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.

The actual atomic mass of `._(20)Ca^(40)` is 39.96259 amu. Find the bi

1.	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» 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`

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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.

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