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How many head-on elastic collisions must a neutron have with deuterium nuclei to reduce it energy from `6.561 MeV` to `1 keV`?

Monjur Alahi

How many head-on elastic collisions must a neutron have with deuterium nuclei to reduce it energy from `6.561 MeV` to `1 keV`?
A. 4
B. 5
C. 8
D. 3


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Salim Ahmed Kawsar

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Correct Answer - A
`("Energy loss")/("Initial KE") = (4m_(1)m_(2))/((m_(1) + m_(2))^(2)) = (4(1)(2))/((1+2)^(2)) = 8/9`
After `1^(st)` collision `DeltaE_(1) = 8/9 E_(0)` After `2^(nd)` collision `DeltaE_(2) = (8)/(9)E_(1)`, After `n^(th)` collision `DeltaE_(n) = (8)/(9)E_(n1)`
Adding all the losses
`DeltaE = DeltaE_(1) + E_(2) + "......." DeltaE_(n) = 8/9 (E_(0) + E_(1) + "......"E_(n-1))`: here `E_(1) = E_(0) - DeltaE_(1) = E_(0) - 8/9E_(0) = 1/9 E_(0)`
`E_(2) = E_(1) - DeltaE_(2) = E_(1) (8)/(9) E_(1) = (1)/(9)E_(1) = (1/9)^(2)E_(0)` and so on
`rArr DeltaE = 8/9[E_(0) + (1/9)^(2)E_(0)+(1/9)^(2)E_(0)+"......"+(1/9)^(n-1)E_(0)] = 8/9 [(1-1/(9^(n)))/(1-1/9)]E_(0) = (1-(1)/(9^(n)))E_(0)`
`E_(0) = 6.561 MeV, DeltaE = (6.561 - 0.001)MeV rArr (6.561 - 0.001)/(6.561) = 1 - (1)/(9^(n)) rArr (1)/(6561) = (1)/(9^(n)) rArr n = 4`

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