Question

Determine the binding energy (in MeV) per nucleon for the nuclides:

a) O-16/8

b) O-17/8

c) Fe-56/26

d) U-235/92

Answer 1

Solution :-

1). O- 16/8

Number of proton = 8

Number of neutron = 8

Mass defect = avg atomic mass - sum of mass of nulcie

Mass of nuclie = [8(1.00728) amu] + [8(0.0005486) amu] + [8(1.00867) amu]

                           = 16.13199 amu

Average atomic mass of oxygen is 15.999 amu

Mass defect = 16.13199 – 15.999 amu

                        = 0.13299 amu

BE = 0.13299 amu * 931.5 MeV / 1 amu = 123.8 MeV

So the binding energy for the oxygen is 123.8 MeV

2) O- 17/8

Number of proton = 8

Number of neutron = 9

Mass defect = avg atomic mass - sum of mass of nulcie

Mass of nuclie = [8(1.00728) amu] + [8(0.0005486) amu] + [9(1.00867) amu]

                           = 17.14066 amu

Average atomic mass of oxygen is 15.999 amu

Mass defect = 17.14066 amu – 15.999 amu

                        = 1.141659 amu

BE = 1.141659 amu * 931.5 MeV / 1 amu = 1063 MeV

So the binding energy for the oxygen is 1063 MeV

3) Fe – 56/26

Number of proton = 26

Number of neutron = 30

Mass defect = avg atomic mass - sum of mass of nulcie

Mass of nuclie = [26(1.00728) amu] + [26(0.0005486) amu] + [30(1.00867) amu]

                           = 56.46364 amu

Average atomic mass of Fe is 55.845 amu

Mass defect = 56.46364 – 55.845 amu

                        = 0.618644 amu

BE = 0.618644 amu * 931.5 MeV / 1 amu = 576 MeV

So the binding energy for the iron is 576 MeV

4) U – 235 / 92

Number of proton = 92

Number of neutron = 143

Mass defect = avg atomic mass - sum of mass of nulcie

Mass of nuclie = [92(1.00728) amu] + [92(0.0005486) amu] + [143(1.00867) amu]

                           = 236.96 amu

Average atomic mass of U is 235.043924 amu

Mass defect = 236.96 amu - 235.043924 amu

                        = 1.916117 amu

BE = 1.916117 amu * 931.5 MeV / 1 amu = 1785 MeV

So the binding energy for the U is 1785 MeV