Question
It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is pd; while for its similar collision with carbon nucleus at rest, fractional loss of energy is pc. The values of pd and pc are respectively :
(0,1)
(.89,.28)
(.28,.89)
(0,0)
Solution
B.
(.89,.28)
For collision of a neutron with deuterium:
Applying conservation of momentum:
mv + 0 = mv1 + 2mv2 .....(i)
v2 -v1 = v ...... (ii)
Therefore, Collision is elastic, e = 1
From equ (i) and equ (ii) v1 = -v/3
Now, for the collision of neutron with carbon nucleus
Applying conservation of momentum
mv + 0 = mv1 + 12mv2 ....; (iii)
v = v2-v1 ....(iv)
