Question

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. Calcutate the energy carried by each `alpha` particle.

Answer 1

Correct Answer - `1.8225xx10^(-12)J`
The nuclear reaction involved is
`._(3)Li^(6)+._(1)H^(2)to._(2)He^(4)+._(2)He^(4)+Q`.
Total initial mass `=6.0155+2.0141`
`=8.0296a.m.u.`
Total final mass `=2xx4.0026=8.0052 a.m.u.`
Mass defect, `Deltam=8.0296-8.0052`
`=0.0244 a.m.u.`
Energy released `=(Deltam)c^(2)`
`=[0.0244xx1.66xx10^(-27)](3xx10^(8))^(2)J`
`=3.645xx10^(-12)J`
`:.` Energy carried by each `alpha` particle
`=3.645/2xx10^(-12)J=1.8225xx10^(-12)J`