Question

The distance of closest approach of an alpha particle from a nucleus when the alpha particle moves towards a nucleus with a kinetic energy 'E' is 'x'. The distance of closest approach when the alpha particle approaches the same nucleus with kinetic energy 0.4E is

• A. 3.5 x
• B. 5 x
• C. 2.5 x
• D. 4 x

Hint:

Remember $d \propto 1/E$ for closest approach in Coulomb scattering. Alpha particle charge is 2e. Use consistent units for constants. Apply to nuclear size estimation in Rutherford experiment.

Answer 1

The Correct Option is C

1. The distance of closest approach in Rutherford scattering occurs when kinetic energy is fully converted to electrostatic potential energy.
2. The formula is $x = \frac{1}{4\pi \epsilon_0} \frac{(2e)(Ze)}{E} = \frac{2 Z e^2}{4\pi \epsilon_0 E}$, where 2e is alpha charge, Ze nucleus charge.
3. Thus, $x \propto \frac{1}{E}$, meaning the distance is inversely proportional to the kinetic energy.
4. For new energy $E' = 0.4 E$, the new distance $x' = x \times \frac{E}{E'} = x \times \frac{1}{0.4} = 2.5 x$.
5. This assumes non-relativistic speeds and point-like particles.
6. Therefore, the correct option is (3) 2.5 x.