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Question and Answer

A neutron, with kinetic energy KE and moving with velocity \(v\), collides head on with a nucleus of mass number \(A\) perfectly elastically. The fraction of total energy possessed by the neutron transferred to nucleus is
(A) \(\dfrac{4A}{(A+1)^2}\)
(B) \(\left(\dfrac{A-1}{A+1}\right)^2\)
(C) \(\left(\dfrac{A+1}{2A}\right)^2\)
(D) \(\left(\dfrac{A-1}{2A}\right)^2\)

Answer

Answer: A
\(Conservation\quad of\quad linear\quad momentum\\ mv=Am{ v }_{ 1 }+m{ v }_{ 2 }\\ and\quad defintion\quad of\quad coefficient\quad of\quad Restitution\\ \frac { { v }_{ 2 }-{ v }_{ 1 } }{ v-0 } =-1\Rightarrow { v }_{ 2 }-{ v }_{ 1 }=-v\\ solving\quad these\quad equations\quad we\quad get\\ { v }_{ 1 }=\frac { 2 }{ A+1 } v\\ KE\quad of\quad the\quad neutron\quad =\frac { 1 }{ 2 } Am{ \left( \frac { 2 }{ A+1 } v \right) }^{ 2 }=\frac { 4A }{ { \left( A+1 \right) }^{ 2 } } KE\)
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