The periodic table of chemical elements was invented as an analytical tool
in the late 1860s.
The chemical elements were sorted into order of ascending atomic weight,
with a few minor adjustments to align similar chemical properties.
Immediately, this begged the question that there might be a function, A(Z),
that maps the atomic weight on to the position in the table (the atomic number).

The functions that are proposed on this page, therefore, are hardly a new idea.
However, it was still considered interesting
to put them down on paper, just for further inspection.

## Closest fit polynomials

The most striking feature of a plot of atomic weight, A, against atomic number, Z,
is the tight proximity of all the points to a smooth curve.
Not only are the isotopes of each element restricted to a very narrow band of values,
but also the central value appears not to deviate far
from what appears to be a very simple function.

Moreover, from _{4;2}He to _{40;20}Ca,
the relationship even promises to start as a very linear one.
For each proton added to the nucleus (to increase Z by 1 unit),
a neutron also needs to be added,
thereby increasing A by a total of two units,
to help glue the cluster of positively charged protons together
more strongly than the electrostatic repulsion between them is trying to push them apart.

As Z is increased further than this, though, it appears that each proton
needs fractionally more than just one extra neutron to bind it,
and the curve deviates from the simple straight line of A=2.Z.

It is inevitably tempting to consider adding a Z^{2} correction term,
in the same way the thermal expansion of solids is, at first,
assumed to be modellable as a linear function of temperature,
but then with extra polynomial terms added for greater precision.

The closest fitting quadratic
is: A = 0.00521*Z^{2} + 2.11*Z - 1.05.
Over the range Z=1 to 92,
the widest deviations, from the observed values of mean A for each element,
are +6.11 and -4.74, and the standard deviation is 1.60.

## Assuming that beta decay is proportional to neutron excess

The above quadratic function simple treats the points on the curve as blind statistical data,
and comes up with a clostest fit polynomial, with an associated standard deviation.
The next step is to attempt to include some reasoning,
using whatever else is known about the function,
in an attempt to reduce the amount of unexpected deviation.

It is assumed, here, that there are two main routes
for arriving at _{90}Th from the decay of _{238;92}U:
the direct route of losing a single alpha particle from _{92}U,
losing 4 units of atomic weight;
or else a longer route of emitting an alpha particle then a beta particle,
another alpha and a final beta,
thereby losing a total of 8 units of atomic weight.

Somewhat naively, these two processes (the loss of two or one alpha particles)
can be modelled as a linear function, a weighted percentage in fact:

- A
_{92-2i-2} = A_{92-2i} - 4.( 2.(92-2i) + 1.(100-(92-2i)) )/100

That is, weighting the loss of two alpha particles (plus two beta) proportionately highly
when Z is large (where there are more additional neutrons in the nucleus to eliminate by beta decay),
and the loss of one alpha particle when Z is small
(and the gradient of the curve needs to be closer to dA/dZ=2).

Even this naive function gives a surprisingly close fit,
but can be improved on.
Instead, the following function was used:

- A
_{92-2i-2} = A_{92-2i} - 4.( K2.(92-2i) + K1.(Zr-(92-2i)) )/Zr

Firstly, this recognises that nature cares little about percentages,
and works happily with any weighting base, Zr.
Secondly, it recognises that alpha and beta particles have more than rest mass,
and that they carry away a substantial energy, and hence mass, in addition to their rest mass.
Thus K2 and K1 are constants whose values are close to 2 and 1, respectively.
(It is not necessary to parameterise the value for the mass of an alpha particle, 4,
since the choice of K2, K1 and Zr already covers this.)

With a small amount of experimentation,
extra constraints imposed to make sure that the function gives a sensible value
for the atomic weight of Hydrogen,
the expression was found to give the best fit when K1 has the value of 1.049.
When K2 still has the value 2, the expression gives the best fit when Zr has the value 178,
and there was no way to find a better fit with any other value for K2.

## Generating the function

The above expression is a recursive one, but not a particularly complicated one.
It can be rewritten as:

- A
_{Z} = 238 + ( (K2-K1)(92-Z)^{2} - 2.((92+1)K2+(Zr-(92+1))K1)(92-Z) )/Zr

Collecting terms together then gives
the quadratic function: A = 0.00535*Z^{2} + 2.11*Z - 1.08.

The resultant expression is such a good one
that it is almost as good as using blindly statistical curve fitting.
This might be a confirmation
of the validity of "assuming that beta decay is proportional to neutron excess"
(though it might also be an indication that the introduction of K1, K2 and Zr
allow for a solution that is almost as general as the original one).
The worst-case deviations of -4.6 and +6.6, over a sample of 92 data points,
are not inconsistent with a normal distribution
with a standard deviation of 1.64.
However, it is disappointing that the fit is not better,
which would have been a much stronger confirmation of validity of the model.
It is also disappointing
that it does not facitate the extraction of further information from the model:
the function does not appear to extrapolate to any useful values
(for Z=0, or for Z greater than 92, including Z=178),
or to interpolate to any illuminating insights at Z=17, for example.
Most disappointing of all, the value of K1 differs from unity by 0.0049 atomic units,
which corresponds to about 90 MeV,
but has no obvious mapping on to reality.

## Shell model versus liquid drop model

The previous model has assumed that the number of neutrons in the nucleus
is a quadratic function of the number of protons:

- N = 0.00535*Z
^{2} + 1.11*Z - 1.08
- A = Z + N

In an attempt to find a more realistic function for predicting the number of neutrons,
a cubic function could be tried.
Implicitly, the function used above has assumed the liquid drop model for the atomic nucleus,
in which neutrons and protons somehow are mixed together in an amorphous and featureless way.
The shell model for the atomic nucleus might indeed
immediately suggest certain cubic functions.

The shell model recognises that some atomic nucleii are more stable than others,
with the most stable ones involving so-called "magic" numbers of protons, and/or of neutrons.
These magic numbers are: 0, 2, 8, 20, 28, 50, 82, 126, 184.

Taking the last five to be the numbers contained within the shells 3, 4, 5, 6 and 7,
gives the following function:

- N = ( n
^{3} + 3.n^{2} + 8.n + 6 ) / 3

Taking the first four to be the numbers contained within the shells 0, 1, 2 and 3',
gives the following function, instead:

- N = ( n
^{3} + 3.n^{2} + 2.n ) / 3

It was decided to concentrate only on the former function, for the heavier nucleii,
with a similar expression for the protons:

- Z = ( p
^{3} + 3.p^{2} + 8.p + 6 ) / 3

It is the inverse of this function that is needed, here, though; namely:

- p = cbrt(3.Z/2+sqrt(9.Z
^{2}/4+125/27)) + cbrt(3.Z/2-sqrt(9.Z^{2}/4+125/27)) - 1
- where sqrt is the square-root, cbrt is the cube-root

The final step is to find a function that predicts the value of n from any given value of p
(the number of shells, full and partial, of neutrons for any given number of shells of protons).
For this, another cubic function was at first tried:

- n = a
_{3}.p^{3} + a_{2}.p^{2} + a_{1}.p + a_{0}

However, the best fit was obtained
when the coefficients a_{3} and a_{0} have the value 0,
the coefficient a_{2} has the value 1,
and a_{1} has the value 0.0388.
This reduces the original cubic to the following simple quadratic:

Unfortunately, the final result of the exercise is disappointing.
Comparing the predicted values of A against the actual values of A
for each of the elements up to Z=92,
this new function shows a standard deviation of 1.60,
which is hardly better than that for the simpler function, at the start of this page,
which showed a standard deviation of 1.64.

## Counting quarks

Another attempt at finding a closer fitting function
can be made by returning to the liquid drop model,
and recognising that the nucleus is composed of quarks.

Since a proton consists of two up quarks and a down quark,
and a neutron consists of two down quarks and an up quark,
the overall nucleus has u=2.Z+N, and d=2N+Z.
This merely confirms that A=N+Z=(d+u)/3, and that ε=N-Z=d-u.
The former confirms that there are three times as many quarks as there are nucleons,
while the latter expression says that the number of down quarks in excess to the number of up quarks
is the same as the number of neutrons in excess of the number of protons
(which is, of course, the amount by which A(Z) deviates from the straight line A=2.Z).

By way of an example of the reverse calculation,
a nucleus with 55 up quarks and 62 down quarks must have 16 protons
(Z=(u-ε)/3=(d-2.ε)/3) and 23 neutrons (N=Z+ε).
Similarly, a nucleus with 62 up quarks and 55 down quarks
has ε=-7, and hence Z=23 and N=16.
By the same reasoning, adding a down quark
to produce a hyperthetical nucleus with 55 up quarks and 63 down quarks
(where 55+63 is not a multiple of 3)
would lead to a fractional reduction in Z (=15.667)
and a fractional increase in N (=23.667),
while adding an up quark, instead,
to produce a hyperthetical nucleus with 56 up quarks and 62 down quarks
would lead to a fractional increase in Z (=16.667)
and a fractional decrease in N (=22.667).

Indeed,
u/Z+d/N does turn out to be nearly constant (6.11),
since increases in one term tend to be accompanied by corresponding decreases in the other.
But this merely leads to N-Z=0.33*mean(N,Z)
and thence to 5N=7Z (using mean to be the arithmetic mean instead of the geometric mean),
and A=2.4*Z.
So, this just leads to a crude linear approximation that is worse than the quadratic approximation
that was proposed earlier.
Unfortuately, though, treating u/Z+d/N as a linear function of N
(a straight line from 6 to 6.217 for N=2, helium, to N=146, uranium)
produces an expression that is too sensitive to deviations from the curve
(presumably because of the extremely shallow gradient of the straight line)
and produces quite wild results for A(Z).

## Conclusion

At the very least,
this page has shown how easy it is to have an idea that shows great promise,
that involves a not insignificant amount of work,
and that still fails to work out in the end.
Such sorts of negative result rarely get published, of course, since they are failures,
and normally just get scrapped.
This failed result is offered here, therefore, for what it is worth,
as a step to redressing this bias.