the electric potential energy and the distance between two point charges?
How does the kinetic energy of a particle relate to its speed?
What is the expression for the electric force between two point charges?
Was this solution helpful?
Correct