Calculate the binding energy per nucleon of 1735Cl nucleus. Given that mass of 1735Cl nucleus=34.98000 u , mass of proton=1.007825 u , mass of neutron =1.008665 u and 1u is equivalent to931 Mev .

Given,Z=17,A=35mp=1.007825, mn=1.008665M=34.98000Δm=[zmp+(A−Z)mn−M]         =17×1.007825+18×1.008665−34.98000         =17.133025+18.15597−34.98        =35.288995−34.98=0.3089954   BE=Δm×931Mev          =0.308995×931=287.67MevBE/nucleon=287.6735≃8.2Mev