Solve this: 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) Share with your friends Share 6 Decoder_2 answered this Dear student,From conservation of moemntum we havemu =mv1+ 2mv2v2-v1 =u⇒v2=2u3Kinetic energy of deuterium=12×2m×(2u3)2=Loss in energy of neutronFractional loss(pd)=2×49×12mu212mu2=0.89similarly,mu=mv1+12mv2⇒v2=2u13Kinetic energy of carbon atom=12×12m×(2u13)2Fractional loss(pc)=12×4169=0.28Regards 8 View Full Answer