Question
A neutron travelling with a velocity $$v$$ and kinetic energy $$E$$ has a perfectly elastic head-on collision with a nucleus of an atom of mass number $$A$$ at rest. The fraction of total energy retained by the neutron is approximately
A.
$${\left[ {\frac{{\left( {A - 1} \right)}}{{\left( {A + 1} \right)}}} \right]^2}$$
B.
$${\left[ {\frac{{\left( {A + 1} \right)}}{{\left( {A - 1} \right)}}} \right]^2}$$
C.
$${\left[ {\frac{{\left( {A - 1} \right)}}{A}} \right]^2}$$
D.
$${\left[ {\frac{{\left( {A + 1} \right)}}{A}} \right]^2}$$
Answer :
$${\left[ {\frac{{\left( {A - 1} \right)}}{{\left( {A + 1} \right)}}} \right]^2}$$
Solution :
$${{v'}_1} = \frac{{\left( {{m_1} - {m_2}} \right){v_1} + 2{m_2}{v_2}}}{{{m_1} + {m_2}}}$$
As $${v_2}$$ is zero, $${m_2} > {m_1},{{v'}_1}$$ is in the opposite direction.
$$\eqalign{
& {m_1} = 1,{m_2} = A. \cr
& \therefore \left| {{{v'}_1}} \right| = \frac{{\left( {A - 1} \right)}}{{\left( {A + 1} \right)}}{v_1} \cr} $$
The fraction of total energy retained is
$$\frac{{\frac{1}{2}mv'\,_1^2}}{{\frac{1}{2}v_1^2}} = \frac{{{{\left( {A - 1} \right)}^2}}}{{{{\left( {A + 1} \right)}^2}}}$$