A) 90%
B) 45%
C) 28 %
D) 12%
Correct Answer: C
Solution :
The given, Mass of neutron \[=1.67\times {{10}^{-27}}kg={{m}_{1}}\] Mass of carbon \[=1.99\times {{10}^{-26}}kg={{m}_{2}}\] Kinetic energy of before collision \[{{K}_{1}}=\frac{1}{2}{{m}_{1}}u_{1}^{2}\] Kinetic energy of after collision \[{{K}_{2}}=\frac{1}{2}{{m}_{1}}v_{1}^{2}\] Kinetic energy transferred from projectile to target \[\Delta K=\] decrease in kinetic energy \[\Delta E=\frac{1}{2}{{m}_{1}}u_{1}^{2}-\frac{1}{2}{{m}_{1}}v_{1}^{2}=\frac{1}{2}{{m}_{1}}(u_{1}^{2}-v_{1}^{2})\] The percentage of kinetic energy \[=\frac{\Delta K}{{{K}_{1}}}\times 100\] \[=\frac{\frac{1}{2}{{m}_{1}}(u_{1}^{2}-v_{1}^{2})}{\frac{1}{2}{{m}_{1}}u_{1}^{2}}\times 100\] \[=\left[ 1-{{\left( \frac{{{v}_{1}}}{{{u}_{1}}} \right)}^{2}} \right]\times 100\] We know that \[\frac{{{v}_{1}}}{{{u}_{1}}}=\left( \frac{{{m}_{1}}-{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right)\] and \[\frac{\Delta K}{{{K}_{1}}\times 100}=\left[ 1-{{\left( \frac{{{m}_{2}}-{{m}_{1}}}{{{m}_{1}}+{{m}_{2}}} \right)}^{2}} \right]\times 100\] \[=\left[ 1-{{\left( \frac{1.99\times {{10}^{-26}}-0.167\times {{10}^{-26}}}{1.99\times {{10}^{-26}}+0.167\times {{10}^{-26}}} \right)}^{2}} \right]\times 100\] \[=\left[ 1-{{\left( \frac{1.823}{2.157} \right)}^{2}} \right]\times 100\] \[=[1-{{(0.845)}^{2}}]\times 100\] \[=[1-0.714]\times 100\] \[=0.286\times 100=28.6%\cong 28%\]
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