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making use of the tables of atomic masses, find the velocity with which the products of the reaction `B^(10)`(n,alpha)`Li^(7)` come apart, the reaction proceeds via interaction of very slow neutons with stationary boron nuclei.

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The rection is `B^(10)(n,alpha)Li^(7)`. Then energy of the reaction is
`Q=(Delta_(B)^(10)+Delta_(n)-Delta_(alpha)-Delta_(Li)^(7))c^(2)`
`=(0.01297+0.00867-.000260-0.01601)am uxxc^(2)`
`=2.79MeV`
Since the incident neutron is very slow and `B^(10)` is stationary, the final total momentum must also be zero. So the reaction products must emerge in opposite directions. If their speeds are, respectively `v_(alpha)`, and `v_(Li)`
then `4v_(alpha)= 7v_(Li)`
and `(1)/(2)(4v_(alpha)^(2)+7v_(Li)^(2))xx1.672xx10^(-24)=2.79xx1.602xx10^(-6)`
So `(1)/(2)xx4v_(alpha)^(2)(1+(4)/(7))=2.70xx10^(18)cm^(2)//s^(2)`
or `v_(alpha)= 9.27xx10^(6)m//s`
Then `v_(Li)=5.3xx10^(6)m//s`
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