Question

Mass of 17^35 CI nucleus is 34.9800 u. Taking the mass of proton as 1.00783 u and that of neutron as 1.00866 u, find the binding energy per nucleon for 17^35 CI nucleus.[Ans: 8.219 Mev/nucleon]

Answer 1

Dear Student,

◆ Answer -

BE/nucleon = 8.2238 MeV

◆ Explanation -

Mass defect of Cl is calculated by -

∆m = (mp.np + mn.nn) - M

∆m = (17×1.00783 + 18×1.00866) - 34.98

∆m = 0.309 u

Binding energy per nucleon is calculated as -

BE/nucleon = 931.5 ∆m/(nm+nn)

BE/nucleon = 931.5 × 0.309 / (17 + 18)

BE/nucleon = 8.2238 MeV

Hope this helps you...