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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 = 1.007825 u, mass of neutron = 1.008665 u and 1 u is equivalent to 931 Mev.

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Correct option is D

Mcl=34.98000 amu ∆m=ZMp+A-Zmn-Mcl= [17(1.007825)+18(1.008665)]–34.98= [17.133025 + 18.15597]–34.98∆m = 0.308995 amu BE=∆m x 931Mev B.E= 0.308995 x 931 = 287.6743MeVB.EA=287.674335=8.2 MeV

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Calculate the binding energy per nucleon of Cl 1735 nucleus. Given that mass of Cl 1735 nucleus = 34.98000 u, mass of proton = 1.007825 u, mass of neutron = 1.008665 u and 1 u is equivalent to 931 Mev.

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Ready to Test Your Skills?

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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 = 1.007825 u, mass of neutron = 1.008665 u and 1 u is equivalent to 931 Mev.