The nuclear reaction `n+._5^(10)B rarr ._3^7Li + ._2^4He` is observed to occur even when very slow-moving neutrons `(M_ (n)=1.0087 am u)` strike a boron atom at rest. For a particular reaction in which `K_(n) =0` , the helium `(M_(He) =4.0026 am u)` is observed to have a speed of `9.30 xx 10^(6) m s^(-1)`. Determine (a) the kinetic energy of the lithium `(M_(Li) =7.0160 am u)` and (b) the `Q` value of the reaction.

Answer

(a)`1.02 MeV` , (b)`2.82 MeV`

Explanation

Since the neutron and boron are both initially at rest, the total momentum before the reaction is zero and afterward also it is zero. Therefore, `M_(Li) n_(Li) =M_(He) v_(He)`.
We solve this for `v_(Li)` and substitute it into the equation for kinetic energy. We can use classical kinetic energy with little error, rather than relativistoc formulas, because `n_(He) =9.30 xx 10^(6) m s^(-1)` is not close to the speed of light `c` and `n_(Li)` will be even less since `M_(Li) gt M_(He)`. Thus, we can write
`K_(Li)=(1)/(2)M_(Li)v_(Li)^(2)=(1)/(2)M_(Li)((M_(He)v_(He))/(M_(Li)))^(2)=(M_(He)^(2)v_(He)^(2))/(2M_(Li))`
We put in numbers, changing the mass in u to kg and recalling that `1.60 xx10^(-13) J=1MeV`
`K_Li =((4.0026)^(2)(1.66 xx 10^(-27))(9.30 xx 10^(6))^(2))/(2(7.0160)(1.66 xx 10^(-27)))`
`=1.64 xx 10^(-13) J=1.02 MeV`
(b) We are given the data `K_(alpha) =K_(X) =0`, so `Q =K_(Li) +K_(He)`, where
`K_(He) =(1)/(2)M_(He) V_(He^(2))`
` =(1)/(2) (4.0026)(1.66 xx 10^(27))(9.30 xx 10^(6))^(2)`
`=2.87 xx 10^(-13)J =1.80 MeV`
Hence, `Q=1.02 MeV +1.80 MeV =2.82 MeV`