Answer :
To solve this problem, we need to determine the closest distance that an alpha particle (α-particle) with a kinetic energy of 5.48 MeV can get to a gold (Au) nucleus to be stopped purely by the electrostatic repulsion between the positively charged α-particle and the nucleus.
### Step-by-Step Solution
1. Identify the Given Values:
- Atomic number of Americium (Am): \( Z_{\text{Am}} = 95 \)
- Atomic number of Gold (Au): \( Z_{\text{Au}} = 79 \)
- Coulomb constant: \( k = 8.98755 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2 \)
- Charge of an electron: \( e = 1.60219 \times 10^{-19} \, \text{C} \)
- Energy of the α-particle: \( E_{\alpha} = 5.48 \, \text{MeV} \)
2. Convert Energy of the α-particle to Joules:
- The provided energy in MeV needs to be converted to Joules because the other values are in standard SI units.
- \( 1 \, \text{eV} = 1.60219 \times 10^{-19} \, \text{J} \)
- Therefore, \( E_{\alpha} = 5.48 \times 10^6 \, \text{eV} \times 1.60219 \times 10^{-19} \, \text{J/eV} = 8.7800012 \times 10^{-13} \, \text{J} \)
3. Calculate the Closest Distance Using Coulomb's Law:
- The closest distance (r_min) can be found using the balance between the kinetic energy of the α-particle and the electrostatic potential energy at the point where the particle stops.
- Coulomb potential energy between two point charges \( q_1 \) and \( q_2 \) separated by distance \( r \) is given by
[tex]\[ U = \frac{k \cdot q_1 \cdot q_2}{r} \][/tex]
- Here, the charges are \( q_{\text{Am}} = Z_{\text{Am}} \cdot e \) and \( q_{\text{Au}} = Z_{\text{Au}} \cdot e \).
- Set the potential energy equal to the kinetic energy \( E_{\alpha} \)
[tex]\[ E_{\alpha} = \frac{k \cdot Z_{\text{Am}} \cdot e \cdot Z_{\text{Au}} \cdot e}{r_{\text{min}}} \][/tex]
Solving for \( r_{\text{min}} \):
[tex]\[ r_{\text{min}} = \frac{k \cdot Z_{\text{Am}} \cdot Z_{\text{Au}} \cdot e^2}{E_{\alpha}} \][/tex]
4. Substitute the Given Values:
- Plug the values into the equation:
[tex]\[ r_{\text{min}} = \frac{8.98755 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2 \times 95 \times 79 \times (1.60219 \times 10^{-19} \, \text{C})^2}{8.7800012 \times 10^{-13} \, \text{J}} \][/tex]
- Simplify the expression to get \( r_{\text{min}} \):
[tex]\[ r_{\text{min}} = \frac{8.98755 \times 10^9 \times 95 \times 79 \times (1.60219 \times 10^{-19})^2}{8.7800012 \times 10^{-13}} = 1.9720842942047896 \times 10^{-18} \, \text{m} \][/tex]
Therefore, the closest distance the alpha particle can approach a gold nucleus is approximately [tex]\( 1.972 \times 10^{-18} \)[/tex] meters.
### Step-by-Step Solution
1. Identify the Given Values:
- Atomic number of Americium (Am): \( Z_{\text{Am}} = 95 \)
- Atomic number of Gold (Au): \( Z_{\text{Au}} = 79 \)
- Coulomb constant: \( k = 8.98755 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2 \)
- Charge of an electron: \( e = 1.60219 \times 10^{-19} \, \text{C} \)
- Energy of the α-particle: \( E_{\alpha} = 5.48 \, \text{MeV} \)
2. Convert Energy of the α-particle to Joules:
- The provided energy in MeV needs to be converted to Joules because the other values are in standard SI units.
- \( 1 \, \text{eV} = 1.60219 \times 10^{-19} \, \text{J} \)
- Therefore, \( E_{\alpha} = 5.48 \times 10^6 \, \text{eV} \times 1.60219 \times 10^{-19} \, \text{J/eV} = 8.7800012 \times 10^{-13} \, \text{J} \)
3. Calculate the Closest Distance Using Coulomb's Law:
- The closest distance (r_min) can be found using the balance between the kinetic energy of the α-particle and the electrostatic potential energy at the point where the particle stops.
- Coulomb potential energy between two point charges \( q_1 \) and \( q_2 \) separated by distance \( r \) is given by
[tex]\[ U = \frac{k \cdot q_1 \cdot q_2}{r} \][/tex]
- Here, the charges are \( q_{\text{Am}} = Z_{\text{Am}} \cdot e \) and \( q_{\text{Au}} = Z_{\text{Au}} \cdot e \).
- Set the potential energy equal to the kinetic energy \( E_{\alpha} \)
[tex]\[ E_{\alpha} = \frac{k \cdot Z_{\text{Am}} \cdot e \cdot Z_{\text{Au}} \cdot e}{r_{\text{min}}} \][/tex]
Solving for \( r_{\text{min}} \):
[tex]\[ r_{\text{min}} = \frac{k \cdot Z_{\text{Am}} \cdot Z_{\text{Au}} \cdot e^2}{E_{\alpha}} \][/tex]
4. Substitute the Given Values:
- Plug the values into the equation:
[tex]\[ r_{\text{min}} = \frac{8.98755 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2 \times 95 \times 79 \times (1.60219 \times 10^{-19} \, \text{C})^2}{8.7800012 \times 10^{-13} \, \text{J}} \][/tex]
- Simplify the expression to get \( r_{\text{min}} \):
[tex]\[ r_{\text{min}} = \frac{8.98755 \times 10^9 \times 95 \times 79 \times (1.60219 \times 10^{-19})^2}{8.7800012 \times 10^{-13}} = 1.9720842942047896 \times 10^{-18} \, \text{m} \][/tex]
Therefore, the closest distance the alpha particle can approach a gold nucleus is approximately [tex]\( 1.972 \times 10^{-18} \)[/tex] meters.
