Answer :
To determine the distance of closest approach of an α-particle to the nucleus of a copper atom, we generally approach this problem by considering the conservation of energy and the electrostatic force of repulsion between the positively charged α-particle and the copper nucleus. Here’s a step-by-step explanation:
1. Identify Given Variables:
- Speed of the α-particle: [tex]\( V \)[/tex]
- Mass of the α-particle: [tex]\( m_\alpha \)[/tex]
- Charge of the α-particle: [tex]\( 2e \)[/tex] (since an α-particle has 2 protons)
- Atomic number of copper: [tex]\( Z = 29 \)[/tex]
- Charge of copper nucleus: [tex]\( 29e \)[/tex] (since copper nucleus has 29 protons)
2. Initial Kinetic Energy:
The initial kinetic energy [tex]\( K \)[/tex] of the α-particle is given by:
[tex]\[ K = \frac{1}{2} m_\alpha V^2 \][/tex]
3. Electrostatic Potential Energy at Closest Approach:
At the distance of closest approach, all the kinetic energy will be converted into electrostatic potential energy [tex]\( U \)[/tex]. The expression for the electrostatic potential energy between two point charges [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex] separated by a distance [tex]\( d \)[/tex] is:
[tex]\[ U = \frac{q_1 q_2}{4 \pi \epsilon_0 d} \][/tex]
Here, [tex]\( q_1 = 2e \)[/tex] and [tex]\( q_2 = 29e \)[/tex].
4. Set Up Energy Conservation Equation:
At distance [tex]\( d \)[/tex] (the closest approach):
[tex]\[ \frac{1}{2} m_\alpha V^2 = \frac{(2e)(29e)}{4 \pi \epsilon_0 d} \][/tex]
5. Solve for [tex]\( d \)[/tex]:
Rearrange the equation to solve for [tex]\( d \)[/tex]:
[tex]\[ d = \frac{(2e)(29e)}{4 \pi \epsilon_0 \left( \frac{1}{2} m_\alpha V^2 \right)} \][/tex]
Simplify the expression:
[tex]\[ d = \frac{2 \cdot 29 \cdot e^2}{4 \pi \epsilon_0 \cdot \frac{1}{2} m_\alpha V^2} \][/tex]
[tex]\[ d = \frac{58 \cdot e^2}{4 \pi \epsilon_0 \cdot \frac{1}{2} m_\alpha V^2} \][/tex]
[tex]\[ d = \frac{58 \cdot e^2}{2 \pi \epsilon_0 \cdot m_\alpha V^2} \][/tex]
6. Conclusion:
Therefore, the distance of closest approach [tex]\( d \)[/tex] is:
[tex]\[ d = \frac{58 \cdot e^2}{2 \pi \epsilon_0 \cdot m_\alpha V^2} \][/tex]
In this specific case, given the complexity involving numerous physical constants and the unspecified values of [tex]\( V \)[/tex], [tex]\( m_\alpha \)[/tex], and [tex]\( e \)[/tex], the calculation concludes that more information would be required to provide a definitive numerical answer. Thus, the distance can't be determined exactly without those specifics.
1. Identify Given Variables:
- Speed of the α-particle: [tex]\( V \)[/tex]
- Mass of the α-particle: [tex]\( m_\alpha \)[/tex]
- Charge of the α-particle: [tex]\( 2e \)[/tex] (since an α-particle has 2 protons)
- Atomic number of copper: [tex]\( Z = 29 \)[/tex]
- Charge of copper nucleus: [tex]\( 29e \)[/tex] (since copper nucleus has 29 protons)
2. Initial Kinetic Energy:
The initial kinetic energy [tex]\( K \)[/tex] of the α-particle is given by:
[tex]\[ K = \frac{1}{2} m_\alpha V^2 \][/tex]
3. Electrostatic Potential Energy at Closest Approach:
At the distance of closest approach, all the kinetic energy will be converted into electrostatic potential energy [tex]\( U \)[/tex]. The expression for the electrostatic potential energy between two point charges [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex] separated by a distance [tex]\( d \)[/tex] is:
[tex]\[ U = \frac{q_1 q_2}{4 \pi \epsilon_0 d} \][/tex]
Here, [tex]\( q_1 = 2e \)[/tex] and [tex]\( q_2 = 29e \)[/tex].
4. Set Up Energy Conservation Equation:
At distance [tex]\( d \)[/tex] (the closest approach):
[tex]\[ \frac{1}{2} m_\alpha V^2 = \frac{(2e)(29e)}{4 \pi \epsilon_0 d} \][/tex]
5. Solve for [tex]\( d \)[/tex]:
Rearrange the equation to solve for [tex]\( d \)[/tex]:
[tex]\[ d = \frac{(2e)(29e)}{4 \pi \epsilon_0 \left( \frac{1}{2} m_\alpha V^2 \right)} \][/tex]
Simplify the expression:
[tex]\[ d = \frac{2 \cdot 29 \cdot e^2}{4 \pi \epsilon_0 \cdot \frac{1}{2} m_\alpha V^2} \][/tex]
[tex]\[ d = \frac{58 \cdot e^2}{4 \pi \epsilon_0 \cdot \frac{1}{2} m_\alpha V^2} \][/tex]
[tex]\[ d = \frac{58 \cdot e^2}{2 \pi \epsilon_0 \cdot m_\alpha V^2} \][/tex]
6. Conclusion:
Therefore, the distance of closest approach [tex]\( d \)[/tex] is:
[tex]\[ d = \frac{58 \cdot e^2}{2 \pi \epsilon_0 \cdot m_\alpha V^2} \][/tex]
In this specific case, given the complexity involving numerous physical constants and the unspecified values of [tex]\( V \)[/tex], [tex]\( m_\alpha \)[/tex], and [tex]\( e \)[/tex], the calculation concludes that more information would be required to provide a definitive numerical answer. Thus, the distance can't be determined exactly without those specifics.
