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 A neutron traveling with a velocity v and K'E. E collides perfectly elastically head-on with the nucleus
of an atom of mass number A at rest. The fraction of total energy retained by neutron is

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a
1−AA+12
b
A+1A−12
c
A−1A2
d
A+1A2

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detailed solution

Correct option is A

Let the final velocity of the neutron be v1 Thenv1=m1−m2m1+m2u1+2m2m1+m2u2 Hero u1=u  and  u2=0, m1=1 and m2  = A v1=1−A1+Au For neutron, final K.E. = E' ∴ E′=12(1)v12=121−A1+A2u2  Initial K.E. =E ∴ E=12(1)u2 From eqs. (i) and [ii), we have E′E=1−A1+A2 E′=1−A1+A2E


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