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The nuclear mass of 141Ba is 140.883 amu. Calculate the binding energy per nucleon for 141Ba.

• Binding energy per nucleon…..

The binding energy is the energy released if a nucleus was converted to its component particles. It is equal to the energy equivalent of the “missing mass”, the difference in mass between the mass of the particles and the mass of the isotope. E = mc² …. where E is in J, when m is in kg and c is in m/s.

Atomic number = 56

Mass number = 141

Mass of isotope = 140.91441 amu

56 x 1.0073 + 85 x 1.0087 = 142.1483 amu

“missing mass” = 142.1483 – 140.9144 = 1.2339 amu

Binding energy per nucleon = 1.2339 amu x (1.66054×10^-27kg / 1 amu) x (2.99792×10^8 m/s)² / 141

Binding energy per nucleon = 1.3060×10^-12 J

Binding energy per nucleon = 1.3060×10^-12 J x (6.242×10^+12 Mev / 1J) = 8.1522 MeV

• 141Ba has 141 nucleons. Barium is element number 56. This means the nucleus contains 56 protons.

141 –56 = 85 neutrons

According to a table in my physics book, the mass of one proton is 1.007276 amu. The mass of one neutron is 1.008665 amu.

Total mass of protons = 56 * 1.007276 = 56.407456 amu

Total mass of neutrons = 85 * 1.008665 = 85.736525 amu

Total mass = 56.407456 + 85.736525 = 142.143981 amu

To determine the total binding energy, subtract the mass of 141 Ba from this mass.

Q = 142.143981 – 140.833 = 1.310981 amu

To determine the binding energy per nucleon, divide by 141

Q/n = 1.310981 ÷ 141

This is approximately 0.0093 amu per nucleon.

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