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

Answer 1

`._(3)^(6)Ll+._(1)^(2)Drarr._(2)^(4)He+._(2)^(4)He+Q`
`m(._(3)^(6)Li)=6.0155u`
`m(._(1)^(2)H)=2.0141 u`
Total initial mass `=8.0296 u`
TOtal final mass `=2m(._(2)^(4)He)=2xx4.0026`
`=8.0052u`
Mass defect, `Delta m=8.296-8.0052`
`=0.0244u=0.024xx1.66xx10^(-27)kg`
Energy released,
`Q-Deltamxxc^(2)=0.244xx1.66xx10^(27)xx(3xx10^(8))^(2)`
`=3.645xx10^(-12)J`
Energy of each `alpha`-particle=`1.8225xx10^(-12)J`