Question

Enter your answer in scientific notation. Be sure to answer all parts. Calculate the nuclear binding energy (in J) and the binding energy per nucleon for the following isotope: Nuclear binding energy = [ ] 10 J Binding energy per nucleon = [ ] x10 J/nucleon

Answer 1

For ₈₃Bi⁰⁹

no of protons in ₈₃Bi²⁰⁹ = 83

no. of neutrons = 209- 83 = 126

mass of proton = 1.00728 amu

mass of 83 protons = 83 x 1.00728 = 83.60424 amu

mass of neutron = 1.00867 amu

mass of 126 neutrons = 126 x 1.00867 = 127.09242 amu

total mass of ₈₃Bi²⁰⁹ = 83 protons mass + 126 neutrons mass

                                = 83.60424 amu + 127.09242 amu

                                 = 210.69666 amu

Actual mass of ₈₃Bi²⁰⁹ = 208.9804 amu.

mass defect = calculated mass - actual mass

                    = 210.69666 amu - 208.9804 amu.

                    = 1.71626 amu

Mass defect = 1.71626 amu

Nuclear Binding Energy = mass defect in amu x 931.5 MeV

                      = 1.71626 amu x 931.5 MeV

                           = 1598.6962 MeV

1 eV = 1.6 x 10^-19 Joules

1598.6962 MeV = (1598.6962 x 10^6) x 1.6 x 10^-19

                          = 2.56 x 10^-10 J

Nuclear Binding Energy = 2.56 x 10^-10 J

Nuclear binding energy per nucleon = Binding Energy/ A                   ( A = mass number = protons + neutrons = 209)

                                                        = 1598.6962 / 209

                                                         = 7.6492 MeV

                                                         = 7.6492 x 10^6 x 1.6 x 10^-19 J

                                                         = 1.22 x 10^-12 J / nucleon

Nuclear binding energy per nucleon = 1.22 x 10^-12 J / nucleon