In alpha particle scattering experiment, if \( v \) is the initial velocity of the particle, then the distance of closest approach is
- \( 4 \, d \)
- \( 2 \, d \)
- \( d \)
- \( \frac{d}{2} \)
The Correct Option is D
Solution and Explanation
In the alpha particle scattering experiment, the distance of closest approach is calculated using the relationship between the initial velocity of the particle and the distance \( d \) between the centers of the alpha particle and the nucleus:
\( r = \frac{d}{2} \)
This formula shows that the distance of closest approach is half the distance between the centers of the two particles when considering the initial velocity of the alpha particle.
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