Answer
$28.4\%$
Work Step by Step
Please see the attached image first.
For an elastic collision, we can write the equation below.
$V_{2}=\frac{2m_{1}}{m_{1}+m_{2}}u_{1}+\frac{(m_{1}-m_{2})}{m_{1}+m_{2}}u_{2}-(1)$
In this case, $m_{1}=U,\space m_{2}=12U,\space u_{2}=0$
From $(1)=\gt$
$V_{2}=\frac{2m_{1}u_{1}}{(m_{1}+m_{2})}-(2)$
Kinetic energy of Neutron $(K_{1})=\frac{1}{2}m_{1}u_{1}^{2}$
Kinetic energy of Neutron $(K_{2})=\frac{1}{2}m_{2}V_{2}^{2}$
Percentage of transferred kinetic energy $=\frac{K_{2}}{K_{1}}\times 100\%$
$\frac{K_{2}}{K_{1}}\times 100=\frac{\frac{1}{2}m_{2}V_{2}^{2}}{\frac{1}{2}m_{1}u_{1}^{2}}\times100-(3)$
$(3)=\gt(2)$
$\frac{K_{2}}{K_{1}}\times 100=\frac{12U(\frac{2Uu_{1}}{13U})^{2}}{Uu_{1}^{2}}\times100\%=\frac{12\times4u_{1}^{2}}{169}\times\frac{1}{u_{1}^{2}}\times100\%$
$\frac{K_{2}}{K_{1}}\times 100=28.4\%$
