InterviewSolution
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InterviewSolution
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A neutron collides elastically with an initially stationary deuteron. Find the fraction of the kinetic energy lost by the neutron (a)in a head-on collision (b) in scatterting at right angles. |
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Answer» Solution :In a head on collision `sqrt(2mT)=p_(d)+p_(n)` `sqrt(2mT)=p_(d)+p_(n)` `T=(p_(d)^(2))/(2M)+p_(n^(2))/(2m)` where `p_(d)` and `p_(n)` are the momenta of deuteron and neutron after the collision. Squaring `p_(d)^(2)+p_(n)^(2)+2p_(d)p_(n)=2mT` `p_(n)^(2)+(m)/(M)p_(d)^(2)= 2mT` or since `p_(d)=0` in a head on collisions `P_(n)= -(1)/(2)(1-(m)/(M))p_(d)` Going BACK to energy CON where `p_(d)` and `p_(n)` are the momenta of deuteron and neutron after the collision. Squaring `p_(d)^(2)+p_(n)^(2)+2p_(d)p_(n)=2mT` `p_(n)^(2)+(m)/(M)p_(d)^(2)= 2mT` or since `p_(d)=0` in a head on collisions `P_(n)= -(1)/(2)(1-(m)/(M))p_(d)` Going back to energy conservation `(p_(d)^(2))/(2M)[1+(M)/(4m)(1-(m)/(M))^(2)]=T` so `(p_(d)^(2))/(2M)=(4mM)/((m+M)^(2))=(8)/(9)` (b) In this case neutron is scattered by `90^(@)`. Then we have from the DIAGRAM `vec(p)_(d)= p_(n)hat(j)+sqrt(2mT)hat(i)` Then by energy conservation `(p_(n)^(2)+2mT)/(2M)+(p_(n)^(2))/(2m)=T` or `(p_(n)^(2))/(2m)(1+(m)/(M))=T(1-(m)/(M))` or `(p_(n)^(2))/(2m)=(M-m)/(M+m).T` The energy LOST by neutron in then `T-(p_(n)^(2))/(2m)=(2m)/(M+m)T` or fraction of energy lost is `eta=(2m)/(M+m)=(2)/(3)` 
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