Question

In Geiger-Marsden experiment, if the initial speed of \(\alpha\) particle is doubled, then the closest distance of approach of the \(\alpha\) particle from the gold nucleus - (Assume \(\alpha\) particle is projected straight towards the gold nucleus)

• A. becomes doubled
• B. becomes quadrupled
• C. becomes half
• D. becomes one-fourth

Hint:

The closest distance of approach in Rutherford scattering is inversely proportional to the initial kinetic energy of the \(\alpha\) particle. If speed is doubled, kinetic energy becomes four times, reducing the approach distance to one-fourth.

Answer 1

The Correct Option is D

Step 1: Understanding Closest Distance of Approach The closest distance of approach \( r_{\text{min}} \) of an \(\alpha\) particle in Rutherford scattering is given by: \[ r_{\text{min}} = \frac{1}{4\pi \epsilon_0} \frac{2 Z e^2}{K} \] where: 
- \( Z \) is the atomic number of the nucleus, 
- \( e \) is the charge of the electron, 
- \( K \) is the initial kinetic energy of the \(\alpha\) particle. 

Step 2: Effect of Doubling Speed Kinetic energy is related to speed as: \[ K = \frac{1}{2} m v^2 \] If the initial speed of the \(\alpha\) particle is doubled: \[ K' = \frac{1}{2} m (2v)^2 = 4K \] Since \( r_{\text{min}} \) is inversely proportional to \( K \): \[ r_{\text{min}}' = \frac{r_{\text{min}}}{4} \] Thus, the closest distance of approach becomes one-fourth of its original value. Thus, the correct answer is \( \mathbf{(4)} \) becomes one-fourth.