Nuclear Physics: Nuclear Models Mahatsab Mandal Government General Degree College at Kalna-I [email protected] 1 Introduction One of the primary goals of nuclear physics is to account the properties of the nucleus in terms of models of their structure and internal motion. However, due to lack of knowledge of the exact mathematical form of the nuclear interaction, it has not been possible to develop a suitable theory of the structure of the nucleus which is made of the large number of protons and neutrons. Due to the lack of a satisfactory theory of nuclear structure, different models have been proposed to explain the properties o nuclei. Over the decades, various models were developed. Among all the nuclear models, the important ones which are going to be discussed here are the following : (i) Liquid drop model, (ii) Shell model (iii) Fermi gas model. 2 Liquid Drop Model The liquid drop model was first model developed by N. Bohr and F. Kalckar in 1937. They observed that the saturation properties of nuclear interaction in analogy with the attractive forces between the constituents of a liquid drop. There exits many similarities between the nucleus of an atom and a liquid drop [1]: (i) The attractive nuclear force near the nuclear surface is similar to the force of surface tension on the surface of the liquid drop. (ii) As in the case of a liquid drop, the density of the nuclear matter is independent of its volume. The nuclear radius R / A1=3, where A is the atomic mass number. Hence the volume is V / A. Since the nuclear mass M ∼ A, the density of the nuclear matter is ρm = M=V , which is independent of A. This also suggests saturation of the nuclear force. (iii) The evaporation of a liquid drop corresponds to the radioactive properties of the nucleus. (iv) The constant binding energy per nucleon of a nucleus is analogous to the latent heat of vaporisation of liquid. (v) The formation of a short lived compound nucleus by the absorption of a nuclear particle in a nucleus during a nuclear reaction is analogous to the process of condensation from the vapour to the liquid phase in case of the liquid drop. Binding energy for the nucleus is 2=3 EB = avA − asA (1) The first term represents the volume energy of all nucleons and the send term for the surface energy similar to the surface tension of the liquid. 2.1 Bethe-Weizsacker Formula On the basis of the liquid drop model, C. V. Weizsacker explain the total binding energy of the nucleus. The semi-empirical binding energy or mass formula gives a connection between the theory of the nuclear matter with the experimental information. The binding energy EB for the nucleus (Z,A) can be expressed as Z(Z − 1) (A − 2Z)2 E = a A − a A2=3 − a − a + E (2) B v s c A1=3 a A p (i) Volume Energy: The density of the nuclear matter is independent of its volume. The nuclear radius 1=3 R / A . Hence the volume is V / A. So, the volume energy, Ev / A = avA, where av = 0:016919 u is a constant. (ii) Surface energy: The nucleus is assumed to resemble a spherical liquid drop, the nucleons at the 1 Mahatsab Mandal 2 LIQUID DROP MODEL surface of the nucleus are not completely surrounded by other nucleus. It result in reducing the total binding energy due to the nucleons on the surface. This surface force is proportional to the surface area 2 1=3 2=3 of the nucleus i.e., 4πR . But R / A . So we can write the total surface energy as Es = −asA , where as = 0:016114 u is a constant. (iii) Coulomb Energy: The Coulomb repulsion between the protons in the nucleus also tends to weaken the nuclear binding. Since each protons repulses all of the other charged particles, this term would be directly proportional to the possible number of combination for a given Z, which is Z(Z − 1)=2. The energy of interaction between protons is inversely proportional to the distance of separation R, so the Coulomb repulsion is Z(Z − 1) Z(Z − 1) Z(Z − 1) E / − / − = −a (3) c R A1=3 c A1=3 Calculation for ac: Total nuclear charge Q = +Ze and the charge density Ze 3Ze ρ = = (4) m 4πR3 4πr3A 3 0 4 3 Suppose the sphere has been build up to a radius r, the total change on it is q = 3 πr ρm. The work done in bringing the charge dq in the suffer of the sphere is 2 q dq 4 3 1 2 1 16π 2 4 dW = = πr ρm 4πr ρmdr = ρmr dr (5) 4π"0r 3 4π"0r 4π"0 3 Hence, the total work done in forming the charge sphere of radius R is Z 2 R 2 1 16π 2 4 3 Q W = ρm r dr = (6) 4π"0 3 0 5 4π"0R 3 Q2 3 (Ze)2 3 (Ze)2 ) E = − = − = − (7) c 1=3 5 4π"0R 5 4π"0R 5 4π"0r0A This leads to 3 e2 ac = (8) 5 4π"0r0 (iv) Asymmetry Energy: Nuclear data for stable nuclei indicate that for lighter nuclei, N ' Z. As we go to the heavier nuclei, the symmetry of the proton and neutron number is lost and the number of neutrons exceeds that of protons to maintain the nuclear stability. This excess of neutrons (N − Z) is the measure of the asymmetry and it decreases the stability. The asymmetry energy, Ea, is directly proportional to (i) the neutron excess, N − Z = A − 2Z and (ii) the fraction of nuclear volume in which the excess neutrons are present. As the volume of the nucleus is proportional to A, the fractional volume of the nucleus in which excess neutrons are present will be proportional to (N − Z)=A. (A − 2Z) (A − 2Z)2 ) E / −(A − 2Z) = −a (9) a A a A (v) Pairing Energy: Nuclear data for stability and binding energy indicates that e-e nuclei are most stable whereas o-o nuclei are least stable. The binding energy of nuclei with o-e or e-o lies between that 1 of e-e and o-o nuclei. Each of the protons and neutrons having spin 2 form with parallel and anti-parallel spins in e-e nuclei giving them a stable configuration. But in o-o nuclei, one unpaired proton and neutron are left to make the nuclei less stable. So, the pairing energy, Ep, is given by (−1)N + (−1)Z a E = p (10) p 2 A1=2 where ap is a constant. For o-e and e-o nuclei Ep = 0. The atomic mass of an isotope can be written as M(A; Z) = ZMH + (A − Z)Mn − EB(A; Z) Z(Z − 1) (A − 2Z)2 = ZM + (A − Z)M − a A − a A2=3 − a − a + E (11) H n v s c A1=3 a A p This formula is known as the semi-empirical mass formula. 2 Mahatsab Mandal 2 LIQUID DROP MODEL 2.1.1 Mass Parabola Neglecting Ep, we can write the semi-empirical mass formula as 2 M(A; Z) = fA + p Z + q Z (12) where 2=3 fA = A(Mn − av + aa) + asA p = −4aa − (Mn − MH ) 1 q = (a A2=3 + 4a ) (13) A c a Eq. (12) is the equation to a parabola for a given A and is known as the mass parabola. The lowest @M point of the parabola Z = ZA can be obtained by @Z = 0. This leads to A p (M − M + 4a )A p + 2q Z = 0 ) Z = − = n H a (14) A A 2=3 2q 2(acA + 4aa) p p 2 ) M(A; Z ) = f + p Z + q Z2 = f + p − + q A A A A A 2q 2q p2 = f − (15) A 4q Hence, 2 p p 2 M(A; Z) − M(A; Z ) = + pZ + qZ2 = q Z + = q(Z − Z )2 = + ve quantity (16) A 4q 2q A This proves that the mass parabola for a given isobar (A=constant) has the lowest point at Z = ZA. Since M(A; ZA) has the smallest value for a given A, the nucleus has the largest binding energy and is the most stable among the isobar for given A. Putting the numerical value of MH ;Mn; ac; aa, we get A Z = (17) A 1:98 + 0:015A2=3 which does not usually given an integral value for ZA. In most cases, the value of Z nearest to ZA corresponds to the actual stable nucleus for a given A. As an example for A=63, Eq (17) gives ZA = 28:4. Actually the isotope 63Cu at Z = 29 is found to be stable. If we taken account the pairing energy, the mass parabola for different isobars fall into two groups depending on if A is odd or even. For odd A we get a single parabola for each A, since Ep = 0. For even A, we get two parabola for the same A, one for even Z (e-e) and other for odd Z (o-o). Since Ep is subtracted for o-o nuclei and added for e-e nuclei, the parabola for o-o nuclei is above that for e-e nuclei for same A by an amount 2Ep. 2.1.2 β-disintegration energy of mirror nuclei Mirror nuclei are pairs of isobaric nuclei in which the proton and neutron numbers are interchanged and differ by one unit so that Z − N = 1.