bei48482_FM

Nuclear Structure 407

The last correction term arises from the tendency of proton pairs and neutron pairs

to occur (Sec. 11.3). Even-even nuclei are the most stable and hence have higher bind-

ing energies than would otherwise be expected. Thus such nuclei as^42 He,^126 C, and^168 O

appear as peaks on the empirical curve of binding energy per nucleon. At the other

extreme, odd-odd nuclei have both unpaired protons and neutrons and have relatively

low binding energies. The pairing energyEpis positive for even-even nuclei, 0 for

odd-even and even-odd nuclei, and negative for odd-odd nuclei, and seems to vary

with Aas A^3 ^4. Hence

Pairing energy Ep(, 0) (11.17)

The final expression for the binding energy of a nucleus of atomic number Zand

mass number A, which was first obtained by C. F. von Weizsäcker in 1935, is

Eba 1 Aa 2 A^2 ^3 a 3

 a 4 (, 0) (11.18)

A set of coefficients that gives a good fit with the data is as follows:

a 1 14.1 MeV a 2 13.0 MeV a 3 0.595 MeV

a 4 19.0 MeV a 5 33.5 MeV

Other sets of coefficients have also been proposed. Equation (11.18) agrees better

with observed binding energies than does Eq. (11.13), which suggests that the

liquid-drop model, though a good approximation, is not the last word on the

subject.

Example 11.6

The atomic mass of the zinc isotope^6430 Zn is 63.929 u. Compare its binding energy with the

prediction of Eq. (11.18).

Solution

The binding energy of^6430 Zn is, from Eq. (11.7),

Eb[(30)(1.007825 u)(34)(1.008665 u)63.929 u](931.49 MeV/u)559.1 MeV

The semiempirical binding energy formula, using the coefficients in the text, gives

Eb(14.1 MeV)(64)(13.0 MeV)(64)^2 ^3 

561.7 MeV

The plus sign is used for the last term because^6430 Zn is an even-even nucleus. The difference

between the observed and calculated binding energies is less than 0.5 percent.

33.5 MeV



(64)^3 ^4

(19.0 MeV)(16)



64

(0.595 MeV)(30)(29)



(64)^1 ^3

a 5



A^3 ^4

(A 2 Z)^2



A

Z(Z 1 )



A^1 ^3

a 5



A^3 ^4

Semiempirical

binding-energy

formula

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