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.
Top Questions on Nuclear physics
- (a) Consider the so-called ‘D-T reaction’ (Deuterium-Tritium reaction).
In a thermonuclear fusion reactor, the following nuclear reaction occurs: \[ \ ^{2}_1 \text{H} + \ ^{3}_1 \text{H} \longrightarrow \ ^{4}_2 \text{He} + \ ^{1}_0 \text{n} + Q \] Find the amount of energy released in the reaction.
% Given data Given:
\( m\left(^{2}_1 \text{H}\right) = 2.014102 \, \text{u} \)
\( m\left(^{3}_1 \text{H}\right) = 3.016049 \, \text{u} \)
\( m\left(^{4}_2 \text{He}\right) = 4.002603 \, \text{u} \)
\( m\left(^{1}_0 \text{n}\right) = 1.008665 \, \text{u} \)
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