question_answer

A neutron moving with a velocity v, collides elastically with an atom of mass number A. The collision is head on. If the initial kinetic energy of a neutron is e, then the kinetic energy of neutron after the collision is :

A) \[{{\left( \frac{1+A}{1-A} \right)}^{2}}E\]                             

B) \[{{\left( \frac{1-A}{1+A} \right)}^{2}}E\]

C)  \[\left( \frac{1+A}{1-A} \right)E\]                           

D) \[\left( \frac{1-A}{1+A} \right)E\]

Correct Answer: B

Solution :

                 In elastic collision, kinetic energy and momentum both are conserved. Let\[{{v}_{1}}\]be velocity after collision \[{{v}_{1}}=\left( \frac{{{m}_{1}}-{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right){{u}_{1}}+\left( \frac{2{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right){{u}_{2}}\] Given, \[{{u}_{1}}=v,\,{{u}_{2}}=0,\,{{m}_{1}}=1,\,{{m}_{2}}=A\] \[\therefore \]  \[{{v}_{1}}=\left( \frac{1-A}{1+A} \right)v\] Initial kinetic energy of neutron                 \[E=\frac{1}{2}{{m}_{1}}v_{1}^{2}=\frac{1}{2}\times 1\times {{\left( \frac{1-A}{1+A} \right)}^{2}}\] \[\therefore \]  \[\frac{E}{E}={{\left( \frac{1-A}{1+A} \right)}^{2}}\] \[\Rightarrow \]               \[E={{\left( \frac{1-A}{1+A} \right)}^{2}}E\] Final KE of neutron \[E=\frac{1}{2}{{m}_{1}}v_{1}^{2}=\frac{1}{2}\times 1\times {{\left( \frac{1-A}{1+A} \right)}^{2}}\] \[\therefore \]  \[\frac{E}{E}={{\left( \frac{1-A}{1+A} \right)}^{2}}\] \[\Rightarrow \]               \[E={{\left( \frac{1-A}{1+A} \right)}^{2}}E\]