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Calculate the binding energy per nucleon...

Calculate the binding energy per nucleon of `._(20)^(40)Ca`. Given that mass of `._(20)^(40)Ca` nucleus = 39.962589 u, mass of a proton = 1.007825 u,, mass of Neutron = 1.008665 u and 1 u is equivalent to 931 MeV.

Text Solution

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`A =40, Z = 20 , A-Z=20`
`Delta m = { Z m _(p ) + (A-Z) m_(n) }-M_(n)`
`= {(20x 1.007825 + (20x1. 008665)} - 39.962589`
`= 40. 329800- 39.962589`
`Deltam = 0/367211`
Binding energy per nucleon `= (Delta m xx 931)/(A)`
` = (0.367211 xx 931)/(40) = 8.547 MeV.`
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