Westonci.ca is the premier destination for reliable answers to your questions, brought to you by a community of experts. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
Sagot :
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.
Visit us again for up-to-date and reliable answers. We're always ready to assist you with your informational needs. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.
