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,,
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.
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For `._(20)^(40)Ca, A= 40, Z= 20, m_(P)=1.007825 u`
`m_(n)=.008665 u , M = 39.962589 u`
(i) `Delta M = [Zm_(P)+(A-Z)m_(n)-M]`
`= [(20) (1.007825) + (40-20) (1.008665) - 39.962589]`
`= [(20xx1.007825)+(20xx1.008665)-39.962589]`
`= [40.3298-39.962589]=0.3672 u`
(ii) `BE= Delta Mc^(2)=0.3672xx931 MeV`
`= 341.86 MeV`
(iii) B.E per nucleon `= (B.E.)/(A)=(341.86)/(40)`
`= 8.547 MeV`
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