Question 15.3: Using the semiempirical formulas, calculate the binding ener...
Using the semiempirical formulas, calculate the binding energy per nucleon and mass of the isotopes ^{56}_{ 26}F e and ^{ 208}_{ 82} P b.
For ^{56}_{ 26}F e, A = 56, Z = 26, and N = 30. Substituting these values into Eq. (15.6) with the values of the parameters given by Eq. (15.7), we obtain the binding energy B(N , Z) = 487.1696;MeV. The binding energy per nucleon is B(N , Z)/56 = 8.699 MeV.
B(N , Z) = \left[ Zm(^{1}_{ 1}H ) + Nm_{n} − m(N , Z)\right] c^{2} , (15.5)
B(N , Z) = aA − bA^{2/3} −\frac{ dZ^{2}}{ A^{1/3}} − s\frac{ (N − Z)^{2}} {A} −\frac { δ}{ A^{1/2} }, (15.6)
a = 15.835 MeV
b = 18.33 MeV
d = 0.714 MeV
s = 23.20 MeV (15.7)
The mass of ^{56}_{ 26}F e can then be calculated using Eq. (15.5). We obtain
m(N , Z) = Zm(^{1}_{ 1}H ) + Nm_{n} − B(N , Z)/c^{2} = 26 × 1.007825 + 30 × 1.008665 −\frac{487.1696 MeV}{931.5 MeV/u} 55.9404 u. (15.10)
This value agrees fairly well with the experimental value, 55.9349 MeV.
In exactly the same way, the binding energy of ^{208}_{ 82} P b is found to be 1624.7481 MeV or 7.811 MeV per nucleon. The mass of the ^{208}_{ 82} P b isotope obtained by using the calculated value of the binding energy together with Eq. (15.5) is 207.9892123, which agrees well with the experimental value of 207.976627. Notice that the binding energy per nucleon we have obtained for ^{208}_{ 82} P b is less than the binding energy per nucleon of ^{56}_{ 26}F e. The isotope of iron with A = 56 is the most stable isotope.