Question

What is the nuclear binding energy per nucleon of potassium-40? Particle Mass (amu) K-40 39.9632591 Neutron 1.008701 Proton 1.007316 Electron 0.000549 (1 kg = 6.022 × 1026 amu; NA = 6.022 × 1023 mol–1; c = 2.99792458 × 108 m/s) 1.33 × 10–12 J/nucleon 5.33 × 10–11 J/nucleon 5.64 × 10–11 J/nucleon 1.41 × 10–12 J/nucleon 2.97 × 10–12 J/nucleon

Answer 1

Total number of nucleons in K-40 = 40

Potassium has 19 protons in its nucleus.

Number of neutrons in K-40 = (40 - 19) = 21

Mass of 1 proton = 1.007316 amu

Therefore, the mass of 19 protons = 19 x 1.007316 = 19.139004 amu

Mass of 1 neutron = 1.008701 amu

Therefore, the mass of 21 neutrons = 21 x 1.008701 = 21.182721 amu

Theoretical combined mass of nucleons = 19.139004 + 21.182721 = 40.321725 amu

Experimental mass of K-40 = 39.9632591 amu

Mass defect = Theoretical mass - Experimental mass

                    = (40.321725 - 39.9632591)

                    = 0.3584659 amu

Now, 1 amu = 1/(6.022 x 10²³) = 1.66 x 10^-24 g = 1.66 x 10^-27 kg

Therefore, mass defect (m) = 0.3584659 x 1.66 x 10^-27 kg

This mass defect is what manifested as the binding energy of the nucleus. Mass and energy are related with the Einstein's mass and energy equivalence relation.

Binding energy = mc²                   (where c= speed of light in vacuum)

                         = 0.3584659 x 1.66 x 10^-27 x (3 x 10⁸)²

                         = 5.355480546 x 10^-11 J

Therefore, the binding energy per nucleon = (5.355480546 x 10^-11)/40

                                                                     = 1.33 x 10^-12 J

Therefore, the binding energy = 1.33 x 10^-12 J/nucleon