# Atomic Physics - NIST Atomic Ionization Data Patterns

## Surprising Patterns in the NIST Data Regarding Atomic Ionization

First placed on the Internet in June 2007

Many amazing statistical patterns exist in the highly respected NIST data regarding the Ionization Potentials for the elements! It is really obvious, and unavoidable, but it seems to contradict all conventional explanations! It even appears to contradict Coulomb's Law!

Consider this: Coulomb's Law says that the electrical attraction between two oppositely charged particles is DIRECTLY proportional to the charge of each of the two (as well as a dependence on the inverse square of the distance between them.) But in 1913, a scientist named Moseley discovered that the radiation created by different elements is proportional to the SQUARE of the electrical charge of the nucleus. Consider the simplest possible situation inside an atom, a single negativelly-charged electron orbiting a (very tiny) nucleus which has a single positive charge in it. This is a standard hydrogen atom. There is obviously some energy of attraction between the opposite charges, and we will call that 13.59 electron-Volts. Now, let's consider the same situation, but where the nucleus contains TWO positive charges, but there is still one electron orbiting it. This is an atom of helium which has been once ionized. Coulomb's Law indicates that the attraction between the electron and nucleus should be TWICE as great, right? But when this is actually measured, it turns out it is actually FOUR times as great, extremely accurately (54.42 electron-volts).

We might think that was just a coincidence, because we KNOW that Coulomb's Law is true! So we now examine a twice-ionized Lithium atom, where a single electron orbits a nucleus with THREE charges in it. Is the attraction three times? No. You probably already guess that it is experimentally shown to be NINE times, again, extremely accurately (122.45 eV). And a triply-ionized Beryllium atom (one electron circling a nucleus with four charges in it) is experimentally shown to have 16 times the binding energy (217.72 eV). And a quadruply-ionized Boron atom has 25 times (340.23 eV). These are all remarkably accurate multiples of the hydrogen binding energy.

This is NOT an accident! Researchers have managed to check 32 different elements where there is a single electron orbiting the nucleus, and they all very accurately follow this same pattern, where even a 31-times-ionized Germanium atom has 322 or 1024 times as great a binding energy of that single electron (14,119.4 eV).

This all absolutely contradicts the basic claim of Coulomb's Law! And not just in random ways but in very precise ways.

These findings also seem to provide an actual physical meaning behind the equations used for the Balmer series of lines of Hydrogen and the Rydberg equation, as well as suggesting how Rydberg could be extended to essentially any element in any ionized state!

They also conclusively prove that the (alleged) Quantum Defect, a number that has always believed to be a random correction, is actually a specific and precise number, accurately specified by equations! No one else has ever noticed that before!

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This presentation was first placed on the Internet in June 2007.

This analysis shows that the concepts can also be extended to describing atoms with multiple electrons, with all the equations still accurately applying. And further, the entire structure of the atomic shells and sub-shells, are all clearly predicted by surprisingly simple mathematical equations.

It certainly appears that Coulomb was wrong regarding the electrical attractions inside atoms!

Each neutral element has equal numbers of protons in the nucleus and electrons orbiting it. We are going to look at this situation in the old-fashioned traditional way, where discrete electrons were actually seen as orbiting the nucleus.

Atoms can be ionized in two distinct ways. The most common experimental way is where external energy is provided and an electron is kicked up into a higher energy state orbital (of many possible choices) and then it soon drops back down and in that process radiates away the exact amount of energy difference between the two energy states. The wavelength of that emitted light photon can be used to determine the exact amount of energy involved in such transitions. NIST has data for all neutral elements and many already ionized elements, based on the wavelengths emitted. In some cases, there are as many as 780 different possible energy states in a given element, and so an immense number of different wavelengths of light can be emitted as an electron falls from one of those states to a lower energy one.

The second distinct way is where the electron is sent to an infinite distance from the nucleus, in other words, entirely removed from the atom, and which is the greatest possible amount of energy which can be inserted in that electron. Effectively, this is the common situation when chemical reactions occur.

The exact wavelengths of the resulting radiation from all elements have been extremely accurately determined. This means that the exact amount of energy transfer regarding any of these ionizations is also known extremely accurately.

There are an immense number of ways that the NIST data might be studied. The transitions from one energy state to a lower energy state must certainly include many valuable insights, but there are a number of variables involved, including for example a very large number of possible target energy states. The single electron in a hydrogen atom is known to have at least 187 raised energy levels available to it (which is known by the fact that there are at least 187 different wavelengths of radiation emitted when the electron later falls back to its neutral state). This also enables a hydrogen atom to emit many thousands of different wavelengths of light, as the electron falls from one of those states to another of them. For now, we choose to avoid this complexity and only consider electrons that are ionized to an infinite distance.

The electrons in atoms are in many different quantum orbitals and sub-orbitals, and the total effect of many of them on each other figures to be quite complex, so we wish to temporarily eliminate this complexity as well. It turns out that we can do that! It turns out that researchers have discovered and experimentally analyzed 32 different atoms which each have a single electron orbiting the nucleus.

The hydrogen atom has one electron. NEUTRAL helium has two, but if the helium is singly-ionized, it also has one electron. NEUTRAL lithium has three electrons, but doubly-ionized lithium has one. NEUTRAL beryllium has four electrons but triply-ionized beryllium has one. And so on. The discussion here will stop at element 32, Germanium, which is 31-times-ionized, such that it too only has one remaining electron.

Therefore, we have simplified the situation into one where there only is a single electron (we will not now specify any shell or sub-shell) orbiting around various nuclei which have (attractive) (integer) positive charges of from one to thirty-two.

If we look at Coulomb's Law regarding these 32 atoms, which are essentially identical except for the charge of the nucleus, we might then assume that the attractive force acting should be in the ratio of 1:2:3:...:32. That is NOT the case!

The EXPERIMENTALLY MEASURED NIST data on these atoms has an amazing pattern in it! Consider for a moment only the H I and He II atoms. The NIST ionization potentials, in electron-Volts, are 13.5984340 and 54.4177630, meaning that these values represent the actual binding energy of the single electron for these two cases, as compared to zero, the potential at infinite distance.

This is quite close to being a 1:4 proportion, within 1/20 of one percent! We might them look at the Lithium II value of 122.454353 and see that it is remarkably close to NINE times the hydrogen value, again around a match of around 1/20 of one percent. Could this be some sort of fluke, where the universally accepted experimental data seems to have a remarkable match with the square of the electric charge of the nucleus? No! The match continues, consistently amazingly accurate, throughout the entire sequence of these 32 elements.

Here is the NIST data (in electron-volts) and the nucleus-charge-squared multiple of hydrogen's.

 nuclear charge NIST ionization value multiple of hydrogen 1 13.5984340 13.5984340 2 54.4177630 54.393736 3 122.454353 122.385906 4 217.718572 217.574944 5 340.225993 339.96085 6 489.99312 489.543624 7 667.04602 666.323266 8 871.40969 870.299776 9 1103.1171 1101.473154 10 1362.1986 1359.843 11 1648.70105 1645.4105 12 1962.6642 1958.1745 13 2304.1401 2298.1353 14 2673.1807 2665.2931 15 3069.84143 3059.648 16 3494.1877 3481.199 17 3946.2907 3929.947 18 4426.2226 4405.8926 19 4934.0439 4909.0347 20 5469.8614 5439.3736 21 6033.7551 5996.9094 22 6625.81 6581.6421 23 7246.1196 7193.5716 24 7894.80 7832.70 25 8571.94 8499.02 26 9277.6874 9192.54 27 10012.1 9913.26 28 10775.4 10661.17 29 11567.612 11436.28 30 12388.928 12238.59 31 13239.4881 13068.10 32 14119.4287 13924.80

This is very troublesome! By doubling the charge in the nucleus, and being careful not to change any other variables, standard understandings in Physics do NOT allow any conclusion where the Binding Energy becomes four times as great!

However, this information is in exact accordance with the Balmer:

and Rydberg:

with the special case for Hydrogen Balmer being:

mathematical equations for spectral lines. (These equations are presented in their traditional form, for solving for the wave number nu. By multiplying that value by Planck's constant, the energy is determined, so we could have presented these to solve for energy in electron-Volts by that simple multiplication.)

This investigation seems to give a physical reality to the mathematics of those equations. A critical point is that it has generally been assumed that all the numbers in the denominators were DISTANCES (because of the well-known inverse-square distance rule of electromagnetic phenomena), but this investigation seems to definitely prove that the a and b parameters are rather the charge in the nucleus! The fact that these quantities of charge need to be squared for these equations appears to contradict Coulomb's Law. (The Rydberg constant is 109,677.8/cm or 13.5978 eV equivalent, the ionization potential of neutral hydrogen.).

For clarification, the first term of the Rydberg equation describes the initial energy state of the electron and the second term describes the final energy state. The initial state is therefore described by R / (n + a)2. For Hydrogen, we are describing here that a is the reciprocal of the nuclear charge of one (which is still one), and the first n is zero, such that we have the energy become 13.6 eV / (12) or 13.6 eV. Regarding the final state being at infinite radius, we have m is infinite and the energy is 0 eV.

Here are these values graphed, which shows the parabolic shape of the curve.

And with our calculated parabola of EXACT INTEGERS in yellow:

However, this seems to represent an immense problem! It seems indisputable that this is a remarkably pure parabola. That means that the BINDING ENERGY is in extremely consistent relationship with the SQUARE of the total charge of the nucleus. No other variable exists in this problem, since we made sure to only have a single electron orbiting the nucleus. We are remembering that the nucleus is extremely tiny as compared to the electron orbits, being on the general scale of 1/10,000 the diameter. This is equivalent to the nucleus being the size of an American dime and the single electron orbiting around a block away. Other than the charge of the nucleus, not many other aspects of it can have significant effect in this problem. Gravitational effects are impossibly small.

Traditional thought would say that with double the central electrostatic attraction, then the attractive force, that is, the centripetal force must also be exactly doubled. But then all the traditional analysis of energy and angular momentum do NOT result in the energy going up as the square of that centripetal force! But it clearly must, per this parabola!

Here is the general reasoning that has always been applied, for each of planets in a gravitational field and electrons in an electrostatic field.

Coulomb's Law says that the force of attraction between two charged objects is given by F = k * q1 * q2 / r2, where the q's are the amounts of charge and r is the distance separating them (with k a constant depending on the system of measurements). Newton's Law similarly says that the force of attraction between two massive objects is given by F = G * m1 * m2 / r2, where the m's are the amounts of mass and r is the distance separating them (with G, the Gravitational constant, depending on the system of measurements).

Newton's Law F = m * a for central force for circular motion is F = m1 * v2 / r. This can also be written F = I * omega2 or I * v2 / r2.

It might also be speculated that the radius of the orbit of the electron would adapt itself to account for this different force. However, calculations show that that is not a viable explanation. Yes, the orbital radius WOULD change to some extent, but it is physically impossible that it could change enough to account for this full effect. An easy proof of that is that the Binding Energy is known to be in the inverse proportion to the orbital radius. We have just been discussing a Germanium XXXII and a Hydrogen I, which each have one electron. The NIST data shows that the Ge atom has a binding energy of over 1,000 times that of the Hydrogen, meaning that its orbital radius would necessarily be 1/1,000th the radius. That electron would be extremely close to being inside the nucleus!

But the standard calculations do not even result in this solution! There is NO conventional way where multiplying the charge of the nucleus by some integer can possibly result in the Binding Energy increasing by the square of that integer! At least, not if Coulomb's Law is assumed to still apply.

## Ions that have Two Electrons

 nuclear charge NIST ionization value Second NIST value 2 24.5873876 3 5.3917191 4 18.211153 136.80273 5 37.930620 205.99 6 64.49390 7 97.89013 8 138.1196 9 185.1868 10 239.0969 11 299.864 1407.705 12 367.497 1698.61 13 441.999 2017.0 14 523.4203 2362.8411 15 611.74 2736.31 16 707.01 3137.35 17 809.2129 18 918.3861 19 1034.5 20 1158 21 1288.0 22 1425.4 23 1569.6399 24 1721.1 25 1879.8 26 2045.7391 27 2218.9

## Ions that have Six Electrons

 nuclear charge NIST ionization value 6 11.26030 7 14.53413 8 35.12111 9 62.7084 10 97.1168 11 138.40 12 186.76 13 241.76 14 303.5381 15 372.13 16 447.46 17 529.2761 18 618.73 19 714.6 20 817.7 21 927.5 22 1043.947 23 1167.931 24 1299 25 1437 26 1575.5911 27 1735 28 1894 29 2060.6172 30 2234.1951 31 2418.9316 32 2665.6601

When an r-squared analysis was done of these various data sets, it seems that a "pure" parabolic shape was very close to the absolute best curve-fit. Of around a hundred common simple equations, the y = a + b * (x-c)2 (a pure parabola) was generally the best, and even when it was not, the r-squared was still incredibly close to 1.

Here are the three graphs above repeated, and some other numbers of electrons, but with the statistical data and the parameters now included and the parabolic curve-fit included:

And here are the Residuals for each of these curve-fits, which each show interesting patterns:

It seems appropriate to show the data for six-electrons again. You probably noted that the Residuals for six- had a large excursion outlier at a nuclear charge of 31. Look at the red dot on the following graph of six- again, to see that even that extreme outlier is actually quite close to the parabolic curve. This demonstrates just how good these parabolas are.

## Apparent NIST Errors in the Data

Some of the analysis curves Residuals seem to show errors in specific values of the NIST data. Some examples are obvious in the Residuals graphs above. Here are some more examples. They not only show that a specific value must be wrong, but they also suggest what the correct value should have been! There are also missing data in the NIST records, such as for the elements Astatine(33) and Bromine(35). These analysis graphs seem to provide accurate values for those missing NIST data points.

Look at element 26 and element 32.

Look at element 26.

Look at element 20.

The parabolic equation y = a + b * (x-c)2 has three variables a, b, and c. a is simply the vertical offset of the bottom of the parabola from the x-axis. c is simply the horizontal offset of the centerline of the parabola from the y-axis, which is essentially the number of electrons. So the b parameter is the really important one, which defines the shape (width) of the parabola.

These graph analyses present some amazing results! We will collect numbers from these parabolic graphs. For non-statisticians, the r2 value is an indication of how well a mathematical curve fits the data, with 1.000 meaning a perfect fit. These are mighty close, meaning that the data is truly extremely close to parabolic!

electronsabcr2
one 19.313.938338 0.210.99999693
.
two 6.3 3.53120931.940.99999095
three -1.5-3.44428062.220.99999944
four -2.3-3.46427553.210.99999939
five -4.4-3.44782633.840.99999941
six -12.8 3.38795344.310.99996958
nine -18.1 3.41651066.640.99999867
.
ten 4.1 1.601446 8.600.99999826
eleven 8.4 1.57284558.810.99998215
twelve-10.01.54217359.860.9999791
seventeen -24.9 1.5293853 12.960.99997388
.
eighteen -15.7 -1.6134294 15.790.99955473
nineteen -9.0 -1.7198515 17.200.99981609
twenty -1.6 -2.1111569 19.140.99991352

All of the electron numbers from two through nine have a b parameter very close to 3.45. These match up to the Quantum level 2 electrons, and it is surprising that they share that same parameter. Starting with ten electrons, that b parameter suddenly and consistently drops to around 1.57, which seems to be true for the Quantum level three electrons. The single electron situation with the b parameter of 13.9 matches up to the Quantum level 1 of Hydrogen. Note that TWO electrons seems to belong with the next quantum level, implying that the Periodic Chart may be wrong! It could be that each shell starts out with an EMPTY state as the first entry, meaning the Noble Gases might belong on the LEFT side of the Periodic Table rather than the right where they have always been shown (as being FILLED shells). In the same way, TEN electrons appears to belong with the next shell!

These results show parabolas that are NOT tangent to the x-axis, having a vertical differential equal to the a parameter. The graphs were curve-fitted with a requirement that there was that tangent requirement, but every such curve was substantially worse than the ones presented above. Here is one example, for 12-electrons.

The r2 value is good, but not nearly as good as the parabola presented well above which would tangent around the -10 eV line. That seems to suggest that it is not appropriate to force the parabolas to be tangent to the x-axis.

This might suggest that there is some physical importance in the fact that the a parameter is not zero. In fact, in progressing from 2-electrons to 9-electrons, it is clear from the chart above that the a value constantly becomes more negative. The same trending toward negative a seems to exist starting with 10-electrons. There may be a physical meaning for this.

It is probably silly to note a potential relationship between these b parameters. If we say that the first value is 14, then the second shell parameter is pretty close to 14/22 or 3.5 and the third is pretty close to 14/32 or 1.56. It will be interesting to see if the next shell (possibly beginning with 18 electrons) might have a b parameter of 14/42 or 0.875, and the shell after that (beginning with 36 electrons) having 14/52 or 0.56

There does not seem to be any logical reason for this to be the case, and I am merely noticing a possible pattern in the data numbers.

Certain errors in the NIST data for specific ions probably cause some variation and the consistency may even be better than that. The amazingly high values for r2 for each graph show the very good curve-fit.

Therefore, it seems that a simple equation certainly exists which takes the (number of electrons) and (number of nuclear charges) and gives the Ionization Potential in electron-Volts!

For real ions, the number of electrons cannot be greater than the number of nuclear charges, although this equation seems to allow certain limited exceptions. (notice the Five- and Four-electron parabolas which clearly extend to the left of the physically possible numbers). This might aid in explaining how and why some chemical molecules can form. The upward curve of the graphs at the left might be describing the REAL situation when an atom is in a compound. In salt, a neutral Chlorine has 17 electrons, but Chemistry has taught us that the single electron that a Sodium atom has handy can "complete the shell" for the Chlorine atom. That is, the Chlorine atom technically has an extra electron in that compound. These graphs might provide some insight into which compounds are possible and what energy amounts are required for each bonding.

I don't know if this is credible reasoning or not, but it seems an interesting possibility. It might help explain a lot about electrochemical processes.

The 10- data actually includes three separate curves! They have very different b parameters. This might indicate the distinction of whether the electron emitted is a 1s electron, or a 2s or a 2p electron.

Keep in mind that most of these atoms have already been multiply ionized, meaning that some electrons are already gone. The point being made here is that it seems possible to know the SEQUENCE of where the electrons are emitted from! A Ca XI ion has already given up ten electrons, suggesting that all the 1s, 2s, and 2p electrons are already gone. Therefore, the only electrons that should still remain would be 3-level electrons, and the graph (10 C) seems to show that.

There are clearly complex changes that occur within the electron movements, which apparently are needed to try to describe how S VI would be able to expel a 1s electron which would normally be expected to have been long gone.

Looking again at the form of the Rydberg Equation, the first term was described as E = 13.6 / (n + a)2 (in eV). We see that a is the repiprocal of the charge of the nucleus. For the first several terms of our first (one-electron) graph, we have 13.6/(1/12) or 13.6; 13.6/(1/22) or 13.6/(1/4) or 54.4; 13.6/(1/32) or 13.6/(1/9) or 122.4; 13.6/(1/42) or 13.6/(1/16) or 217.6; 13.6/(1/52) or 13.6/(1/25) or 340

The present analysis has shown the physical reality of the a parameter in the Rydberg Equation, being the nuclear charge. The n and m parameters appear to still be distance/radius, which brings up an interesting question regarding how an equation could mix and match charge and distance! There appears to be a level of understanding still beyond us all!

This presentation was first placed on the Internet in June 2007.

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C Johnson, Theoretical Physicist, Physics Degree from Univ of Chicago