A 0.02 Mev alpha particle is scattered from a gold nucleus (Z=79) at an angle of 90o. Calculate the impact parameter.
Rutherford Scattering Formula:
The formula for impact parameter (\(b\)) is derived from the Rutherford scattering formula:
\[ \tan\left(\frac{\theta}{2}\right) = \frac{k \cdot Z \cdot e^2}{4\pi\epsilon_0 \cdot E \cdot b} \]
Where:
- \( \theta \) = scattering angle (90 degrees in this case)
- \( k \) = Coulomb's constant (\( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \))
- \( Z \) = atomic number of the gold nucleus (79)
- \( e \) = elementary charge (\( 1.6 \times 10^{-19} \, \text{C} \))
- \( \epsilon_0 \) = vacuum permittivity (\( 8.85 \times 10^{-12} \, \text{F/m} \))
- \( E \) = kinetic energy of the alpha particle
- \( b \) = impact parameter
Given:
- Scattering angle (\( \theta \)) = 90 degrees
- Atomic number of gold (\( Z \)) = 79
- Kinetic energy (\( E \)) of alpha particle = 0.02 MeV (\( 0.02 \times 10^6 \) eV)
First, convert the energy to joules:
\[ 0.02 \times 10^6 \text{ eV} \times \frac{1.6 \times 10^{-19} \, \text{J}}{1 \, \text{eV}} = 3.2 \times 10^{-13} \, \text{J} \]
Next, let's rearrange the Rutherford scattering formula to solve for the impact parameter (\( b \)):
\[ b = \frac{k \cdot Z \cdot e^2}{4\pi\epsilon_0 \cdot E \cdot \tan\left(\frac{\theta}{2}\right)} \]
Now, plug in the known values:
\[ b = \frac{8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \cdot 79 \cdot (1.6 \times 10^{-19} \, \text{C})^2}{4\pi \times 8.85 \times 10^{-12} \, \text{F/m} \cdot 3.2 \times 10^{-13} \, \text{J} \cdot \tan\left(\frac{90^\circ}{2}\right)} \]
Solving this equation will give us the impact parameter (\( b \)). Let's calculate it.
\[ b \approx 1.41 \times 10^{-14} \, \text{m} \]
Therefore, the impact parameter is approximately \( 1.41 \times 10^{-14} \, \text{m} \).
Also Read : Write the definition of μ-space.
No comments: