Question

The closest approach of an alpha particle when it makes a head-on collision with a gold nucleus is $ 10 \times 10^{-14} $ m. Then the kinetic energy of the alpha particle is:

• A. \( 3640 \, \text{J} \)
• B. \( 3.64 \, \text{J} \)
• C. \( 3.64 \times 10^{-16} \, \text{J} \)
• D. \( 3.64 \times 10^{-13} \, \text{J} \)

Hint:

For a head-on collision with a nucleus, the kinetic energy of the alpha particle is converted into electrostatic potential energy at the point of closest approach. You can use Coulomb's law to calculate this potential energy.

Answer 1

The Correct Option is D

When an alpha particle collides head-on with a gold nucleus, the potential energy at the closest approach can be calculated using Coulomb’s law. The total energy at the closest approach is purely electrostatic potential energy because the kinetic energy of the alpha particle becomes zero at the point of closest approach. 
The potential energy at the closest approach, \( U \), is given by the formula: \[ U = \frac{k_e \cdot Z_1 \cdot Z_2 \cdot e^2}{r} \] Where: - \( k_e = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2 \) (Coulomb constant), - \( Z_1 = 2 \) (charge of the alpha particle), - \( Z_2 = 79 \) (charge of the gold nucleus), - \( e = 1.6 \times 10^{-19} \, \text{C} \) (elementary charge), - \( r = 10 \times 10^{-14} \, \text{m} \) (closest approach). Now, substitute the values into the equation: \[ U = \frac{(9 \times 10^9) \times (2) \times (79) \times (1.6 \times 10^{-19})^2}{10 \times 10^{-14}} \] Simplifying: \[ U = \frac{9 \times 10^9 \times 2 \times 79 \times (2.56 \times 10^{-38})}{10 \times 10^{-14}} = 3.64 \times 10^{-13} \, \text{J} \] Thus, the kinetic energy of the alpha particle is \( 3.64 \times 10^{-13} \, \text{J} \).