Question

Calculate the binding energy per nucleon (in J) for 3He and 4He. The atomic masses are 3.016029 u for 3He, and 4.002603 u for 4He. (Enter unrounded values. Assume that the mass of 11H = 1.007825 u, mp = 1.007275 u, mn = 1.008666 u, and me = 0.000549 u, respectively.)

Answer 1

3He: 3He has 2 protons, 2 electrons and 3-2 = 1 neutron.

Mass of all constituent particles = 2*mp + 1*mn + 2*me = 2*1.007275 u + 1.008666 u + 2*0.000549 u = 33.024314 u

Given mass of 3He = 3.016029 u

Mass defect, $\Delta$m = (Mass of all constituent particles) - (mass of 3He)

=> $\Delta$m = 3.024314 u - 3.016029 u = 0.008285 u

Total binding energy =  $\Delta$m * 931.5 MeV / u = 0.008285 u * 931.5 MeV / u

=> Total binding energy = 7.7174775 MeV

Total number of nucleons (p+n) in 3He = 2+1 = 3

=> Binding energy per nucleon = 7.7174775 MeV / 3 = 2.5724925 MeV

=> Binding energy per nucleon (in J) = 2.5724925 MeV *(1.60218*10^-13 J / 1MeV)

=> Binding energy per nucleon (in J) = 4.121587*10^-13 J (Answer)

4He: 4He has 2 protons, 2 electrons and 4-2 = 2 neutrons.

Mass of all constituent particles = 2*mp + 2*mn + 2*me = 2*1.007275 u + 2*1.008666 u + 2*0.000549 u = 4.03298 u

Given mass of 4He = 4.002603 u

Mass defect, $\Delta$m = (Mass of all constituent particles) - (mass of 4He)

=> $\Delta$m = 4.03298 u - 4.002603 u = 0.030377 u

Total binding energy =  $\Delta$m * 931.5 MeV / u = 0.030377 u * 931.5 MeV / u

=> Total binding energy = 28.2961755 MeV

Total number of nucleons (p+n) in 4He = 4

=> Binding energy per nucleon = 28.2961755 MeV / 4= 7.074043875 MeV

=> Binding energy per nucleon (in J) = 7.074043875 MeV *(1.60218*10^-13 J / 1MeV)

=> Binding energy per nucleon (in J) = 1.133387*10^-12 J (Answer)