Atomic radii

  • Thread starter elas
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  • #1
Can anyone tell me where to find Atomic Radii for Isotopes?
(All the tables I have found so far only list atomic radii for the 92 elements shown on the main table.)
 

Answers and Replies

  • #2
MrCaN
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you can calculate them, roughly by multiplying the contents radii
 
  • #3
You want atomic radius, not nuclear radius, right? That's almost exactly the same for all isotopes of a given element.
 
  • #4
Yes, because the charge of nucleus is the same for all isotopes of given element, wave function of electrons spreads same way.
 
  • #5
Perhaps I should be a little more frank. I have found a formula for producing Atomic Radii from mass and as Emsley,s 'Elements' and all the tables on the net do not give Atomic radii for isotopes I am hoping someone can give me an accurate figure so I can check my predictions.
As far as I am aware there is no formula for producing atomic radii accurate (to 4 decimal places), but as I am an amateur with no formal training I would appreciate comfirmation of that.
I seek to explain why element 92 is almost 300 times the mass of element 1 but only three times the size, according to Enc. Brit. this is one of the unsolved prolems of particle physics.
 
  • #6
marcus
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Originally posted by elas

I seek to explain why element 92 is almost 300 times the mass of element 1 but only three times the size, according to Enc. Brit. this is one of the unsolved prolems of particle physics.

I just consulted the CRC "Handbook of Chemistry and Physics" table of "Ionic Radii of the Elements"

this is not exactly what you were asking about because an ion, being charged, is not quite the same as a neutral atom

but it has at least some relation to your interest, I suppose.

All the figures were around one angstrom but they varied in what seemed like a quite irregular fashion.

A closer look showed more regularity if one considers only those ions with the same charge. For example uranium(+4) is the uranium atom missing 4 electrons and its radius is 0.97
and this can be compared with carbon(+4) also missing 4 electrons, which has radius 0.16

Here is a little table for you with some sizes of (+4 charged) ions:

carbon (+4) 0.16

silicon (+4) 0.42

germanium (+4) 0.53

tin (+4) 0.71

lead (+4) 0.84

uranium (+4) 0.97



Please quote an exerpt from the Encyclopaedia Britannica article which
you are using. I am curious to know the exact wording.
If the Britannica says something like this in connection with
particle physics (as contrasted with the physics of the atom)
then it suggests that they might have be talking about the
size of the NUCLEUS---damgo raised this issue.

Sorry i don't have information on the size of neutral atoms----tho I think they are all around one angstrom. Good luck in your search!
 
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  • #7
Originally posted by elas As far as I am aware there is no formula for producing atomic radii accurate (to 4 decimal places), but as I am an amateur with no formal training I would appreciate comfirmation of that.



I am curious if atomic or ionic "size" can even be defined to 3 digits accuracy. It is just a spread of outer electron wave function which strictly speaking is infinite.



I seek to explain why element 92 is almost 300 times the mass of element 1 but only three times the size...


Just because its outer electron does not see U nucleus as a charge +92, (thus U atom is not 92 times SMALLER than H one), but only as a charge about +2 - +3 or so.

If not Fermi repulsion of electrons, U atom would actually be 2-3 times smaller than H atom.



according to Enc. Brit. this is one of the unsolved prolems of particle physics.

Well, it is hard to calculate a wave function of an electron in multielectron atom with high precision. Too many of other electrons and of their messy mutual interactions to account for.
 
  • #8
marcus Many thanks, this is getting close but unfortunately I had isotopes with mass but not radii, you have given radii but not mass.
Although it is not quite what I wanted it is in an area that I intended to tackle next because it shows the relationship between electrons and atomic radii. The change is in the expected direction but larger than expected. This supports those who claim that atomic nuclei sizes are related to the number of nucleons. Therefore changes in atomic radii are due mainly to changes in electron numbers. Which is another way of saying that quarks and electrons play the major part while the other constituents of the nucleons do no more than provide the basic structure.
 
  • #9
marcus
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Originally posted by Alexander
I am curious if atomic or ionic "size" can even be defined to 3 digits accuracy. It is just a spread of outer electron wave function which strictly speaking is infinite.
Just because its outer electron does not see U nucleus as a charge +92, (thus U atom is not 92 times SMALLER than H one), but only as a charge about +2 - +3 or so.
If not Fermi repulsion of electrons, U atom would actually be 2-3 times smaller than H atom.
Well, it is hard to calculate a wave function of an electron in multielectron atom with high precision. Too many of other electrons and of their messy mutual interactions to account for.

Alexander these are thoughtful and interesting comments!
As for the ability to define the ionic radius----it is an experimentally measured dimension specific to ionic crystals
as I understand it.

One measures a crystal and one counts the atoms in it and one determines the lattice spacings----then somehow one decides
on the ionic radius.

And in that case it must vary some from one sort of crystal to another. this at least is how I imagine the thing is defined and measured.

It seems significant to me that instead of calculating the ionic radius from QED and worrying about the fuzziness of the electron clouds one can simply put a bunch of atoms together in a crystal and SEE how close they pack and therefore---in a practical sort of way---how big they are
 
  • #10
Well, both ways you get some fuzzines. Experimentally, say, Na in its own metal crystal and Na in table salt crystal have completely different wave function of outer (S) electron and slightly different p electron wave function, thus slightly different "size" of ion no matter how you define this size. Simply because it depends on environment. So I doubt that experimental "size" is more stable better than 2 digits.

Theoretical "size" (say, distance from nucleus ro which splits volume integral square of wave function 50/50 (inside/outside)) is calculable well (to 3+ digits) only for H atom (and H-like ions). Even He wave functions are so messy to calculate that various methods give various results in 3rd digit.
 
  • #11
This is where my ignorance shows up, some tables show the measurement with a decimal point and some show the same number without the decimal point, which I assume means they are using different units of measurement.
However the key point is that, as far as I am aware, the measurements are found by experiment and no explanation has been put forward as to why most increases in the number of sub-atomic particles in an atom lead to a reduction in radius and only a few lead to an increase in radius.
I believe I am also correct in saying that to date there has been no explanation of the relationship between mass and radius but only between mass and density and I would like comfirmation of this please.
The relationship I discovered follows on from my discussions on the physics theory pages and from the work of rduncan and cannot be transferred to the physics pages which are correctly restricted to the standard model. That said it is necessary to come to the physics pages for advice and I greatly appreciate the advice received so far. Further clarification is requested on the points raised.
Many thanks
elas
 
  • #12
I should add that I am using The Elements by John Emsley. Radii are given in three forms 'atomic', 'covalent', and 'van der Waals'
 
  • #13
Originally posted by elas


...However the key point is that, as far as I am aware, the measurements are found by experiment and no explanation has been put forward as to why most increases in the number of sub-atomic particles in an atom lead to a reduction in radius and only a few lead to an increase in radius.


There is explanation, and you don't have to go futher than a good physical chemistry or good quantum chemistry textbook.

Explanation is quite simple - interplay of coulomb attraction to partially screened by other electrons nucleus and of Pauli repulsion of electrons.

In horizontal direction (from left to right) in Mendeleev's periodic table screening is decreasing (say, outer s- electron of Na atom sees nucleus as +1, but outer p-electrons of Cl see nucleus as +3 due to much less screening by other p- neigbor electrons than by inner electrons. That is why, by the way, Cl atom sucks electron so strongly that it tears it almost from any other atom - and we call such behavior of Cl (and F, Br, O) as "strong oxidizing".

So atoms gradually become smaller and smaller as you move from right side of periodic table to left side - due to increasing Coulomb force on outer electron.

When the shell is full (noble gases), then next electron has to start new shell and Pauli repulsion from electrons in previous shell makes its wave function way bigger. So from top to bottom of periodic table you have gradual increase in atomic radii.

Competition for lowest energy between pairs of p and d subshells, and d and f subshells belonging to different shells contributes into non-monotomy of radii charge of heavy atoms.
 
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  • #14
Alexander Thanks for reply. I checked with local library and it is not in the books I have access to.
I gather from the general tone of the replies that there is no firm statement of Isotope radii. As Emsley gives the radii for set conditions I assume that any isotope radii I produce using Emsley's data, would be subject to those conditions and I intend to see how far I can go in that direction.
I will post the results (or lack of results) within two weeks.
elas
 
  • #15
Tyger
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Condider this Elas

The size of an atom is determined by the outermost electron orbitals and is independant of the nuclear composition. The isotope data isn't given because it doesn't affect the size and therefore isn't important. That should answer your question.
 
  • #16
Tyger
You have made the one statement that I can prove to be incorrect, is this your own opinion or is it the accepted view? If it is the considered opinion of experts can you please supply a reference.
elas
 
  • #17
Tyger
398
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It is correct

Maybe it could be worded a little better. The number of neutrons in the atoms does not affect the size of the outer orbitals, which is what determines the size of atoms in a crystal lattice. Simple as that. An atom of U238 is the same size as U233 or U235, and that is why the data isn't supplied.
 
  • #18
For all neutral atoms, the outer electron has roughly the same distance from the nucleus (about 1-2 Bohr radii). This is because if you have for instance an atom with 50 protons and 50 electrons, 49 of the electrons will shield 49 protons, so the situation is similar to hydrogen with 1 electron 1 and proton. The exact value varies within a factor 2 or so because of the interaction of the electrons (however, for highly excited states (where the outer electron is very far away from the nucleus) the coincidence is almost exact).

The same should also hold for a given degree of ionization, i.e. atoms that have the same number of electrons missing, although here the outer electron will be closer because of the stronger Coulomb force.

In any case, the number of neutrons (i.e. the isotope) can not possibly have an effect on the atomic electrons as they don't interact electrostatically.
 
  • #19
Tyger
As this is 'Physics Forum' and not the 'New Theory Forum' I cannot go into my reasons other than to say I am making a case for a different interpretation of the accepted mathematical theory (I do not challenge the maths).
So far I have work on elements 1 to 19 and I am convinced your statement is incorrect. I will look at the elements you quote before going further.
Meanwhile I would very much appreciate a clear statement as to whether this is your personal answer or the establishment answer. I need to know if I am treading on new ground?
 
  • #20
Elas, think: neutrons are NEUTRAL. How can they change wave function of electron? They can't. So why shall isotopes of same element have different wave function of outer electron?
 
  • #21
I realize that I have ignored those who mention waves.
My earlier work indicated a relationship between mass and waves, the current work indicates (no more than that) a relationship between radius and waves.
It would appear that mass radius and wavelength are related in groups of eight divided into two sub-groups of four. I have only in the last hour, realized that this sub-grouping applies to elements and isotopes as well as waves (which I was aware of earlier), I will have to revise the work done so far.
This means taking a break from the discussion and getting on with the article. I will post it on PF as soon as I am finished
 
  • #22
I am not talking about waves in general, but about a wavefunction of outer electron. Basicly by "size of atom" we mean a "size" of outer electron's wave function, right?
 
  • #23
Alexander
I believe I know what you are telling me but, I use the atomic radius because this is a half of the distance between two atoms in a crystal and therefore is a precise distance providing all elements (or as many as possible) are measured under the same experimental conditions. This is one measurement that does not require the introduction of energy and is therefore made with less disturbance of the atoms than is the measurement of wavelengths.
The electron wavelength is I believe a concentric measurement as is (as far as I am aware)all atomic wave measurements. This ignores the case for a perpendicular wave structure that of course, cannot be directly observed, but can be deduced indirectly.
I believe that the (undisturbed) atomic radius can be combined with mass and the known radii of the elements to predict the radii of isotopes which can then be tested by experiment, if found to be correct my case is made.
The work is very simple but also very tedious. Having jump the gun to early on several occassions I am determined to say nothing about the details until I have finished, checked, and double checked.
I did not intend my request for information to lead to a full scale debate because I am clearly not ready and also on the wrong forum for a new theory debate. None the less I am very grateful for the advise and comments received,
elas
 
  • #24
I hope that you at least realize that this "radius" does not depend on isotope mass (=all isotopes of the same element have the same "radius").
 
  • #25
Alexander
Appreciate your help and understand your reply but I am looking for the origin of what at present appears to be an assumption.
Given that the electron orbit is determined by magnetic force, why should the addition of a neutral particle alter the orbit but, as mass is being added, why should the volume remain the same?
This can of course be explained as an increase in density but I cannot find any reference to such an increase and even and if I accept a change in density and no change in radii this begs the question 'why does a change in density not alter the electron orbit'?
Given that some isotopes are radioactive (when their associated element is not); clearly there are occassions when the addition of a neutron does alter the magnetic relationship.
 
  • #26
Chemicalsuperfreak
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Originally posted by elas
Alexander
Appreciate your help and understand your reply but I am looking for the origin of what at present appears to be an assumption.
Given that the electron orbit is determined by magnetic force, why should the addition of a neutral particle alter the orbit but, as mass is being added, why should the volume remain the same?
This can of course be explained as an increase in density but I cannot find any reference to such an increase and even and if I accept a change in density and no change in radii this begs the question 'why does a change in density not alter the electron orbit'?
Given that some isotopes are radioactive (when their associated element is not); clearly there are occassions when the addition of a neutron does alter the magnetic relationship.

If a hydrogen atom was the size of a football field, the proton would be the size of (roughly), a pea. If you put another pea in the field you've doubled the mass but the football field doesn't get any bigger.

You've answered your own second question. The size of an electron's orbit is measured by the electomagnetic force, the mass of the nucleus doesn't have anything to do with it. I don't see the source of your confusion.

What do unstable nuclei have to do with the electromagnetic force and orbitals?
 
  • #27
Your analogy is poor, remove all the footballs and you still have an unchanged football field; remove all the nucleons and there is no atomic field. Regardless of the difference in size the atomic field is part of the nucleons and as the nucleons decrease so does either the field radii and/or the mass.
If the Earth escaped from the gravitational field of the sun it would take the Earth's gravity field with it; but you cannot say that an atom's field is a distortion of spacetime without saying spacetime is magnetic and clearly that is ridiculous. The alternative is that nucleons take something with them when they leave. That is only possible if you think in terms of force fields and not solid objects such as footballs.
 
  • #28
elas, do you know what isotop is and what element is?

Do you know, for example, what is the difference between U238, U235, Pu238, Pu235?
 
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  • #29
Three neutrons!, but it seems the difference between element and isotope is really a matter of commonality and that is hardly a good scientifc reason for giving things different names. In reality they are atoms and is is by taking all atoms (element or isotope) into one table that we see the fundamental structure of all atoms.
The beauty of ignoring clasification and using an all atoms table is that it is then possible to see the structural link between atoms and sub-atomic particles and also between mass and waves. The only problem is the tedious work of writing up the tables, but I have done enough to see that I am on the right track.
I realize that I lack the training to explain myself in the proper terms, that willl come out in the wash, the beauty is in the simplicity of the numbers and I hope to get to the end soon.
 
  • #30
Ok, so all isotopes of the same elment have SAME number of protons, thus SAME electric field everywhere (except small difference inside nucleus), thus SAME conditions for electrons. Thus, electrons in these similar condidions behave similar (=have SAME wave function for all isotopes of SAME element). Thus radii of all isotopes of the same element are SAME.
 
  • #31
So how do you allow for the difference in mass?
 
  • #32
You use the reduced mass, m_e*m_nucleus/(m_e+m_nucleus) . Since the electron is so much lighter than the nucleus, this is a very small correction; and so the correction for different isotopes is a *very*-small second-order effect. Even for say hydrogen vs. tritium, where it will be most noticeable, you get

hydrogen: m_e*(1-0.00054)
tritium: m_e*(1-0.00018)
 
  • #33
I agree the difference is very small, but I found a method of determining the difference in radius caused by this small difference in mass. When I tried to discover if I had something new I was met with several replies stating that there is no change in radius. I still have not had any authoritive reference for this claim.
Meanwhile I am continueing with my work as I hope other structural relationships will come to light, but all replies are much appreciated,
elas
 
  • #34
well, because the difference between psi-function of various isotopes of the same element is in 5-th digit only, make sure that your semi - empirical formula can fetch that far.
 
  • #35
Creator
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Originally posted by elas

I seek to explain why element 92 is almost 300 times the mass of element 1 but only three times the size,...

Elas; there are several considerations that need to be addressed in trying to develope an empirical formula for atomic size, which make it almost prohibitive.

Ist, element 92, for example, has 92 positive charges accelerating each electron, which, you would think, should actually shrink the orbits. However, there is something else that you forgot to take into account.
There is something called "shielding".

This, in effect, is due to successive 'layers' of electron orbitals at various distances from the nucleus. The innermost electrons actually 'shield' successive outer electrons from the full electrostatic effect of the nucleus. This effect, the combined effect(and its magnitude) is probably the hardest thing to quantify especially at higher atomic numbers.

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