Dear Vishal:

unit cell of Wurzit / hexagonal crystal system is only based on 2 neighboring triangles of the hexagon area ( see 1st link) and not on the full one (consisting of 6 symmetrical triangles).

Thus unit cell volume is V = a²c*sin(60°) and not 3*a²c*sin(60°) as you did. Here is the factor '3' you are searching for.

Wurzit unit cell (3-rd and 4-th picture):

https://en.wikipedia.org/wiki/Wurtzite_crystal_structure

Unit cell Volume (have a look at volume of hexagonal crystal system):

http://webmineral.com/help/CellDimensions.shtml

let's distinguish between the "crystal structure or system" and the "unit cell" of the wurtzite. the crystal structure is hexagonal and it contains 6 atoms of each type. the unit cell is one third of the hexagonal and it contains 2 atoms of each type. thus, there is no error or mismatch.

Thank You Gerhard Martens and Imad Disher.

I understand here that the unit cell is in the shape of a parallelogram which is 1/3rd of a hexagon. So these three parallelograms form a hexagon when arranged in a specific pattern.

However, repeating pattern of this parallelogram or unit cell (without change in orientations/angle) does not result in the hexagonal structure.

Instead, a parallelogram equivalent to area of a hexagon (as shown in the link below) when repeated forms a hexagon pattern, which is the reason why I considered the volume of the unit cell as equivalent to area of the hexagon structure.

http://www.chemtube3d.com/solidstate/_wurtzite(final).htm

Please let me know your thoughts on this.

Derar Vishal;

you are right when you state, that repeating the hexagon in all directions the whole 2D space will be occupied.

But the 'unit cell' is defined as the smallest cell which, when repeated in all directions, is able to completely occupy the space (here the 2D area) without holes or even overlay .

You state above: "However, repeating pattern of this parallelogram or unit cell (without change in orientations/angle) does not result in the hexagonal structure".

That is not correct: repeatedly arranging of the parallelograms in all direction will fill the full area by the footprints, without the necessity of rotations.

Please have a look at my attached drawing.

I must admitt that some of the footprints of the parallelograms cross the borders of the hexagons. But that doesn't matter.

The main thing here is that the crystall lattice, which is represented by the 3D arrangement of lattice points (not the connecting bars between them) is reproduced by multiple arrangement of unit cells.

Thank You Gerhard Martens for the explanation.

In the attached drawing where sides of hexagon are bolded or thicker, there are some parallelograms (say group A) which contain two of sides of hexagon while there are some parallelograms (say group B) which do not contain any hexagon side (bold sides) as one or two of its edges. If atoms in the structure are present dominantly at the vertices of the hexagon (with thick edges), then won't parallelogram or unit cell from group A be different from group B ? If yes, would it still be a unit cell for the structure ?

Please let me know your thoughts on this.

Dear Vishal,

the crystall lattice is characterized by the 3D arrangement of the crystall lattice points, not on the bars between them. For the hexagonal lattice the lattice points show up in the arrangement which is displayed in the attachment of my answer above ((Hexagonal&ParallelogramStructure_GM.pdf).

However the lattice points for the hexagonal case are not only positioned at the hexagon circumference but the point at the hexagon center also belongs to. This is very essential.

So next please have a look at the attachment belonging to this answer (HexagonAllOver_GM.pdf). Here in addition to the black hexagons I have placed blue ones around a lot of crossing points.

You now see that the whole hexagonal structure is now (at the center of the drawing) filled up by thick bars, which connect the lattice points. Thin (connecting) bars do not show up here any more.

For the green colored parallelograms now you cannot distinguish two groups A and B.

These parallelograms represent the 'unit cell'.

The atomic arrangement relative to the lattice points is named as the ' basis'.

So "the crystal structure consists of the same group of atoms, the basis, positioned around each and every lattice point."

You can find this sentence under the section 'bravais lattice' of the following link.

https://en.wikipedia.org/wiki/Crystal_structure

Thank You Gerhard Martens for the explanation.

I came across a paper " First-principles total-energy calculation of gallium nitride" by B.J. Min, C. T. Chan, and K. M. Ho. In this paper, the volume of GaN wurtzite unit cell is stated in a range of 100-150 a.u. As per my calculations, the volume come out as 46.62 A^3 for GaN wurtzite unit cell.

Please let me know your thoughts on this.

Your number of about 46,62A³ for the unit cell is ok (based on parallelogram).

But I am not able to evaluate how the above mentioned authors have calculated the volume of their unit cell with volume numbers of about 100 to 150 a.u. in fig. 3 of the paper, as you mentioned above.

In the following link (at section 'derived atomic units' first line in that table)

https://en.wikipedia.org/wiki/Atomic_units

there is the Bohr radius ao mentioned (ao~0,53A)(https://en.wikipedia.org/wiki/Bohr_radius)

When applying that Bohr radius as atomic unit length (1 a.u.~ao ~0,53A) onto the lattice constants given in table II of the mentioned paper, then you will get at least the lattice constants a and c as they can be found in

https://en.wikipedia.org/wiki/Gallium_nitride.

But when applying that a.u.-conversion onto a volume, for example 100 a.u ~ 100 *ao³ ~ 100*0,15A³~15A³, we get a number which is far too small.

They might have used another conversion factor for the a. u. volume.

Maybe 1 a.u.= 4*pi/3*ao³; sphere with Bohr radius; but the resulting volume numbers do not fit at all.

I do not know that conversion factor, sorry.