Question

The actual atomic mass of `._(20)Ca^(40)` is 39.96259 amu. Find the binding energy for this nuclide, using 1.008665 amu for the mass of a neutron and 1.007825 amu for the mass of atomic hydrogen. Also calculate the binding energy per nucleon.

Answer 1

Mass of `._(20)Ca^(40)` atom
= Mass of 20 atoms of hydrogen + Mass of 20 neutrons
`= 20 xx 1.007825 + 20 xx 1.008665 = 40.32980` amu
Acutal mass of `._(20)Ca^(40)` atom = 39.96259 amu (given)
`:.` Mass defect `(Delta M) = 40.32980 - 39.96259`
= 0.36259 amu
For a mass defect of 0.36721 amu, energy released
`= 931.5 xx 0.36721 MeV = 342.06 MeV`