Question

Calculate the binding energy (kJ/nucleon) for a ⁶³Cu nucleus. ¹H = 1.00785 g/mol, ¹n = 1.008665 g/mol, ⁶³Cu = 62.93960 g/mol

Question 6 options:

A)

8.7

Answer 1

Mass defect , ?m = ( ZM_H + (A-Z) M _n ) - M

Where

Z = atomic number = 29

(A-Z) = No. of neutrons = 63-29 = 34

M_H = mass of Hydrogen atom = 1.00785 amu

M _n = mass of Neutron = 1.008665 amu

M = real mass = 62.93960 amu

Plug the values we get ?m = [( 29x1.00785)+ (34x1.008665 )] - 62.93960

                                          = 0.58266 amu

So nuclear binding energy , BE = ?m x 931.5 MeV

                                               = 0.58266 x931.5 MeV

                                               = 542.75 MeV

                                               = 542.75 x 10⁹ eV

                                               = 542.75 x 10⁹ x 1.6x10^-19 J

                                               = 8.684 x 10^-8 J

Binding energy per nucleon = BE / atomic mass number

                                     = 8.684 x 10^-8 / 63 J / necleon

                                      = 1.4 x 10 ^-9 J/nucleon

                                         = 1.4x10^-12 kJ/nucleon

Therefore option (c) is correct.