Taking the value of atomic masses from the tables, calculate the energy per nucleon which is liberated in the nuclear reaction `Li^(6)+H^(2)rarr2He^(4)`. Compare the obtained magnitude with the energy per nucleon liberated in the fission of `U^(235)` nucleus.

Explanation

The energy realesed in the reaction
`Li^(6)+H^(2)rarr2He^(4)`
is `Delta_(Li^(6))+Delta_(H^(2))-2Delta_(He^(4))`
`=0.01513+0.01410-2xx0.00260am u`
`=0.02403 am u= 22.37MeV`
(This resut for change in `B.E`. Is correct because the contribution of `Delta_(n) & Delta_(H)` cancels out by conservation law for protons & neutrons).
Energy per nucleon is then
`(22.37)/(8)= 2.796MeV//n ucl eon`.
This should be compared with the value `(200)/(235)=0.85MeV//"nucleon"`