A plot of binding energy per nucleon Eb A versus the mass number (A) shows that nuclei with a small mass number have a small binding energy per nucleon, as the mass number increases the binding energy per nucleon increases, and the value for the binding energy per nucleon has a maximum value for nuclei with a mass number around 60. Verify that this is the case by determining the binding energy per nucleon for each of the following four nuclei. (Let the mass of a proton be 1.0078 u, the mass of a neutron be 1.0087 u, the mass of 6He be 6.0189 u, the mass of 8Li be 8.0225 u, the mass of 62Ni be 61.9283 u, and the mass of 115In be 114.9039 u. Enter your answers in MeV and to at least three significant figures.) (a) 6He MeV (b) 8Li MeV (c) 62Ni MeV (d) 115In MeV

6He = 4.90 MeV/Nucleon

8Li = 5.18 MeV / Nucleon

62Ni = 8.82 MeV / Nucleon

115 In = 8.54 MeV / Nucleon

Explanation:

Our strategy here is to remember that when the mass of a given nuclei is calculated from the sum of the mass of its protons and neutrons, this mass is greater than the actual value. This is the mass defect.

Now this mass defect we can convert to energy by utilizing Einstein's equation, E = mc². This is the binding energy.

For 6He with actual mass 6.0189 u (He has Z = 2, that is 2 protons)

mass protons = 2 x 1.0078 u = 2.0516 u

mass neutrons = 4 x 1.0087 u = 4.0348 u

predicted mass = (2.0516 + 4.0348) u = 6.0504 u

mass defect = (6.0504 - 6.0189) u = 0.0315 u

Now we need to convert this mass expressed in atomic mass units to kilograms (1 u = 1.66054 x 10⁻²⁷ Kg)

0.0315 u x 1.66054 x 10⁻²⁷ Kg = 5.231 x 10⁻²⁹ Kg

E = 5.231x 10⁻²⁸ Kg x (3 x 10⁸ m/s) ² = 4.707 x 10⁻¹²J

Finally we will convert this energy in Joules to eV

E = 4.707 x 10⁻¹² J x 6.242 x 10¹⁸ eV/J = 2.94 x 10⁷ eV = 29.4 MeV

E per nucleon for 6He = 29.4 MeV / 6 = 4.90 MeV / Nucleon

Now the calculations for the rest of the nuclei are performed in similar manner with the following results:

8Li = 5.18 MeV / Nucleon

62Ni = 8.82 MeV / Nucleon

115 In = 8.54 MeV / Nucleon