Question

Calculate the binding energy per nucleon (in J) for ²H and ³H. The atomic masses are 2.014102 u for ²H, and 3.016049 u for ³H. (Enter unrounded values. Assume that the mass of ¹₁H = 1.007825 u,m_p = 1.007275 u, m_n = 1.008666 u, and m_e = 0.000549 u, respectively.)

Answer 1

Atomic mass of 2H = 2.017102 u

Atomic mass of 3H = 3.016049u

Mass of proton (mp) =1.007275 u

Mass of neutron (mn) =1.008666u

Mass of electron (me) =0.000549u

Binding energy per nucleon =?

Calculated mass of 2H = mass of 1proton +mass of 1 electron +mass of 1 neutron

=1.007275 u +0.000549u +1.008666u = 2.01649u

Actual mass of 2H =2.014102 u

Mass defect (delta M) = calculated mass - actual mass

=2.01649u - 2.014102u =0.002388u

For a mass defect of 1amu energy released =931.5Mev

For a mass defect of 0.002388u, energy released =931.5MeV× 0.002388=2.224422 MeV

Thus binding of the nuclide =2.224422MeV

Binding energy per nucleon = 2.224422MeV /2=1.112211MeV

1Mev = 1.602×10^-13joules

1.112211MeV =1.112211×1.602×10^-13= 1.7816×10^-13 Joule

Binding energy of 2H =1.7816×10^-13joule

For 3H calculations are as follow:

Calculated mass of 3H = mass of 1proton +mass of 1electron +mass of 2neutrons

=1.007275 u+0.000549u +2×1.008666= 3.025156

Actual mass of 3H = 3.016049u

Mass defect = calculated mass - Actual mass

= 3.025156 u- 3.016049u = 0.009107u

For a mass defect of 1u energy released =931.5MeV

For a mass defect of 0.009107u, energy released =0.009107×931.5MeV =8.4831705MeV

Binding energy of nuclide = 8.4831705 MeV

Binding energy per nucleon = 8.4831705MeV/3 =2.8277235MeV

1MeV = 1.602×10^{-13 joule}

^{​​​​​​}2.8277235MeV = 2.8277235×1.602×10^-13 = =4.53×10^-13joule

Thus binding energy of 3H =4.53×10^-13 joule

For calculating binding energy per nucleon , divide the binding energy with number of nucleon (sum of neutron and proton which is equal to mass number) for 2H nucleon =2and for 3H nucleon =3