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:

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Accepted Answer

(89, -28). For a head-on elastic collision the fractional energy loss is 4*m1*m2/(m1+m2)^2. Neutron-deuteron: 4*1*2/9 = 8/9 ~ 89%. Neutron-carbon: 4*1*12/169 = 48/169 ~ 28%. So (Pd, Pc) = (89, 28).

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:

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