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.

Taking the value of atomic masses from the tables, calculate the energ

1.	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.
Answer» 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"`

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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.

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