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Q. It is found tha if a neutron suffers an elastic - collinear collision with deuterium at rest, fractional loss of its energy is Pc. The values of Pd and Pc are respectively:

(1) (0,1)
(2) (.89, .28)
(3) (.28, .89)
(4) (0,0)

Dear student,From conservation of moemntum we havemu =mv1+ 2mv2v2-v1 =uv2=2u3Kinetic energy of deuterium=12×2m×(2u3)2=Loss in energy of neutronFractional loss(pd)=2×49×12mu212mu2=0.89similarly,mu=mv1+12mv2v2=2u13Kinetic energy of carbon atom=12×12m×(2u13)2Fractional loss(pc)=12×4169=0.28Regards

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