calculating the binding energy per nucleon for boron, berylium and lithium

FOR LITHIUM -6, the binding energy calculated using

lithium six has 3 protons and 3 neutron

thus actual mass is given by

 3(1.007825) + 3(1.008665)=6.04947

the measured mass is given by 6.015122

mass defects will be 0.034348amu

and the binding energy 3.09132 x 10⁹ MeV

binding energy per nucleon for  lithium can be calculated as follows ,lithium has 6 nucleons , thus3.09132 x 10⁹  /6  = 5.15220X 10⁸ MeV.

FOR beryllium , it has p nucleons , 4 protons and  neutrons

 the actual mass is given by

4(1.007825) + 5(1.008665)=9.07463amu

 the measured  mass is 9.012182

mass defect is equal to 0.0624430amu

 the binding energy is given by

0.0624430 x (3 X 10⁸ )²  m/s

5.61987 x 10⁹ MeV

 To find the binding energy per nucleon , beryllium has 9 nucleons , thus 5.61987 x 10⁹ MeV/9 = 9.36645 x 10⁸  MeV

NEXT IS BORON

 boron has  two isotopes boron 10 and boron 11,

 the binding energy for boron-10 can be calculated as follow.

boron-10 has 5 protons and 5 neutrons

5(1.007825) + 5(1.008665)= 10.0825

the actual mass is given by 10.012937

 thus mass defect is 0.0695130

binding energy for boron  0.0695130x (3 X 10⁸ )² m/s

6.25617x 10⁹ MeV

6.25617x 10⁸ MeV per nucleon