Liquid Drop Model: Nuclear Stability (Contd.)
| Now, the above two analysis clearly show that the B.E./A, which is supposed to be constant (as per equation m1.14) decreases for lower and higher A values because of surface effect and Coulomb repulsion effect respectively. However, nuclear physicists have observed that the change in B.E./A value for higher A is slightly more than that expected from Coulomb repulsion effect alone. The liquid drop model does not have any explanation for this. An explanation based on single particle excitation model1 , where nucleons are considered to occupy only certain energetic orbital (you will understand it better after reading lecture 7) has been adopted. It has been realized that as A increases an asymmetry in the nucleon number develops, i.e. number of neutrons dominate over proton numbers. What I mean is that a symmetry distribution Z = N = A/2 proves to be the energetically most favoured and any other repartition, N = (A/2) + a, Z = (A/2) – a will involve lifting particles from occupied into empty orbitals (see Fig. m1.10), If the average energy separation between adjacent orbitals amounts to Δ , replacing k nucleons will cost an energy loss of 2 | ||
Δ Ebinding = α ( Δ α /2) ——————- (m1.35) | ||
| and, with α = (N–Z)/2, this becomes | ||
Δ Ebinding = (N–Z) 2 Δ /8 —————— (m1.36) | ||
| It has been seen by theoreticians that the average energy spacing between the single particles, Δ should vary inversely proportional to A. Therefore a correction to B. E. known as asymmetry energy term is expressed as | ||
B4 = –α 4(N–Z) 2/A ———————– (m1.37) | ||
| The minus sign again indicates the weakening of binding energy caused by asymmetry in N and Z. | ||
| Regarding nuclear stability another important point was noticed, i.e. even-even (even Z and even N) nuclei are most stable, even-odd and odd-even nuclei are less stable and odd-odd nuclei are most unstable. This is known as pairing effect in nuclear physics. To take into account of this pairing effect, an additional term is incorporated into the binding energy expression: | ||
B 5 = + δ = 0 = – δ | for even – even nuclei for even – odd nuclei and odd – even nuclei ——m(1.38) for odd – odd nuclei | |
| where δ is empirically found to be given by | ||
δ = a 5A -3/4 —————– (m1.39) | ||
| Now considering all the above mentioned terms, the B. E. of a nucleus can be expressed as: | ||
B. E. = B1 + B2 + B3 + B4 + B5 = a1A -a2A 2/3 – /equ9.gif){kind=small} –a 4(N–Z)2/A+a5A-3/4 ———- (m1.40) | ||
| Now the B.E./A vs. A can be better depicted as shown in Fig. m1.10. The effect of Coulomb repulsion and the nucleon asymmetry for lower and higher A values respectively clearly demonstrated in this figure. Overall, the resultant binding energy is to be considered, which is same as Fig. m1.10. | ||
/m1.10.png) | ||
FIGURE m1.10 B.E./A versus A showing various energy contribution to B. E. The resultant B. E. is also clearly indicated. | ||
| Considering the above formulation of B. E. the famous nuclear physicists Weizsacker and Bethe developed a semi- empirical equation to express the mass of a nucleus as: | ||
/equ10.png) | ||
| This is popularly known as Weizsacker’s semi empirical mass formula and has a great implication in nuclear physics. This equation can make a clear picture about nuclear stability, one of which is the stability against beta particle disintegration (see tutorial section 2). | ||
| Using fits of known masses to this final equation one can determine the coefficients a1 to a5 as shown by Wapstra and Audi4 : | ||
a1 = 15.85 MeV a2 = 18.34 MeV a3 = 0.71 MeV a4 = 23.21 MeV a5 = 12.0 MeV | ||
| The volume energy amounts to almost 16 MeV per nucleon: the various corrections than steadily brings this value towards the empirical value as observed throughout the nuclear mass region. | ||