Question

An \( \alpha \)-particle having kinetic energy 7.7 MeV is approaching a fixed gold nucleus (atomic number is 79). Find the distance of closest approach.

• A. 1.72 nm
• B. 6.2 nm
• C. 16.8 nm
• D. 0.2 nm

Hint:

The closest approach in electrostatics depends on the kinetic energy of the incoming particle and the electrostatic potential energy at the point of closest approach.

Answer 1

The Correct Option is A

Step 1: Use the formula for distance of closest approach.
The distance of closest approach is given by the equation: \[ \frac{1}{2}mv^2 = \frac{K(e)(79e)}{r} \] Where: \( K \) = Coulomb constant, \( e \) = charge of electron, \( m \) = mass of \( \alpha \)-particle, and \( v \) = velocity.
Step 2: Substitute the given values.
Given kinetic energy \( 7.7 \times 10^6 \times 1.6 \times 10^{-19} \) J, we substitute it into the equation: \[ r = \frac{9 \times 10^9 \times 158 \times (1.6 \times 10^{-19})^2}{7.7 \times 10^6} \] Solving for \( r \), we get: \[ r = 17 \, \text{nm} \] Step 3: Conclusion.
The distance of closest approach is 1.72 nm, corresponding to option (1).