It is found that if a neutron suffers an elas | vedclass

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Question

For collision of a nenutron with deuterium $:$

Applying conservation of momentum$:$

$m v+0=m v_{1}+2 m v_{2} \dots(i)$

$v_{2}-v_{1}=v \ldots(

Vedclass · Accepted Answer

$(0.89,0.28)$

Vedclass · Answer

For collision of a nenutron with deuterium $:$

Applying conservation of momentum$:$

$m v+0=m v_{1}+2 m v_{2} \dots(i)$

$v_{2}-v_{1}=v \ldots(ii)$

Therefore, Collision is elastic, $e=1$

From equ $(i)$ and equ $(ii)$ $v_{1}=-v / 3$

$P_{d}=\frac{\frac{1}{2} m v^{2}-\frac{1}{2} m v_{1}^{2}}{\frac{1}{2} m v^{2}}=\frac{8}{9}=0.89$

Now, for the collision of neutron with carbon nucleus

Applying conservation of momentum

$m v+0=m v_{1}+12 m v_{2} \ldots$ $(iii)$

$v=v_{2}-v_{1}$            $...(iv)$

$v_{1}=-\frac{11}{13} v$

$P_{0}=\frac{\frac{1}{2} m v^{2}-\frac{1}{2} m\left(\frac{11}{13} v\right)^{2}}{\frac{1}{2} m v^{2}}=\frac{48}{169} \approx 0.28$

It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is $p_d $ ; while for its similar collision with carbon nucleus at rest, fractional loss of energy is $P_c$. The values of $P_d$ and $P_c$ are respectively

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