Question

The distance of closest approach when an alpha particle of kinetic energy 6.5 MeV strikes a nucleus of atomic number 50 is

• A. 0.221 fm
• B. 1.101 fm
• C. 0.0221 fm
• D. 4.42 fm

Hint:

Remember to convert energy units to joules when using the equation for the closest approach.

Answer 1

The Correct Option is C

The distance of closest approach \( r \) for an alpha particle can be determined using the formula derived from the electrostatic potential energy at the point of closest approach: \[ E_{\text{kin}} = \frac{1}{4\pi\epsilon_0} \times \frac{2z_1z_2e^2}{r} \] where: - \( E_{\text{kin}} \) is the kinetic energy of the alpha particle, - \( z_1 = 2 \) is the charge of the alpha particle, - \( z_2 = 50 \) is the charge of the nucleus, - \( e \) is the elementary charge (\( 1.602 \times 10^{-19} \, \text{C} \)), - \( \epsilon_0 = 8.854 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \) is the permittivity of free space. Given: \[ E_{\text{kin}} = 6.5 \, \text{MeV} = 6.5 \times 10^6 \times 1.602 \times 10^{-13} \, \text{J} = 1.042 \times 10^{-6} \, \text{J} \] Now, substituting the values into the formula: \[ 1.042 \times 10^{-6} = \frac{(2)(50)(1.602 \times 10^{-19})^2}{4 \pi \times 8.854 \times 10^{-12} \times r} \] Solving for \( r \): \[ r = \frac{(2)(50)(1.602 \times 10^{-19})^2}{4 \pi \times 8.854 \times 10^{-12} \times 1.042 \times 10^{-6}} = 0.0221 \, \text{fm} \]
Thus, the distance of closest approach is \( 0.0221 \, \text{fm} \).