calculate binding energy per nucleon following isotopes of barium

barium has 4 isotopes, ¹³²Ba, ¹³⁴Ba, ¹³⁵Ba and ¹³⁶Ba, we will start with

for barium-130  the actual mass is given as follows

 barium has 130 number  of nucleons, 56 protons and 74 neutrons,

 the actual mass 56 x 1.007825 + 74 x 1.008665 =131.1 amu.

the actual mass is given by 129.906310

the mass defect is given as 1.2 amu

1.2 x (3 x 10⁸)² = 10.807182 x 10¹⁶Joules

and the binding energy per nucleon is given by

8.1257 x 10¹⁴ Joules

¹³²Ba,  Barium 132 has 56 protons and 76 neutrons

56 x 1.007825 + 76 x 1.008665 = 133.1

the measured mass is given as 131.905056

the mass defect is therefore  1.2 amu

1.2 x (3 x 10⁸)² = 10.807182 x 10¹⁶ joules

and the binding energy per nucleon is given by

8.1257 x 10¹⁴ Joules

for ¹³⁴Ba, the actual mass is given as follows

56 x 1.007825 + 78 x 1.008665 = 135.1

the actual mass is given as follows 133.904503

mass defect is therefore 1.2 amu

therefore we compute the binding energy as follows  1.2 x (3 x 10⁸)² = 10.807182 x 10¹⁶ joules

and the binding energy per nucleon is given by

8.1257 x 10¹⁴ Joules

 it seems for all isotopes of barium they tend to have same binding energy per nucleon,

 according to sir Albert Einstein, the mass defect is not lost but converted to force that hold the constituent protons and neutrons  to form the atomic nuclei.