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

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