The binding energy per nucleon of ²⁰⁹₈₃Bi is _______ MeV. [ Take m(²⁰⁹₈₃Bi) = 208.980388 , u, , m_p = 1.007825 , u, , m_n = 1.008665 , u, , 1 , u = 931 , MeV /c^2]

Step-by-step solution

For the nucleus ²⁰⁹₈₃Bi , the number of protons is Z = 83 and the number of neutrons is N = 209 - 83 = 126 . Mass of 83 protons = 83 1.007825 = 83.649475 , u Mass of 126 neutrons = 126 1.008665 = 127.091790 , u Total mass of the nucleons = 83.649475 + 127.091790 = 210.741265 , u The mass defect m is given by: m = Total mass of nucleons - Mass of nucleus m = 210.741265 - 208.980388 = 1.760877 , u The total binding energy ( BE ) is: BE = m 931 , MeV BE = 1.760877 931 = 1639.376 , MeV The binding energy per nucleon is