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Calculate the binding energy per nucleon...

Calculate the binding energy per nucleon of `._(17)^(35)Cl` nucleus. Given that mass of `._(17)^(35)Cl` nucleus = 34.98000 u, mass of proton = 1.007825u, mass of neutron = 1.008665u and 1 is equivalent to 931 MeV.

Text Solution

Verified by Experts

For `._(17)^(36)Cl, A=35, Z = 17`,
`m_(P)=1.007825 u`
`m_(n)=1.008665 u, M = 34.98 u`
(i) `Delta M= [Zm_(P)+(A-Z)m_(n)-M]`
`= [17xx1.007825+(35-17)(1.008665)-34.98]`
`= [ 17.13303+18.15597-34.98]`
`Delta M = [ 35.289-34.98] = 0.3089 u`
(ii) `BE = Delta Mc^(2)`
`= 0.3089xx931 MeV = 287.5859 MeV`
`therefore` BE per nucleon
`= (B.E)/(A)=(287.5859)/(35)=8.21 MeV`
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