question_answer

It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is\[{{\text{P}}_{\text{d}}}\]; while for its similar collision with carbon nucleus at rest, fractional loss of energy is\[{{\text{P}}_{\text{c}}}\]. The values of \[{{\text{P}}_{\text{d}}}\]and \[{{\text{P}}_{\text{c}}}\]are respectively:  [JEE Main Online 08-04-2018]

A)  \[\text{(0, 0)}\]                       

B)  \[\text{(0, 1)}\]

C)  \[\text{(}\cdot 89\text{, }\cdot \text{28)}\]                

D)  \[\text{(}\cdot 28\text{, }\cdot \text{89)}\]

Correct Answer: C

Solution :

 Conservation of momentum \[m{{v}_{0}}+0=m{{v}_{1}}+2m{{v}_{2}}\] \[{{v}_{0}}=({{v}_{1}}+2{{v}_{2}}).....(i)\] \[e=\frac{{{v}_{2}}-{{v}_{1}}}{{{v}_{0}}}(e=1)\] \[{{v}_{2}}-{{v}_{1}}={{v}_{0}}...(ii)\] \[{{v}_{0}}-2{{v}_{1}}=2{{v}_{0}}+{{v}_{1}}\] \[3{{v}_{1}}=-{{v}_{0}}\Rightarrow {{v}_{1}}=\frac{-{{v}_{0}}}{3}\] Fractional loss of its K.E. \[=\frac{\frac{1}{2}m{{v}_{0}}^{2}-\frac{1}{2}m{{\left( \frac{{{v}_{0}}}{3} \right)}^{2}}}{\frac{1}{2}m{{v}_{0}}^{2}}=\frac{8}{9}=0.88\approx 0.89\] Neutron colliding with carbon Conservation of momentum \[m{{v}_{0}}=m{{v}_{1}}+12m{{v}_{2}}\] \[{{v}_{1}}+12{{v}_{2}}={{v}_{0}}...(i)\] \[e=1=\frac{{{V}_{2}}-{{V}_{1}}}{{{V}_{0}}}\] \[{{V}_{2}}-{{V}_{1}}={{V}_{0}}...(ii)\] \[\Rightarrow {{V}_{1}}+12{{v}_{1}}={{v}_{0}}-12{{v}_{0}}\] \[=\frac{-11{{v}_{0}}}{13}\] Fractional loss in K.E. \[=\frac{\frac{1}{2}m{{v}_{0}}^{2}-\frac{1}{2}m{{\left( \frac{11}{13}{{v}_{0}} \right)}^{2}}}{\frac{1}{2}m{{v}_{0}}^{2}}=1-\frac{121}{169}=\frac{48}{169}\] \[\approx 0.28\]