It is found that it a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is `p_d` , while for its similar collision with carbon nucleus at rest, fractional loss of energy is `p_c`. The values of `p_d and p_c` are respectively

Options

A. 0,0
B. 0,1
C. 0.89,0.28
D. 0.28,0.89

Answer

C

Explanation

In the first case,
`mu_n=mv_n+2mxxd_D`……(1)
[`u_n and v_n` are repectivley initial and final velocity of neutron , `v_D` is the final velocity of deuterium]
`u_n=(v_D-v_n) [because e=1]` …….(2)
From (1) and (2) we get,
`v_n=-(u_n)/(3)`
`therefore p_d=(DeltaE)/(E)=(1/2mu_n^2-1/2mv_n^2)/(1/2mu_n^2) =8/9=0.89`
In the second case,
`mu_n=mv_n^.+(12m)xxv_C`.......(3)
[`u_n and v_n^.` are respectively initial and final velocity of neutron, `v_C` is the final velocity of carbon atom ]
`u_n =v_c-v_n^.[ because e=1]`.........(4)
From (3) and (4) we get,
`v_n^. =-(11)/(13)u_n`
`therefore p_c =(DeltaE)/E=(1/2 mu_n^2-1/2m(v_n6.)^2)/(1/2mu_n^2) =(48)/(69) =0.28`