When a deuteron of mass 2.0141 a.m.u and negligible K.E. is absorbed by a Lithium `(._(3)Li^(6))` nucleus of mass 6.0155 a.m.u. the compound nucleus disintegration spontaneously into two alpha particles, each of mass 4.0026 a.m.u. Calculate the energy carried by each `alpha` particle.

Explanation

`m(""_(3)Li^(6)) = 6.0155` amu
`m(""_(1)H^(2)) = 2.0141` amu
`m(""_(2)He^(4)) = 4.0026` amu

The given reaction is
`""_(3)Li^(6) + ""_(1)H^(2) to ""_(2)He^(4) + ""_(2)He^(4) + Q`
Total initial mass of reactants `(""_(3)Li^(6) "and " ""_(1)H^(2))`
`=(6.0155 + 2.0141)` amu
=8.0296 amu
Total final mass of products (two `""_(2)He^(4))`
Mass defect, `Deltam = 8.0296 - 8.0052 = 0.0244` amu
Energy released in the disintegration is
`Q = Deltam xx c^(2)`
`= 0.0244 xx 166 xx 10^(-27) xx (3 xx 10^(8))^(2)`
`= 3.645 xx 10^(-12)` J
`therefore` Energy of each `alpha`-particle `=Q/2 = (3.645 xx 10^(-12) J)/2`
`=1.8 225 xx 10^(-19)` J