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Draw the graph to show variation of binding energy per nucleon with mass number of different atomic nuclei. Calculate binding energy per nucleon of " "_(20)^(40)Ca nucleus. Given : mass of " "_(20)^(40)Ca = 39.962589 u, mass of proton=1.007825 u, mass of neutron=1.008665 u and 1 u=931 MeV/c^(2). |
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Answer» Solution : Binding ENERGY of `" "_(20)^(40)Ca` nucleus `E_(b) =[20m_(p) + 20m_(n)-m (" "_(20)^(40)Ca)] xc^(2)= [20 XX 1.007825 + 20 xx 1.008665 - 39.962589] u xx c^(2)= 0.367211 u xx c^(2) = 0.367211 xx 931 MeV = 341.8 MeV`. `THEREFORE` Binding energy per nucleon `E_(bn) =341.8/40= 8.545 MeV` /nucleon. |
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