Figure 1 in Ivor Lončarić and Vito Despoja, Phys. Rev. B 90, 075414 Benchmarking van der Waals functionals with noncontact RPA calculations on graphene-Ag(111) (available in Researchgate) shows a schematic of the graphene structure on what I believe is a silver (111) surface.

The caption reads:

FIG. 1. (Color online) Geometry of the supercell used for our vdW DFT calculations. Upper part: 2 × 2 graphene lattice matched to √3 × √3 Ag(111) lattice. Lower part: Geometry of the supercell in perpendicular direction.

What exactly is a "√3 × √3 Ag(111) lattice" and how exactly does the square root of three come in to this?

supercell 2 × 2 graphene lattice matched to √3 × √3 Ag(111) lattice

  • $\begingroup$ companion question: Silver (111) surface structure, and is bulk structure body-center or face-center cubic? $\endgroup$
    – uhoh
    Jun 10, 2019 at 3:51
  • 3
    $\begingroup$ The $\sqrt{3}$ is the distance shown as black lines between every other C atom (assuming bond length is one unit). Twice this is the equivalent length between the Ag atoms shown as silver/grey. So I'm assuming this is what they mean by lattice matching? $\endgroup$
    – porphyrin
    Jun 10, 2019 at 10:23
  • 1
    $\begingroup$ See ww2.sljus.lu.se/staff/anders/Compressed/Notation_Edvin.pdf for a general overview (googled 'silver 111 surface reconstruction'). $\endgroup$
    – Jon Custer
    Jun 10, 2019 at 12:53
  • 1
    $\begingroup$ Good news! Classic surface science seems to have faded away if late. Recall that most of the original surface reconstruction information was from techniques like LEED that gave relative symmetry and rotation, but we had to wait for AFM to see actual atom positions. $\endgroup$
    – Jon Custer
    Jun 11, 2019 at 1:20
  • 1
    $\begingroup$ Start at the left bottom most blue atom (crossed by 2 black lines), go north by 1 atom and north east by 1 atom. The distance as the crow flies is $\sqrt{3}$ times bond length, i.e. length of black line connecting atoms. Now look at underlying Ag atoms; twice this length is equivalent length in this lattice so I guessed that this is connection between lattices and why one fits over the other. This could be rubbish, however, as I'm not an expert in surface science. $\endgroup$
    – porphyrin
    Jun 12, 2019 at 11:12

1 Answer 1


"Root 3" or "$\sqrt{3}$" or "$\sqrt{3} \times \sqrt{3}$" are slang for the Wood's notation (1, 2) $ \left( \sqrt{3} \times \sqrt{3}\right)R30$, a commonly occurring configuration of one hexagonal 2D lattice with respect another.

In this example the substrate is the (111) face of silver. Since it is face-centered cubic with lattice constant of about 4.08 Angstroms the (111) surface is hexagonal with a lattice constant 4.08 / $\sqrt{2}$ = 2.88 Angstroms.

The figure caption says "2 × 2 graphene lattice matched to √3 × √3 Ag(111) lattice."

The annotation in the OP's drawing copied below shows solid arrows for the Ag(111) lattice vectors the dashed arrows are rotated by 30 degrees and $\sqrt{3}$ longer, and they match the periodicity of a $2 \times 2$ graphene cell.

The lattice constant of free graphene is about 2.46 Angstroms and double that is 4.92. Compared to $\sqrt{3}$ times 2.88 = 4.99 Angstroms it's a fairly good lattice match.

2x2 graphene as root3 of Ag(111)


This site is temporarily in read-only mode and not accepting new answers.

Not the answer you're looking for? Browse other questions tagged .