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.
Mass Defect and Energy Released in the Fission of \( ^{235}_{92}\text{U} \)
When a neutron collides with \( ^{235}_{92}\text{U} \), the nucleus gives \( ^{140}_{54}\text{Xe} \) and \( ^{94}_{38}\text{Sr} \) as fission products, and two neutrons are ejected. Calculate the mass defect and the energy released (in MeV) in the process.
Given: