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A neutron with velocity v suffers head on elastic collision with the nucleus of an atom of mass number A at rest. The fraction of energy retained by neutron is |
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Answer» `((A+1)/(A))^(2)` `:.` velocity of neutron after collison is `V=((m_(1)-m_(2))u_(1)+2m_(2)u_(2))/(m_(1)+m_(2))=((m-Am)u)/(m+Am)=((1-A)u)/((1+A))` `:.` K.E. of neutron after collison, `1/2 m_(1)v_(2)` or `E_(1)=1/2m_(1)((1-A)/(1+A))^(2)u^(2)` and K.E. before collison, `E_(1)=1/2"mu"^(2)` `:. (E_(1))/(E_(1))=(1/2M((1-A)/(1+A))^(2)u^(2))/(1/2mu^(2))=((1-A)/(1+A))^(2)` |
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