1.

The atomic masses of Li, He and proton are 7.01823 amu, 4.00387 amu and 1.00715 amu respectively. Calculate the energy evolved in the reaction, 3^(Li^(7)) + 1^(HA^(1)) rarr 2 2^(He^(4)) + triangle E Given 1 amu = 931 MeV.

Answer»

Solution :Mass of reactants = mass of Li + mass of H
=7.01823 + 1.00715
=8.02358 amu
mass LOSS during change=(8.02538 -8.00774) amu
=0.01764 amu
`therefore` ENERGY evolvedduringreaction
=0.0176 X 931 MeV
=16.423 MeV


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