Nuclear and Particle Physics - Lecture 20 The shell model

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1 1 Intoduction Nuclea and Paticle Physics - Lectue 0 The shell model It is appaent that the semi-empiical mass fomula does a good job of descibing tends but not the non-smooth behaviou of the binding enegy. Fo this, we need to go to a vey diffeent model of the nucleus, which is based on quantum enegy levels. It is supising that such a adically diffeent pictue can be descibing the same physical system but we shall see that seveal popeties of nuclei ae well descibed by this model. Magic numbes A close look at the discepancies fom the semi-empiical mass fomula is in ode. We saw thee wee paticula values of Z and N fo which the nuclei had a highe binding enegy than would be expected. These stongly bound states occu when Z o N have one of a set of so-called magic numbes. In fact, even fo A < 0, whee the semi-empiical mass fomula is not valid, it is appaent that cetain nuclei, e.g. 4 He with a binding enegy of 8.3 MeV, ae much moe stongly bound that thei neighbous, e.g. thee ae no bound A = 5 nuclei. The magic numbes which ae obseved ove the whole ange of nuclei ae, 8, 0, 8, 50, 8, 16 Some nuclei have both Z and N at magic numbes, such as 4 He (Z =, N = ) and the most common isotope of lead, 08 8Pb (Z = 8, N = 16); these ae called doubly-magic and ae coespondingly even moe stongly bound. The shell model says these magic numbes coespond to filling a quantum enegy level, so giving a paticulaly well-bound nucleus. The magic nuclei ae theefoe equivalent to the inet gases (helium, neon, agon, etc.) in chemisty. While this povides a qualitative explanation, we still need to undestand why the magic numbes have the values they do. 3 Nuclea potentials Ideally, we would wite down the Schödinge equation fo the nuclea foce potential and solve it to calculate the enegy levels, as done fo the hydogen atom. Howeve, this is not as simple as fo hydogen, fo two easons. Fistly, the potential fo the nuclea foce is much moe complicated than the 1/ fo hydogen. Secondly, it is not a cental potential in which ae nucleons move independently; thee is no cental object coesponding to the poton in hydogen but each nucleon feels the foce fom the othes. Hence, we need to make a physical guess fo a easonable potential and compae with the obseved magic numbes. We will conside each nucleon as moving in a potential esulting fom the aveage of the inteactions with all the othe nucleons. What would this potential look like? We aleady saw the shot ange foce means a nucleon is bound to all its neaest neighbous by an equal contibution to the binding enegy fo each nucleon. Inside the nucleus, the numbe of neaest neighbous is equal in all diections so the net foce on any nucleon is in fact zeo. Thus the effective potential is constant within the nucleus and the constant value must be negative to keep the nucleon bound. Outside the nucleus, moe than a few femis away, the shot ange nuclea foce will have died off, so again thee will be no foce and hence a constant effective potential, which we can take as zeo. Finally, as stated peviously, the nucleons nea the suface 1

2 have only neaest neighbou foces into the nucleus as they ae missing the neaest neighbous outside. Hence, they do have a net inwads foce and so a ising potential as the adius inceases. This change to the potential takes place ove a distance of ode the nuclea foce, so aound 1 fm. Hence, we would guess an effective potential would look like 1fm 1/3 0 A This is called the Saxon-Woods potential and is often mathematically expessed as V 0 V () = 1 + e ( a)/d whee a 0 A 1/3 sets the adius and d 1 fm the speed with which the potential ises. While it is possible to solve the Schödinge equation fo this potential, it is not tivial. To give a feel fo the esults, we can look at some simple cases, such as an infinite squae well o a simple hamonic oscillato. These levels can be calculated moe easily and look like the following

3 What do these pedict fo the magic numbes? Each state has l + 1 values of l z and due to the nucleon spin, each can take two potons (and also two neutons) in the two s z states. Hence, the numbe of potons (o neutons) in an l state is (l + 1) = 4l +. l 4l Hence, the fist magic numbe of coesponds to filling the fist state in both cases. The next magic numbe is 8, which is the total numbe of nucleons which fills the fist two states, again in eithe case. The othe numbes given by completing the levels ae shown in the diagams above. They both give 0 but then stat to disagee with the measued values fo the magic numbes. Hence, we can epoduce the fist few but not the highe values. You may think this is just a question of tweaking V () to aange the states to be just ight, but it tuns out it is not possible to get all the coect magic numbes by this method. 4 Spin-obit coupling A new tem is needed in the potential and this is a spin-obit coupling, whee the enegy is l.s, just as happens in atomic physics. This has the effect of splitting some of the 4l + degeneacy and giving new enegy levels. We peviously said each l state has l + 1 values of l z and s + 1 = values of s z. These could equally well be descibed by total angula momentum j and j z, athe than l z and s z. Fo a given l, then thee ae two values of j, namely l ± 1/ and these have j + 1 = (l ± 1/) + 1 = l + and l values of j z, summing to 4l + in total, as equied. Without a spin-obit coupling, both values of j have the same enegy and so ae totally degeneate. Howeve, a spin-obit coupling splits the two j values but leaves the 3

4 j z degeneacy in each one (as is equied by the isotopy of space). To see how this woks, we can use the same tick as we used fo the hypefine splitting in the mesons. The total angula momentum is j = l + s so squaing gives Reaanging, then j = l + s + l.s l.s = 1 [j l s ] In tems of eigenvalues, this is l.s = h [j(j + 1) l(l + 1) s(s + 1)] showing that this tem does indeed depend on the value of j. Since j = l ± 1/, then fo l + 1/, this gives l.s = h [(l + 1/)(l + 3/) l(l + 1) s(s + 1)] = h while fo l 1/, it gives l.s = h [(l 1/)(l + 1/) l(l + 1) s(s + 1)] = h [ ] l + l + 3/4 l l 3/4 = h l [ ] l 1/4 l l 3/4 = h The effect of applying this splitting to the Saxon-Woods potential is shown below (l + 1) 4

5 This model can also coectly pedict the spins and paities of many nuclei whee thee is a single unpaied nucleon eithe alone in a state o missing fom a completed state. It woks paticulaly well fo Z o N close to magic. All even-even nuclei ae J P = 0 +. If Z o N ae both even and one coesponds to a completed level, then adding one exta of this type of nucleon means the total spin of the nucleus must be the angula momentum of this final nucleon, as must its paity. E.g. 17 8O has Z = 8 and N = 9, so the final neuton must be in the next state above the level which gives the magic numbe 8. This is a 1d 5/ level and so has j = 5/ and l =, which gives P = ( 1) l = +1. Hence, this nucleus would be expected to be J P = 5/ +, as obseved. Similaly, emoving one nucleon fom a filled magic state gives a nucleus with total spin exactly opposite to the emoved nucleon (as they sum to give zeo), i.e. the same j value but opposite j z, and also the same paity (as they multiply to give +1). This means it has the same quantum numbes as the unfilled state. Hence, fo example 15 8O, with Z = 8 and N = 7, would have the popeties of the 1p 1/ state, which has l = 1 and hence P = ( 1) l = 1, and so would be expected to be J P = 1/, again as obseved. 5