A neutron traveling with a velocity u and  kinetic energy  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 neutron be v₁  According to theory of collision

$$\begin{gathered} {\mathrm{v}}_1=\left(\frac{{\mathrm{m}}_1-{\mathrm{m}}_2}{{\mathrm{m}}_1+{\mathrm{m}}_2}\right){\mathrm{u}}_1+\left(\frac{2{\mathrm{m}}_2}{{\mathrm{m}}_1+{\mathrm{m}}_2}\right){\mathrm{u}}_2 \\ \text{ Here }{\mathrm{u}}_1=\mathrm{u}\text{ and }{\mathrm{u}}_2=0\text{. Further }{\mathrm{m}}_1=1\text{ and }{\mathrm{m}}_2=\mathrm{A} \\ \therefore {\mathrm{v}}_1=\left(\frac{1-\mathrm{A}}{1+\mathrm{A}}\right)\mathrm{u} \\ \text{ Final K.E. of neutron }=\frac{1}{2}{\mathrm{mv}}_1^2 \\ =\frac{1}{2}\mathrm{m}{\left(\frac{1-\mathrm{A}}{1+\mathrm{A}}\right)}^2{\mathrm{u}}^2 \\ \text{ Initial K.E. of neutron }=\frac{1}{2}{\mathrm{mu}}^2 \\ \therefore \text{ Fraction of energy retained } \\ ={\left(\frac{1-\mathrm{A}}{1+\mathrm{A}}\right)}^2 \end{gathered}$$