A 'madelung Rule' For The Atomic Nucleus

[tetrahedron]

A couple of days ago I discovered how to squeeze a Madelung-type rule out of atomic nuclear structure. In the Left-step periodic table created by the elderly French polymath Charles Janet in the late 1920's, all the periods end in s-block elements, ignoring chemical bonding behavior but reproducing (at least ideally) the sequence of introduction of new blocks in the table. Thus s, ps, dps, fdps, with each structure repeated (so 1s, 2s, 2p3s, 3p4s, 3d4p5s, 4d5p6s, 4f5d6p7s, 5f6d7p8s). 
 

Several workers around the same time started noticing that the sums of the shell number plus the quantum number (m)l value for each of the orbitals within the Janet period structure (N+L) always had the same value:

 

1s= 1+0=1

2s= 2+0=2

2p= 2+1=3, 3s= 3+0=3

3p= 3+1=4, 4s= 4+0=4

3d= 3+2=5, 4p= 4+1=5, 5s= 5+0=5

4d= 4+2=6, 5p= 5+1=6, 6s= 6+0=6

4f=  4+3=7, 5d= 5+2=7, 6p= 6+1=7, 7s= 7+0=7

5f=  5+3=8, 6d= 6+2=8, 7p= 7+1=8, 8s= 7+0=8

 

In the atomic nucleus, on the other hand, under a simple harmonic oscillator model, we are also presented with a Left-step pattern, but with a major difference. Unlike the electronic periods, shells in the harmonic oscillator system have all their orbital components sorted for parity (either all positive (even (m)l) or negative (odd (m)l)). And unlike the electronic system, period lengths by element count are not repeated. Rather the total number of orbitals within the shell are repeated.

 

So 1s, 1p (both with only one orbital); 1d2s, 1f2p (with two orbitals); 1g2d3s, 1h2f3p (with three orbitals); 1i2g3d4s, 1j2h3f4p (with four orbitals).

 

I found that the Madelung-type rule for the harmonic oscillator nucleus with 2N+L works just fine, thus:

 

1s:            1s: (2x1)+0=2

1p:            1p: (2x1)+1=3 

1d2s:        1d: (2x1)+2=4;   2s: (2x2)+0=4

1f2p:         1f:  (2x1)+3=5;   2p: (2x2)+1=5

1g2d3s:    1g: (2x1)+4=6;   2d: (2x2)+2=6;   3s (2x3)+0=6

1h2f3p:     1h: (2x1)+5=7;   2f:  (2x2)+3=7;   3p (2x3)+1=7

1i2f3d4s:   1i:  (2x1)+6=8;  2g: (2x2)+4=8;   3d: (2x3)+2=8;   4s: (2x4)+0=8

1j2h3f4p:   1j:  (2x1)+7=9;  2h: (2x2)+5=9;   3f:  (2x3)+3=9;   4p: (2x4)+1=9

 

The more realistic model which includes the spin-orbit coupling presents a more complicated picture, as the intruder levels which insert themselves into the structure of the previous harmonic oscillator shells have different 2N+L value than that shell (one greater). This raises the question as to whether they are truly incorporated into these receptor shells. First, they have the opposite parity (- adds to +, + adds to -). Secondly, calculations of total shell energies (conserved over ellipsoidal deformation) show that the intruders aren't part of the energy conservation (which was a big surprise for me).

 

Jess Tauber

Edited February 22, 2019 by pascal

[tetrahedron]

I've been looking at the oblate oscillator ratios to see if I can unpack their features with regard to a Madelung-like rule, or as Philip says, a Janet one. Or, as a wag might say, a Brady Bunch rule (that is, Jan Mad).

 

If one lays out the orbital summation within derived shells marking deformed ellipsoid magic numbers as an array, something really interesting happens. 

 

So, for oblate oscillator ratio 1:2 (equatorial extent of the matter wave twice the polar extent) the shells look like this:

 

1s1/2

1p3/2   (one orbital component per shell)

 

1d5/2 1p1/2

1f7/2   1d3/2   (two components per shell)

 

1g9/2 1f5/2 2s1/2   

1h11/2 1g7/2 2p3/2  (three components per shell)

 

1i13/2 1h9/2 2d5/2 2p1/2

1j15/2 1i11/2 2f7/2 2d3/2   (four components per shell)

 

*1k17/2 *1j13/2 2g9/2 2f5/2 3s1/2

*1l19/2  *1k15/2  *2h11/2 2g7/2 3p3/2  (five components per shell)

 

One can easily stagger individual sequences here relative to each other. When you do you can draw '*isospin' lines connecting all the same-spin components. They all slope downwards to the right.  This would be a 'right-step' table of sorts. And at a different angle you'd connect all the orbital components making up the spherical shells as described in my first posting on this topic.

 

With a prolate deformed ellipsoid of oscillator ratio 2:1 (which has the doubled N+L pattern) we get the following (I hope it comes out ok here and doesn't reorganize itself in terms of line justification):

 

//////////////////////////////1s1/2

//////////////////////////////1p1/2   (one orbital component per shell)

 

////////////////////1p3/2 2s1/2

////////////////////1d3/2 2p1/2   (two components per shell)

 

//////////1d5/2 2p3/2 3s1/2

//////////1f5/2  2d3/2 3p1/2   (three components per shell)

 

1f72   2d5/2 3p3/2 4s1/2

1g7/2 2f5/2  3d3/2 4p1/2   (four components per shell)

 

You'll note that THIS array is organized as a kind of left-step table. '*isospin' lines are vertical. Lines connecting orbital components of the same spherical shell are sloping up and to the right, in the opposite direction from those in the oblate 1:2 osicllator ratio array listed above this one.

 

I suspect similar arrays can be made for higher deformations in either oblate or prolate directions, and they will bear some resemblance to the Madelung/Janet diagrams which do this job for the electronic system.

So far I haven't been able to discern what the (N,L) rule is for the oblate system. Note however that in the prolate deformed nuclei under the harmonic oscillator model, the doubled triangular number intervals are used twice for 2:1, thrice for 3:1 and so on, a MULTIPLICATION (I don't yet know how this relates to the (N,L) summations). But the oblates do things in the inverse manner, that is for oscillator ratio 1:2 we find a doubled triangular number interval between EVERY SECOND magic, and for 1:3 between every THIRD, and so on. I have to believe that any Madelung/Janet formulation will be strongly related to these facts for the oblates.

 

Jess Tauber

[tetrahedron]

So, I found another Madelung-like relation. At oscillator ratio 4:1, the relation is N+2L (remember that for 1:1 (the sphere) the relation is 2N+L, and for 2:1 (prolate ellipsoid) the relation is N+L (or perhaps restatable as 2N+2L?).

 

1s1/2: 1+2(0)=1

1p1/2: 1+2(1)=3

2s1/2: 2+2(0)=2

2p1/2: 2+2(1)=4

_____________

 

1p3/2 3s1/2: 1+2(1)=3; 3+2(0)=3

1d3/2 3p1/2: 1+2(2)=5; 3+2(1)=5

2p3/2 4s1/2: 2+2(1)=4; 4+2(0)=4

2d3/2 4p1/2: 2+2(2)=6; 4+2(1)=6

___________________________

 

1d5/2 3p3/2 5s1/2: 1+2(2)=5; 3+2(1)=5; 5+2(0)=5

1f5/2  3d3/2 5p1/2: 1+2(3)=7; 3+2(2)=7; 5+2(1)=7

2d5/2 4p3/2 6s1/2: 2+2(2)=6; 4+2(1)=6; 6+2(0)=6

2f5/2  4d3/2 6p1/2: 2+2(3)=8; 4+2(2)=8; 6+2(1)=8

_________________________________________

 

1f7/2 3d5/2 5p3/2 7s1/2: 1+2(3)=7; 3+2(2)=7; 5+2(1)=7; 7+2(0)=7

1g7/2 3f5/2 5d3/2 7p1/2: 1+2(4)=9; 3+2(3)=9; 5+2(2)=9; 7+2(1)=9

2f7/2 4d5/2 6p3/2 8s1/2: 2+2(3)=8; 4+2(2)=8; 6+2(1)=8; 8+2(0)=8

2g7/2 4f5/2 6d3/2 8p1/2: 2+2(4)=10; 4+2(3)=10; 6+2(2)=10; 8+2(1)=10.

 

Note that these sums do not increase monotonically but involve 2 steps forward and one step back. Other than that they are entirely regular mathematically.

[tetrahedron]

Here's another- a super hyperdeformed harmonic oscillator nucleus- far more deformed in the prolate direction than any real nucleus ever discovered: So far there seems to be a pattern to even-numerator prolate ellipsoidally deformed nuclei with regard to the Madelung-like formula. For 2:1 we have N+2L. For 4:1 we have N+4L. And now for 6:1 the formula is N+3L. The multiplication factor for L is half the oscillator ratio's numerator. Still can't figure out how the odd-numerator oscillator ratio shells pattern in this regard, or any of the oblate systems. Will keep working at it. I'm sure something will come to me.

 

6:1

 

1s1/2  1+3(0)=1

1p1/2  1+3(1)=4

2s1/2  2+3(0)=2

2p1/2  2+3(1)=5

3s1/2  3+3(0)=3

3p1/2  3+3(1)=6

------------------------

1p3/2 4s1/2   1+3(1)=4; 4+3(0)=4

1d3/2 4p1/2   1+3(2)=7; 4+3(1)=7

2p3/2 5s1/2   2+3(1)=5; 5+3(0)=5

2d3/2 5p1/2   2+3(2)=8; 5+3(1)=8

3p3/2 6s1/2   3+3(1)=6; 6+3(0)=6

3d3/2 6p1/2   3+3(2)=9; 6+3(1)=9

------------------------------------------------

1d5/2 4p3/2 7s1/2  1+3(2)=7; 4+3(1)=7; 7+3(0)=7

1f 5/2 4d3/2 7p1/2  1+3(3)=10; 4+3(2)=10; 7+3(1)=10

2d5/2 5p3/2 8s1/2  2+3(2)=8; 5+3(1)=8; 8+3(0)=8

2f5/2  5d3/2 8p1/2  2+3(3)=11; 5+3(2)=11; 8+3(1)=11

3d5/2 6p3/2 9s1/2  3+3(2)=9;  6+3(1)=9; 9+3(0)=9

3f5/2  6d3/2 9p1/2  3+3(3)=12; 6+3(2)=12; 9+3(1)=12

--------------------------------------------------------------------------

1f7/2 4d5/2 7p3/2 10s1/2  1+3(3)=10; 4+3(2)=10; 7+3(1)=10; 10+3(0)=10

1g9/2 4f5/2 7d3/2 10p1/2  1+3(4)=13; 4+3(3)=13; 7+3(2)=13; 10+3(1)=13

2f7/2 5d5/2 8p3/2 11s1/2   2+3(3)=11; 5+3(2)=11; 8+3(1)=11; 11+3(0)=11

2g5/2 5f5/2 8d3/2 11p1/2   2+3(4)=14; 5+3(3)=14; 8+3(2)=14; 11+3(1)=14

3f7/2 6d5/2 9p3/2 12s1/2   3+3(3)=12; 6+3(2)=12; 9+3(1)=12; 12+3(0)=12

3g7/2 6f5/2 9d3/2 12p1/2   3+3(4)=15; 6+3(3)=15; 9+3(2)=15; 12+3(1)=15

----------------------------------------------------------------------------------------------------

 

Jess Tauber

[tetrahedron]

There is a more general equation that covers all prolate and oblate nuclei that I've been able to tease out of the sequences of shells at different harmonic oscillator values. They depend on relative position within the chain of orbital partials that make up the shells.

 

For the prolate nuclei, we count from the left, so that the first orbital partial (with the largest spin) has positional value 0, the second partial has position 1, the third 2 and so on. Then the oscillator ratio's numerator defines a multiplication factor (one less than said numerator) that creates a subtraction term. So for example oscillator ratio 2:1 has a multiplication factor of 1. Then we multiply this factor against the position value.

 

For the oblate nuclei, we count from the RIGHT, where the spin is minimal in that orbital partial.  The multiplication factor is one less than the DENOMINATOR of the oscillator value.

 

With these two rules, all the Madelung-like equations for simple oscillator ratios are solved (so X/1, 1/Y) and all sums are completely consistent.  I'm currently working on the more complex oscillator ratios (such as 3:2, 2:3, etc.). They appear to use very similar rules to the above.

 

Jess Tauber