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Lattice constant from first principles

  1. Dec 3, 2014 #1

    Demystifier

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    I would like to know how to estimate the value (length) of the lattice constant in crystals in terms of fundamental parameters such as electron charge, electron mass, proton mass and Planck constant. (Alternatively, the formula may contain the Bohr radius, since I know how to calculate Bohr radius from the fundamental parameters). I am not interested in an accurate calculation, by DFT or whatever. All I need is a simple formula that estimates the order of magnitude.

    If the formula can be found in a standard textbook, please tell me which textbook and a precise location (page number, equation number, etc.).

    Thanks!
     
  2. jcsd
  3. Dec 3, 2014 #2
    Chemists have tables of ionic radii. You can combine that with space filling factors for different packings of spheres to get a very rough estimate of the unit cell volume. After that you have to make assumptions about the symmetry of the cell (cubic, hexagonal, ...). Note that the results will be a very very crude approximation.

    From Wikipedia, the densities of graphite and diamond:

    diamond: 3.515 g·cm−3
    graphite: 2.267 g·cm−3

    This gives you a taste of what kind of accuracy you can expect from such a formula.
     
  4. Dec 3, 2014 #3

    Demystifier

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    You take the measured densities as an input, but that's not what I want. In a formula, I don't want any non-fundamental parameters such as those measured densities.

    A kind of formula I would accept is something like
    $$a=4\pi a_{Bohr}$$
    where ##a_{Bohr}## is the well known expression for the Bohr radius.
     
  5. Dec 3, 2014 #4

    f95toli

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    I doubt there is a "simple" formula for that. Unless I am missing something this is a complicated optimization problem which it is always done numerically. You don't define what you mean by "crystal", but the crystallography can get very complicated even if you only consider pure elements (carbon would be the obvious example); and unless you know the crystal structure you can't calculate the lattice constants.

    It might be possible if you consider some "simple" element and you assume a simple cubic structure, but I don't remember ever seeing such a formula.
     
  6. Dec 3, 2014 #5

    Demystifier

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    f95toli, can one, at least, make a qualitative theoretical argument that ##a## cannot be much larger than ##a_{Bohr}##? If such an argument can be made (without invoking experimental facts on ##a## for different materials), that would satisfy me.
     
  7. Dec 3, 2014 #6

    TeethWhitener

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    What do you mean by "much" larger? If you're talking about, say, protein crystals that are grown in molecular biology research labs, [itex]a[/itex] can be hundreds of times larger than [itex]a_{bohr}[/itex]. Furthermore, the lattice constants are simply the repeat lengths of a crystal, so if you were to, for example, create regularly spaced defects in a diamond that were a distance of 1mm apart, the lattice constant would technically be very close to 1mm, even though the lattice constant of pure diamond is only a few angstroms.
     
  8. Dec 3, 2014 #7

    DrClaude

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    Asuming you are talking about a crystal of atoms, and you interested in the average interatomic distance (see TeethWhitener's comment), then the only way I can think of is to start from a diatomic molecule. The only case that can be solved exactly is H2+ in the Born-Oppenheimer approximation. You then get that the minimum energy is at about ##2 a_{Bohr}##. You then need to extrapolate by saying that this distance is (possibly) not much affect by the presence of other atoms, including when an atom in bounded to many other, such that in a solid the atoms will be found this order of magnitude apart one from another.
     
  9. Dec 3, 2014 #8

    DrDu

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    You might give W. Thirring,
    Quantum Mathematical Physics: Atoms, Molecules and Large Systems
    a try. He derives some bounds on the stability of extended matter.
     
  10. Dec 4, 2014 #9

    Demystifier

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    I, of course, meant the distance between the nearest neighbor atoms. Otherwise, quasicrystals would have infinite ##a##.
     
  11. Dec 4, 2014 #10

    Demystifier

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    That's close to what I wanted. But in the meantime, I have found my own intuitive qualitative answer which I would like to share.

    What fundamental force binds the atoms in a crystal together? This, of course, bust be some manifestation of the electric force. But atoms are electrically neutral, so the force cannot be the electric monopole force. So it must be some electric multipole force, originating from a non-trivial distribution of charge in the atoms. But the atom (and its charge distribution) looks point-like when seen from a distance much larger than the atom size. Consequently, the multipole force must be negligible at such large distances. Therefore the distance ##a## between the neighboring atoms which act on each other by a mutipole force cannot be much larger than the size of the atom. Since the size of atom is of the order of ##a_{Bohr}##, it implies that ##a## cannot be much larger than ##a_{Bohr}##. Q.E.D.
     
  12. Dec 4, 2014 #11

    DrClaude

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    That completely neglects all the QM processes that lead to stable bonds, and throws under the carpet the fact that electrons do not stay "attached" to one particular atom.

    See van der Waals interaction.

    That is not true: the induced-dipole - induced-dipole attraction that is the van der Waals interaction takes place on length scales much greater than the size of the atoms.

    What I like least about this "demonstration" is that you have just proved that gases (and even liquids) have no cohesion, since the atoms have then much greater mean separations than in solids.
     
  13. Dec 4, 2014 #12

    DrClaude

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    It goes as ##r^{-6}##. It's hard to say at what point it becomes negligible: ngligible with respect to what? Calculating its magnitude requires knowing the polarizability of the atoms, which is not something you get directly from fundamental constants.

    If you do not want to introduce chemical bonding, then I think the simplest answer is just ~2 aBohr, which is what you get when packing atoms-as-spheres together.
     
  14. Dec 4, 2014 #13

    ShayanJ

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    I think temperature becomes important here. Because whatever the energy of the interaction is, if it is less than the thermal vibrations, no bond can form. So here "not negligible" has something to do with temperature.
    But if you assume temperatures very close to absolute zero, then of course there will be a bond with any interaction energy, only the lattice constant should be determined. I think considering a first-neighbour approximation can give a good starting point.
    But such considerations will be classical, so I think its better to be careful about them at very low temperatures.
     
  15. Dec 4, 2014 #14

    DrDu

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    Your argument should somehow hold true for nonpolar atomic and molecular solids. VdW interaction is not simple. There is a long range part with gives rise to surface tension and other effects. You can probably separate the long range forces off somehow. For the short range, you have Pauli repulsion which decays exponentially for distances larger than the Bohr radius and the VdW attraction. So you end up with some Lennard Jones intermolecular potential. The only length scale entering is the Bohr radius.
     
  16. Dec 4, 2014 #15

    Demystifier

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    Well, to get something which has dimension of force, the ##r^{-6}## factor must be multiplied with some constant dimensional factor. This dimensional constant contains something with a dimension of length (with an appropriate power). So what this parameter with a dimension of length can be, if not something of the order of ##a_{Bohr}##?

    I suspect that the length scale is still proportional to ##a_{Bohr}##, but multiplied with something relatively big such as ##(4\pi)^4##.
     
  17. Dec 4, 2014 #16

    Demystifier

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    But for long-range interactions, it must be multiplied with something relatively big as in my post above, right?
     
  18. Dec 4, 2014 #17

    DrDu

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    For me, 4##\pi## is still ##O(a^0)##, so it is neither big nor small.
     
  19. Dec 4, 2014 #18

    Demystifier

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    For me, ##(4\pi)^4\approx 24936\gg1##, so it's big.
     
  20. Dec 4, 2014 #19

    DrDu

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    Not in comparison with the inverse bohrs radius. Where do you take the 4th power from ?
     
  21. Dec 4, 2014 #20

    Demystifier

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    I have not derived the 4th power, but perhaps something like that might originate from the 6th power in ##r^{-6}## compared to the 2nd power in Coulomb law.

    Anyway, it is certainly true that ##(4\pi)^4 a_{Bohr} \gg a_{Bohr}##. My idea is that some relation of this form might be the reason why van der Waals is effective at distances much larger than ##a_{Bohr}##. That's just a conjecture, I don't have a proof.
     
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