Question about calculating interatomic distances in metals

[Snark]

Various content has been redacted and/or removed.

Folks, please be civil.  It's entirely appropriate to engage in lively debate about matters of scientific fact; if you think someone else is incorrect, by all means point that out, and give your reasoning & evidence why.

However, it is never okay to engage in trolling, personal remarks, or insults, per forum rules 2.2.d and 2.2.n.  Please address the post, not the poster; if you can't win your argument or make your point without resorting to name-calling, finger-pointing, or outright mockery, then perhaps best just to stroll on by.

Thank you for your understanding.

[K^2]

6 hours ago, kerbiloid said:

The simple method brought by me has given a ~20%-accurate solution for a qualitative assessment task from the 1st year of the universe after a minute of googling for data and calculations.

(2.55-2.28)/2.55 *100% = 10.5%.

2.55 is minimal distance, not average distance. Your "average" is 10% under the SMALLEST distance there is. You still aren't giving an estimate for what you are promising to estimate.

Again, if you want to see how it fails miserably, run through the exercise with your own example. Place 10 people into a 1000m³ house so that the average distance between people is (1000/10)^1/3. Sure, do it with a 20% error if you want. You literally just need to place 10 dots on the graph to resolve this whole thing.

[mikegarrison]

Wow, this all went in a direction I didn't foresee,

Anyway, with crystalline solids I think @K^2 is right. Obviously with gasses or liquids or other kinds of solids, the answer is different.

@ARS asked about "a metal compound", and that's a problem because of the word "compound". If it was just intended to mean "substance" then perhaps it's a pure metal. If it was intended to mean a compound of more than one metal (or even a non-metal, like carbon), then it's an alloy and in that case sometimes you have atoms from one material replacing atoms of the other material in the crystal, but sometimes you have atoms from one material squeezing into the spaces between the atoms of the other material.

Heat treatment of metals (especially steel) is about forcing one crystal structure or another, or one type of alloy or the other (substutution versus interstitial inclusion). It also affects the size of the individual crystal grains. So even with a metal or alloy that has the same chemical composition, the answer can be different depending on how that particular sample was formed.

Edited December 18, 2020 by mikegarrison

[paul_c]

On 12/17/2020 at 7:00 AM, ARS said:

[Moderator's note:  This topic was originally split off from For Questions That Don't Merit Their Own Thread, since the ensuing discussion turned out to be a lengthy one.]

Is it possible to calculate the distance between molecules on a metal compound purely by mathematical calculation as long as we know the ratio of the metals used, density and molar mass of each metal as well as the resulting shape of the crystal matrix of the compound? Assuming that the metal constituent is evenly distributed in ideal condition

Yes and no.

If this is a copy/paste of an exam question, then it should never have been presented as such - the use of the term "metal compound" is misleading, since mixtures of different metals aren't compounds but are ALLOYS. A metal compound suggests another element, in a chemical reaction, to produce a compound. An alloy isn't a chemical reaction, its a physical structure. So strictly, no, because we would need to know the other element(s) making up the compound.

But I think you meant metal alloy - then its yes (approximately). Its not possible exactly because you would need to know the detail of the crystalline structure down to the atomic level, to know the true range of values of distance to adjacent/nearest atoms (I think that's what you meant when you said "distance between molecules") (and.........there would be a range of values, not just one single value, because of the nature of the crystalline structure).

And to be pedantic.....the structure consists of metal IONS (or atoms) not molecules because a molecule is 2 or more atoms joined by a chemical bond (by definition). 

[K^2]

39 minutes ago, mikegarrison said:

Obviously with gasses or liquids or other kinds of solids, the answer is different.

For an ideal gas, probability that there is at least one other particle within distance r of another is P₁₊(r) = 1 - exp(-(4/3)πρr³), where ρ is number density.

Spoiler

Say there are N particles within some sufficiently large volume region V around the first particle. Then probability of any given particle being within distance r is q = (4/3)πr³/V. Probability that there is at least one of the N particles within that region is P = 1 - (1 - q)^N = 1 - (1 - (4/3)πr³/V)^N. Substitute V = N/ρ, then P = 1 - (-(4/3)πρr³/N)^N. Finally, take the limit using the fact that lim_N->inf (1 + x/N)^N = exp(x) to get P₁₊(r) = 1 - exp(-(4/3)πρr³). QED.

What we actually want is probability that nearest neighbor is distance r away, P_n(r) = d/dr P₁₊(r) = 4πρr² exp(-(4/3)πρr³)

Spoiler

Suppose we're given P_n(r). Then the odds of having at least one particle at a given distance is just cumulative probability of nearest neighbor being up to that point. P₁₊(r) = ∫₀^r P_n(r) dr. Take derivative of both sides to get expression above.

Finally, we want expectation E[r | P_n(r)] = ∫₀^inf r P_n(r) dr = ∫₀^inf 4πρr³ exp(-(4/3)πρr³) dr = Γ(4/3) / ((4/3)πρ)^1/3 ≈ 0.554 ρ^-1/3. (Thanks Mathematica!)

To verify, I found a similar derivation on this page in Wikipedia. They are using Wigner-Sietz radius instead of number density. They are related via (4/3)πρa³  = 1. Substituting that into the above, you get the same expression for average nearest-neighbor distance.

Unsurprisingly, this number is significantly lower than it is for solids. Gases are pretty sparse so you get a lot less density for the same nearest-neighbor distance.

Also unsurprisingly, to me, anyways, just taking ρ^-1/3 is a really bad estimate for gases.

[mikegarrison]

Well, with gasses pressure is a big deal, because pv=nRt and so pressure and volume are inversely related for the same mass, meaning that all you have to do to make the atoms get closer to each other is to squeeze them with more pressure. With liquids or solids that doesn't work.

[paul_c]

PV=nRT is an approximation.

5 minutes ago, mikegarrison said:

Well, with gasses pressure is a big deal, because pv=nRt and so pressure and volume are inversely related for the same mass, meaning that all you have to do to make the atoms get closer to each other is to squeeze them with more pressure. With liquids or solids that doesn't work.

A gas is, by definition, a state of matter which expands to fill the available volume, hence why it exerts a pressure.

I am sure solids are compressible too - to an extent....

[K^2]

7 minutes ago, mikegarrison said:

Well, with gasses pressure is a big deal, because pv=nRt and so pressure and volume are inversely related for the same mass, meaning that all you have to do to make the atoms get closer to each other is to squeeze them with more pressure. With liquids or solids that doesn't work.

Sure, but number density changes as you compress the gas, so that's all accounted for. What's more exciting is that there is no temperature dependence, since you're taking an instant snapshot, and atom distribution in an ideal gas is the same at all temperatures. It's only the velocity distribution that changes.

5 minutes ago, paul_c said:

PV=nRT is an approximation.

17 minutes ago, K^2 said:

For an ideal gas,

 

[mikegarrison]

6 minutes ago, paul_c said:

PV=nRT is an approximation.

A very useful approximation.

[kerbiloid]

7 hours ago, mikegarrison said:

Anyway, with crystalline solids I think @K^2 is right. Obviously with gasses or liquids or other kinds of solids, the answer is different.

He is not.

***

There is no difference, if the atoms are placed chaotically or not.
The average distance depends only on the total amount and the total number of items. That's the idea of the "average" itself.

To get an average height in the classroom, you don't need all pupils be of same height.
In such case you would just don't need the "average".

It's doesn't matter if the volume is filled uniformly or not, it's a blackbox of knnown volume and of known number of atoms.
You neither have, no need their positions, that's all.
You just divide  the volume by the number and get the average height to have the average. volume per atom.

Then as we live in 3d world, where d~V^1/3,  you estimate the average distance by getting the cubic root from the average volume. 
In most cases that's enough, and any book stops here, because you need:
1) a qualitative assessment - you have it;
2) compare different materials - the exact value doesn't play any role -  you need relative values  to compare;
3) compare different ratios of components of the same material -  you need relative values  to compare;
4) compare properties of the same material  at varying conditions-  you need relative values  to compare;

But as by d ~ V^1/3 we actually approximate the volume with finite elements of cubic shape (it has nothing to do with cubic or any other crystal lattice, it's a purely abstract cube), we can obviously  see that
the element diameter (i.e. the longest diagonal, between the opposite corners) 
is (1²+1²+1²)^1/2 = 3^1/2 ~=1.73 times longer than
the element width (i.e. its minimal size of the cube, i.e. its edge length , which is  V^1/3).

We can find exact value by integrating the distance between any two points and find the average coefficient between 1 and 1.73, who wants may do that.
It would be ~1,25 or ~1.5, no matter.

Then we can have more accurate value of the average distance,  say 1.25 V^1/3  or 1.5 V^1/3.
It could make sense if you need such accuracy.
In case of this task, it's absolutely clear an estimation task, so the first approximation is enough good.

Say, 1.25 would give (1-1.25)/1.25 = 0.2, (1.5 - 1) /1,5 = 0.33, i.e. 20..30% accuracy, which is a good result even for measurement, let alone the quick estimation.

Exact positions of the atoms play no role at all. Its crystal lattice as well.
Do you have the atoms placed in cubes, tetrahedrons, stars, flowers, or full mess, plays no matter.
If you place the atoms in otherpositions, neither total volume, nor total count will change. The average distance as well.

8 hours ago, mikegarrison said:

asked about "a metal compound", and that's a problem because of the word "compound".

It's absolutely no difference if the atoms in the volume are alone, paired, mixed, or divided. Total values stay same.

8 hours ago, mikegarrison said:

Heat treatment of metals (especially steel) is about forcing one crystal structure or another, or one type of alloy or the other (substutution versus interstitial inclusion). It also affects the size of the individual crystal grains. So even with a metal or alloy that has the same chemical composition, the answer can be different depending on how that particular sample was formed.

See above. No totals change, you don't need to know anything but total volume and total number get averag volume per atom and thus aberage distance/

***

Gas pressure, exponential distribution, luminosity of Sirius B play no role here and are just artifical overcomplication.

[YNM]

IMO metal alloys are very interesting - even the slightest inclusion of other materials can massively change how useful the resulting product is, ie. pure iron is not very useful, but steel is iron with fairly small amounts of impurities (mainly carbon) added into it and it's already much stronger to break (requires higher stresses).

 

I wonder if the way how a material will behave could be predicted purely by the lattice structure and the amount of impurities...

 

Spoiler

21 minutes ago, kerbiloid said:

The average distance depends only on the total amount and the total number of items. That's the idea of the "average" itself.

What exactly would that be useful for, really ? If I have 30 person in a 3 x 3 m room, and call the average distance at least 1.5 m, can I call that the room was in compliance with physical distancing rules of at least 1 - 2 m ? I know you're probably trying to be precise about the meaning, but your interpretation have barely any useful case.

[kerbiloid]

17 minutes ago, YNM said:

What exactly would that be useful for, really ?

It's useful in most cases.
It's the basics of the "Strength of materials" course, where you study about defects ("dislocations").
it can be used to estimate "how does something depend on the distance between the atoms": free path, optics, mechanical properties, etc.

In most cases you need a relative value, not an absolute one, so any exact coefficient (1.25, 1.5 or whatever) is just divided by itself and disappears.
So, nobody needs anything but d~V^1/3 unless the exact task requires the exact value.

Upd.
It's just a basic culture of experiment, to make as many factors as possible not required and exclude them.
So, in most cases you need y((x - x₀)/x₀) or y(x/x₀) rather than y(x)

Edited December 19, 2020 by kerbiloid

[YNM]

5 minutes ago, kerbiloid said:

It's the basics of the "Strength of materials" course, where you study about defects ("dislocations").

Yeah, but isn't this more to do with the other interpretation of "how things stack up" rather than an "averaged" distance ? And that's what the OP was asking... Like one case that I know is with strain hardening as well as necking.

5 minutes ago, kerbiloid said:

free path

I get about mean free path, but it's only in non-solid stuff (I'm not even sure if it's applicable in liquids, because nothing really shoots around in a straight line in a liquid that's part of the liquid itself).

Edited December 19, 2020 by YNM

[kerbiloid]

4 minutes ago, YNM said:

Yeah, but isn't this more to do with the other interpretation of "how things stack up" rather than an "averaged" distance ? And that's what the OP was asking..

As far as I can see, the OP question just asks about the distance in a compound.
As obviously there can be no "distance" but average, it's just a typical average distance from V/N.

4 minutes ago, YNM said:

but it's only in non-solid stuff (I'm not even sure if it's applicable in liquids).

Got you. What about free path of photon in a transparent material, a fast particle in any material, etc?

(Updated the previous post.)

Edited December 19, 2020 by kerbiloid

[YNM]

25 minutes ago, kerbiloid said:

Got you. What about free path of photon in a transparent material, a fast particle in any material, etc?

But the photon isn't a useful part of the solid or whatever material it is. I'm assuming we're talking material properties and not nuclear physics.

25 minutes ago, kerbiloid said:

As far as I can see, the OP question just asks about the distance in a compound.

If you want to be pedantic, I suppose, you could say that the average distance between concrete aggregate in a block of shear wall to be whatever the size of the shear wall is halved. You could even say the same to the rebars inside of it. It's not a useful measure at all, at least for things that people who make shear walls would be interested in.

 

It's in the same way that most people interested in metallurgical properties won't be interested in that kind of measure. (While I'm not studying metallurgy, properties of material are very important in knowing how construction materials will behave, especially in failures).

Edited December 19, 2020 by YNM

[kerbiloid]

7 minutes ago, YNM said:

But the photon isn't a useful part of the solid or whatever material it is. I'm assuming we're talking material properties and not nuclear physics.

You have manufactured a composite material and measured the free path of X-rays in it.
You want to improve it., and you presume that it depends on the average distance between atoms. 
You don't need to know exact value of that distance. You need to know how does the result depend on its change.
So, you don't need to think if it is 1.25 V^1/3 or 1.5 V^1/3, or whatever. You experiment with (d-d_original)/d_original or d/d_original.

Thus the V^1/3  is absolutely enough for you to plan the experiment.

So, nobody needs that exact coefficient, like nobody needs the internal distribution.

7 minutes ago, YNM said:

If you want to be pedantic, I suppose, you could say that the average distance between concrete aggregate in a block of shear wall to be whatever the size of the shear wall is halved. It's not a useful measure at all, at least for things that people who make shear walls would be interested in.

If you need that concrete to catch radioactive particles, it could be useful.

If you need it to build a house, you don't need the atomic distance.

If you optimize the concrete production, you may (actually, unlikely) need it.

Edited December 19, 2020 by kerbiloid

[YNM]

2 minutes ago, kerbiloid said:

You have manufactured a composite material and measured the free path of X-rays in it.

...

If you need that concrete to catch radioactive particles, it could be useful.

Well, now it's up to the OP to clarify what does it need... Though I can't see why you'd need alloys for radioactive shielding, you just go and find the largest, densest thing.

Edited December 19, 2020 by YNM

[kerbiloid]

5 minutes ago, YNM said:

Well, now it's up to the OP to clarify what does it need...

I guess, he needs a student's task, so as far as can understand, it's the case where it's just no need to go deeper.

[K^2]

56 minutes ago, YNM said:

Yeah, but isn't this more to do with the other interpretation of "how things stack up" rather than an "averaged" distance ? And that's what the OP was asking... Like one case that I know is with strain hardening as well as necking.

He's not actually computing anything like an average distance. It's a meaningless concept. Average distance between atoms in a copper wire that's a mile long includes distance between the two atoms on the very ends, and if you do the math, the average distance between all possible atom pairs in that wire is 1/3 of a mile.

The meaningful concept is average nearest-neighbor distance, which explicitly depends on structure, with results for gas vs metal being different by a factor of 2 after you compensate for density difference. Which is not unexpected at all, but I'm baffled how somebody can continue claiming that they have a formula for "average distance", and that it doesn't depend on anything other than density.

Taking a formula for nearest neighbor distance in a simple cubic lattice, because that's what ρ^1/3 formula counts, and claiming that you can use it for any structure is absolutely absurd. Not only are you implicitly making assumption about structure by using this formula, but you're taking the least likely structure for a metal, which is what OP was asking about. And yeah, it's in the ballpark for average nearest-neighbor distance, because it happens to be a formula for one, just the wrong structure, but he's claiming that it's a formula for "average distance", and that it should just magically work for everything based on some horrible misunderstanding of how average distances work and what a number density means.

 

[kerbiloid]

I don't know how else to explain the term "average" known since elementary school.

So, I have brought the method used in real university practice, who wants to reinvent the Universe, may do that.

Once again: the internal structure of the sample means nothing here, and everything but volume and number is an artificially complicated nonsense.

"Purpose" can be defined by the task. If the task doesn't declare the purpose, it's no need to artificially invent it. Any invented "purpose" is an empty guessing.

Edited December 19, 2020 by kerbiloid

[ARS]

This turns out to be much longer than what I expected. Ok to clarify some stuff, I have a task in my metallurgical department. Let's just say we're measuring the atomic distance of a cupronickel alloy sample (apologize for the confusion about compound and alloy) in material's lab (each group may or may not get the same sample). The goal is to find the distance between atoms in the crystalline structure of the sample, first the distance between the atoms of same types (Cu with Cu, Ni with Ni) and the second is the distance between the atoms of different type (Cu with Ni). Things go as planned before suddenly, in the middle of operation, the test machine broke (can't blame it, it's kinda old anyway). We asked the teacher for the solution for this and basically it boils down to the "if the actual test can't be performed, then use the mathematical calculations to find it out"

Now we could calculate the atomic distance of individual types when it's separated, but for the atomic distance for alloys, things get complicated. The ratio of cupronickel in question is 3:1, the distance from each Ni atoms should be uniform, same case with Cu. The main question is, how to calculate the distance between Ni and the nearest Cu atom inside the cupronickel alloy, and the distance between Ni atoms with each other in cupronickel allow with a ratio of 3:1 (assuming Ni atoms distributed evenly), it doesn't matter if it's exact value or approximate value (if we change the ratio, then the distance between Cu and Ni should be different too, right?)

[YNM]

2 hours ago, K^2 said:

Taking a formula for nearest neighbor distance in a simple cubic lattice, because that's what ρ^1/3 formula counts, and claiming that you can use it for any structure is absolutely absurd.

Actually, wouldn't a differently-packed crystal of the same atoms would result in different density on itself ?

[kerbiloid]

If I had this given, I would try something like this.

***

Cupronickel = ~3 copper : 1 nickel (by mass).

RatioByMass: copper ~= 0.75, nickel ~= 0.25.

Density: ~8 900 kg/m³.

Atomic mass: copper ~= 0.0635 kg/mol, nickel ~= 0.0587 kg/mol.

***

TotalMass = TotalVolume * Density.

NumberOfAtoms_Copper = TotalMass * RatioByMass_Copper * AvogadroNumber / MolarMass_Copper;
(TotalMass * AvogadroNumber * 0.75/ 0.0635 = const * 0.75/ 0.0635 = const * 11.8)

NumberOfAtoms_Nickel = TotalMass * RatioByMass_Nickel * AvogadroNumber / MolarMass_Nickel; 
(TotalMass * AvogadroNumber * 0.25/ 0.0587 = const * 0.25/ 0.0587 = const * 4.3)

As we can see (11.8 > 4.3), there are more atoms of copper than atoms of nickel.
About 2.75 atoms of copper per 1 atom of nickel.

***

Considering all atoms distributed uniformly, calculate the average distance between the atoms of all kinds (as shown above or in any other way).

Of every 3.75 atoms there are 2.75 atoms of copper per one atom of nickel.

(Actually 

***

Estimate average volume per nickel atom:

AverageVolumePerAtom_Nickel
= TotalVolume / (TotalMass * RatioByMass_Nickel * AvogadroNumber / MolarMass_Nickel;)
= TotalVolume / (TotalVolume * Density * RatioByMass_Nickel * AvogadroNumber / MolarMass_Nickel)
= 1 / (Density * RatioByMass_Nickel * AvogadroNumber / MolarMass_Nickel)
= 1 / (8 900 * 0.25 * 6.022*10²³ / 0.0587)
~= 4.38*10^-29 m³ / Nickel atom.

So, you have a volume of 4.38*10^-29 m³ containing 1 blue ball (atom of nickel) and 2.75  red balls (atoms of copper), 3.75 balls in total.

***

Now we have a pure math question: "We have a known volume, 1 blue ball and 3 red balls in it, uniformly/randomly distributed. How to find average distance between the the blue ball and a red bull ball".

(I would ask a math teacher (just to be sure), and then replace 3 with 2.75, but if no, I'll try to calculate.)

12 minutes ago, YNM said:

Actually, wouldn't a differently-packed crystal of the same atoms would result in different density on itself ?

We already have the resulted difference! The actual density is given.

So, any effect of crystal structure, temperature, gravity, whatever has already affected the result, giving that density difference, an is not essential anymore.
It's a pure geometry.

Edited December 19, 2020 by kerbiloid

[K^2]

1 hour ago, YNM said:

Actually, wouldn't a differently-packed crystal of the same atoms would result in different density on itself ?

Sure. But we're looking at atomic sizes as the unknown, so given a measured density, your estimate of the nearest-neighbor distance will change depending on information you are given about the structure. We are not talking about an actual phase transition taking place resulting in a density change.

Though, that's a completely valid way to look at it. In simple cases of monoatomic crystals, the nearest-neighbor distance is usually given by properties of the atom in question, so it won't change a whole lot. So during a phase transition, density will change as result of changing structure.

There are notable exceptions, however. Distance between nearest carbon atoms in diamond is about 1.55A, while in graphite, it's 1.42A. Big part of why carbon is so versatile is that it can form bonds with very different overlaps of orbitals resulting in different interatomic distances. Still, despite nearest neighbors in graphite sitting closer together, the carbon atoms are more spread out overall, and density of graphite is significantly lower than that of diamond, so structure is playing bigger role here than nearest neighbor distances.

2 hours ago, ARS said:

Now we could calculate the atomic distance of individual types when it's separated, but for the atomic distance for alloys, things get complicated. The ratio of cupronickel in question is 3:1, the distance from each Ni atoms should be uniform, same case with Cu. The main question is, how to calculate the distance between Ni and the nearest Cu atom inside the cupronickel alloy, and the distance between Ni atoms with each other in cupronickel allow with a ratio of 3:1 (assuming Ni atoms distributed evenly), it doesn't matter if it's exact value or approximate value (if we change the ratio, then the distance between Cu and Ni should be different too, right?)

This can actually be worked out because of the 3:1 ratio. With copper, this tells you that you are still dealing with FCC lattice, but you have Ni atoms at vertices of the cell and Cu at faces. A single cell will then contain 3 atoms of copper and one of nickel, which is exactly what we're looking for with CuNi25. The FCC formula I've given in the first reply still works, but instead of having 4 * atomic mass, it will be 3 * atomic mass of copper + atomic mass of nickel. I've also found 8.95g/cm³ as density for CuNi25, so I hope that matches your numbers.

Ni-Cu distance = ((3 * 63.546amu + 58.693amu) / (6.022 * 10²³ * 8.95g/cm³))^1/3 / sqrt(2) = 2.54A

Cu-Cu distance = Ni-Cu distance = 2.54A

Ni-Ni distance = sqrt(2) * Ni-Cu distance = 3.59A

Which makes sense. Copper atoms are larger than Nickel, so since they are "touching" each other in this lattice, you expect them to be limiting the density and so the Cu-Cu distance is about the same as it is for pure copper. And then the nearest Ni-Cu and Ni-Ni distance are simply dictated by the lattice.

Edit: Actually, above assumes the 3:1 is ratio by number of atoms, not by weight. If you have a 3:1 ratio by weight, you should work out what the ratio by numbers is. It's not going to give you NEARLY as neat of a lattice, but it should be generally close to the above. Your limiting factor is still going to be copper-copper distances, so it should stay very close to 2.55A until you go way more than 25% nickel.

Edited December 19, 2020 by K^2

[YNM]

32 minutes ago, K^2 said:

We are not talking about an actual phase transition taking place resulting in a density change.

 

1 hour ago, kerbiloid said:

We already have the resulted difference! The actual density is given.

So, any effect of crystal structure, temperature, gravity, whatever has already affected the result, giving that density difference, an is not essential anymore.
It's a pure geometry.

Ah, well yeah. We're using something already given. There isn't anything to compare against either.

Edited December 19, 2020 by YNM