Nuclear Models Liquid drop Model Semi Empirical Mass FormulaShell ModelConcept of Magic number
Nuclear ModelsWhy is the binding energy per nucleon almost constant?Why do certain nuclei emit α- and β-particles through these particles do not exist inside the nucleus?Why are the nuclei containing 2,8,20,28,50,82 nucleons most stable?Like any other quantum mechanical system, a nucleus also exist in its excited states. The most stable state is the ground state in which the nuclei are generally found.Different models:1. Liquid drop model2. Shell model3. Collective model4. Optical model
Nuclear models:1. the “water-drop” model 2. the “shell” model Models describe aspects of the structure of nuclei and how they behave.
Assumptions1. The nuclei of all elements are considered to be behave like a liquid drop of incompressible liquid of very high density.2. In an equilibrium state the nuclei of atoms remain spherically symmetric under the action of strong attractive nuclear forces just like the drop of a liquid which is spherical due to surface tension.3. The density of a nucleus is independent of its size just like the density of liquid which is also independent of its size.4. The nucleons of the nucleus move about within a spherical enclosure called the nuclear potential barrier just like the movement of the molecules of a liquid within a spherical drop of liquid.5. The binding energy per nucleon of a nucleus is constant just like the latent heat of vaporization of a liquid.
As a first approximation, we can think of Volume Energy each nucleon in a nucleus as interacting solely with its nearest neighbors.Energy associated with each nucleon-nucleon bond = UBecause each bond energy is shared bytwo nucleons thereforeeach has a binding energy of ½ U.When an assembly of spheres of thesame size is packed togetherinto the smallest volume, each interiorsphere has 12 other spheresIn contact with it
Volume EnergyHence each interior nucleon in a nucleus has a binding energy of =12(1/2U) = 6U.If all A nucleons in a nucleus were in the interior, the total bindingEnergy would be EV= 6AUOr EV = a1 A Volume energyAlso we know that if nucleus is spherical then 4 R3 A 3
Surface Energy Actually, of course, some nucleons are on the surface of every Nucleus and therefore have fewer than 12 neighbors.Surface energy surface area 4πR2If R = R0 A1/3 thenSurface energy 4πR0 A2/3OrSurface energy = - a2 A2/3Surface energy A2/3
It is important for lighter nuclei since a greater fraction of their nucleonsare on the surface.Coulomb Energy The repulsion between each pair of protons in a nucleus also contribute toward decreasing its binding energy. The coulomb energy EC of a nucleus is the work that must be done To bring together Z protons from infinity into a spherical aggregate The size of the nucleus. e2 V 4 0r Since there are Z(Z-1)/2 pairs of protons, Z (Z 1) Z (Z 1)e 2 1 EC V 2 8 0 r av
1 Where r is the value of 1/r averaged over all proton pairs. av If the protons are uniformly distributed throughout a nucleus of radius R, 1 is proportional to 1/R and hence 1/A1/3 r av Therefore Z ( Z 1) EC a3 1 3 AThe Coulomb energy is negative because it arises from an effectthat opposes nuclear stability.
The total binding energy Eb of a nucleus Surface energies 2 3 Z ( Z 1) Eb Ev ES EC a1 A a2 A a3 1 3 A Volume energies Coulomb energiesThe binding energy per nucleon is EBE a2 Z ( Z 1) a1 1/ 3 a3 4 A A 3 A
Empirical binding energy per nucleon curvetheoretical binding energyper nucleon curve (usingLiquid drop model concept)
Correction to the formula Asymmetry EnergyCorrection 1: When the neutrons in a nucleus outnumber the protons,which means that higher energy levels to be occupied than would bethe case if N and Z were equal. • Neutrons and protons are spin ½ fermions obey Pauli exclusion principle. • If other factors were equal ground state would have equal numbers of n & p. Illustration Neutron and proton states with same spacing ε. Crosses represent initially occupied states in ground state. If three protons were turned into neutrons the extra energy required would be 3 3ε. In general if there are N-Z excess neutrons over protons the extra energy is ((Z-N)/2)2 ε. relative to Z = N.
Correction to the formula Correction 1: When the neutrons in a nucleus outnumber the protons, which means that higher energy levels to be occupied than would be the case if N and Z were equal. ΔE=(number of new neutrons)(energy increase/new neutron) 1 1 Because N=A-Z, (N-Z)2=(A-2Z)2 E N Z N Z 2 2 2 2 N Z 8The greater the number of nucleon in anucleus the smaller is the energy levelspacing ε, with ε proportional to 1/AAsymmetry energy due to differenceBetween N and Z 2 A 2Z Ea E a4 A
Correction to the formula Pairing TermNote: The asymmetry energy is negative because itreduces the binding energy of the nucleusCorrection 2: The correction term arises from the tendency of protonpairs and neutron pairs to occur. Even –even nuclei are the most stableAnd hence have higher binding energies. Therefore nuclei such as 42He, 12 C, 16 O etc. Appear as peak on the empirical curve of BE/A. 6 8The pairing energy Ep ispositive for even-evennucleiThe pairing energy Ep iszero for odd-even andeven-odd nucleiThe pairing energy Episnegative for odd-odd nuclei
Pairing Term• Nuclei with even number of n or even number of p more tightly bound fig.• Only 4 stable o-o nuclei cf 153 Nuclei Pairing e-e.• p and n have different energy term levels small overlap of wave functions. Two p(n) in same e-e +ive level with opposite values of jz e-o 0 have AS spin state sym spatial w.f. maximum overlap o-o -ive maximum binding energy because of short range attraction.
Correction to the formula ……The pairing energy Ep isgiven by the relation a5 EP ,0 3/ 4 A Semi Empirical Mass Formula (SEMF)a1=14.1 MeV Surface Term a3=0.595 MeV Pairing Term Z ( Z 1) ( A 2Z ) 2 a5 Eb ( A, Z ) a1 A a2 A2 / 3 a3 1 a4 ( ,0) 3 / 4 3 A A A Volume Term Coulomb Term Asymmetry Term a2=13.0 MeV a4=19.0 MeV a5=33.5 MeV
Numericals: Numerical 1. The atomic mass of the zinc isotope 6430Zn is 63.929 a.m.u. calculate its binding energy using semi-empirical mass formula and compare the results with direct formula. [Ans: 561.7 MeV]
Numerical 2. Isobars are nuclides that have the same mass number A. Derive a formula for the atomic number of the most stable isobar of a given A and use its to find the most stable isobar of A=25Solution: To find the value of Z for which the binding energy Ebis a maximum which correspond to maximum stability, we mustSolve dEb/dZ = 0 for Z. From the liquid drop mass formula dEb a3 4 a4 1 2Z 1 A 2Z 0 dZ 3 A A 1 1 3 3 a3 A 4a4 0.595 A 76 Z 1 1 3 1 3 1 2a3 A 8a4 A 1.19 A 152 AFor A =25 this formula gives Z = 11.7, from which weconclude that Z = 12 should be the atomic number ofThe most stable isobar of A=25. This nuclei is 2512Mg.
Main Achievements of liquid dropmodel (LDM) 1.It explains binding energy of large number of nuclei. 2.It explains the fusion and fission processes nicely. 3. Explains energies of radioactive decays, fission and fusion.Applications of the water-drop model1. Nuclear fission(very large nuclei break up)2. Nuclear fusion(very small nuclei fuse together)
Main drawbacks of liquid dropmodel (LDM) 1. It is not able to explain the magic numbers. 2. It is not able to explain excited states. 3. It is not able to calculate the nuclear spin.
Shell ModelThe basic assumption of the liquid drop model is that each nucleon ina nucleus interact only with its nearest neighbors, like a molecule in aliquid.In shell model each nucleon interact chiefly with a general force fieldproduced by all the other nucleons.The atoms with 2,10,18,36,54 and 86 electrons have all theirelectron shell completely filled.In the same way, nuclei that have 2,8,20,28,50,82 and 126 neutrons and protons are more abundant than other nuclei of similar massnumbers, suggesting their structures are more stable.
Shell Model ……It has been observed that nuclei having either thenumber of protons Z on number of neutrons N =A- Z equal to one of the numbers 2, 8, 20, 50, 82and 126 are more stable than their neighbours.These numbers are called magic numbers
Main Assumptions ……1. Nucleon forms subshells and shells with in the nucleus1. The shell within the nucleus get closed with a suitable number of nucleons3. Each nucleon is supposed to possess a spin angular momentum of ħ/2 and orbital angular momentum lħ.4. This theory assumes that LS coupling holds only for the very lightest nuclei in which the l values are necessarily small in their normal configuration.5. The heavier nuclei exhibit j-j coupling.
Shell model ……It is assumed that the nucleons move in its orbit within the nucleus,independently of all other nucleons.The orbit is determined by a potential energy function V(r) whichrepresent the average effect of all interaction with other nucleons,and is same for each particle.Each nucleons is regarded as an independent particle and theinteraction between a nucleons is considered to be a small perturbationOn the interaction between a nucleon and the potential field.There is a direct analogy between the theoretical treatment of aNucleus and an electron in an atom.
Problem: Using shell model calculate the spin and parities of the following nucleus 1. 27 13Al 2. 33 16Ar 3. 16 8O Where l is orbital quantum numberParity: (-1)l If l is even parity is even, If l is odd parity is odd The parity of a wave function refer to its behaviour under a simultaneous reflection of the space coordinates i.e. x to –x, y to –y and z to –z. Parity of a nucleus refers to the behaviour of the wave function as a result of the inversion of the coordinates. P ( x, y , z ) ( x, y, z ) Even P ( x, y , z ) ( x, y , z ) Odd
Electric quadrupole moment (Q): measure of how much nuclear chargedistribution depart from sphericity.A spherical nucleus has no quadrupole moement, i.e. Q = 0 while one shaped like a prolate - speroid or egg shaped Q is +veIn an oblate or disc like spheroid (pumpkin shaped) have Q is –ve.Nuclei of magic N and Z are found to have zero quadrupole moment andhence are spherical in nature.Shell model is an attempt to account for the existence of magicnumbers and certain other nuclear properties in terms of nucleonbehavior in a common force fieldNeutron and protons occupy separate sets of sates in a nucleusBecause proton interact electrically as well as through the specificallynuclear charge.
Quadrupole moment 2 2 2 Q zb a 5Prolate (b > a) Sphere (b = a) Oblate (a > b) SpheroidSpheroid Q is 0 Q is -veQ is +ve
Experimental Evidence of Nuclear Magic Numbers1. It has been experimentally observed that the nuclei for whichN and Z are 2,8…….. 42He and 168O are more stable then theirneighbours.2. Stability also related to the natural abundance. It has been observedFrom the experimental data that the nuclei having number of nucleonsas magic number are in abundance as compared to other nuclei inNature.For Example: 16 8O (N, Z = 8), 40 Ca (N, Z = 20), All are abundant in nature 20 206 Pb ( Z = 82,N=126) 82
Experimental Evidence of Nuclear Magic Numbers ……3. Sn (Z = 50) has 10 stable isotopes, more than any other element ,While Ca (Z = 20) has six isotopes. This indicates that elements withZ=50 and Z=20 are more than usually stable.4. No more than 5 isotones occur in nature for any N except N=50,where there are 6 and N = 82, where there are 7. Neutron numbers of82, 50 therefore, indicate particular stability.5. The doubly magic nuclei (N &Z both magic) 42He, 168O, 4020Ca and206 Pb are particularly tightly bound. 826. The binding energy of the next neutron and proton after magicnumber is very small.
Main Achievements of Shell model (SM) 1.It explains Magic numbers. 2. It explains the magnetic moment of some nuclei nicely. 3. It explains successfully the ground state spin. 4. It explains the great stability and high binding energy 5. It explains the phenomenon of nuclear isomerism.Note: Atoms having the same mass number but different and same atomic number, are distinguish by certain difference in theinternal structure of the nucleus are called isomers.
Main Limitation of Shell Model (SM) 1. It fail to explain the stability of four stable nuclei 21H, 63Li, 105B 14 N. 7 2. It does not predict correct values of nuclear spin for certain nuclei 3. The Quadrupole moment calculated using this model in also not in good agreement. 4.The magnetic moment is also shows some deviation from observed values.