Nuclear Binding Energy

Using the semi-empirical binding energy formula with the pairing term in the lectures, calculate the total binding energy, atomic mass, neutron and proton separation energies for 183Ta. (∆p=7.2889 MeV, ∆n=8.0713 MeV)

183Ta means A= 183 (odd), Z =73 (odd) – protons, N= 110 (even) -neutrons
The total binding energy is
$B(A, Z) =15.36*A -16.32*A^{(2/3)} -90.45*(A/2 -Z)^2/A -0.6928*Z^2/A^{(1/3)}$
$B(183,73) =2810.88 -526.042 -169.161 -650.284 = 1465.393 Mev$

This binding energy is calculated without the term E(odd-odd) (Z odd, A odd) and corresponds to a nuclei even-odd.
$E(odd-odd) = -11.18/A^{(1/2)} =0.826 MeV$
Hence the neutron separation energy (number of neutrons is even) is
$\Delta(n) = B(183,73)/A =1465.393/183 =8.0076 Mev$
and the proton separation energy is (number of protons is odd)
$\Delta(p) = B(183,73)/A – E(odd-odd) =8.0076 -0.826 =7.18 Mev$
The corresponding mass correction to the binding energy is
$\Delta(m) = B(A,Z)/c^2 =1465.393*10^6*1.6*10^(-19)/(3*10^8)^2 =$
$= 2.60*10^{-27} Kg =2.60/1.66 = 1.566 amu$ (atomic mass units)
Hence the atomic mass of the nuclei is simple
Atomic mass =$ Z*Mp +N*Mn – \Delta(m) =73*1+110*1 -1.566 =183-1.566 = 181.43 amu$
Now having the atomic mass one can compute again the neutron and proton separation
$\Delta(n) = B(A,Z)/atomic mass =1465.393/181.43 =8.076 MeV$
$\Delta(p) = B(A,Z)/atomic mass – E(odd-odd) =8.076-0.826 = 7.25 MeV$


valentin68

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