A hexagonal closed packing (hcp) unit cell has an ABAB type of packing. For calculating the packing fraction we require the volume of the unit cell.

Volume of hcp lattice = (Base area) $\cdot$ (Height of unit cell)
Each hexagon has a side = $2\cdot r$
Base area = $6$ (Area of small equilateral triangles making up the hexagon)
$$=6 \cdot \frac{\sqrt{3}}{4}\times(2r)^2$$ $$=6 \cdot \sqrt{3} \cdot r^2$$

Hence, volume $= 6 \cdot \sqrt{3} \cdot r^2 $ (Height of unit cell)

This is the point where I am stuck. How do I find out the height of the unit cell?

I searched in textbooks and found out that height $= 4r \cdot \sqrt{\frac{2}{3}}$. Can you please explain why is this so?


We shall try it using the similarities between hcp and ccp. Here, we know that $hcp$ and $ccp$ have similar lattice except the fact that $hcp$ is ABAB type whereas $ccp $ is ABCABC type. Hence we also know that their packing fraction $(\phi) $ is same and $$\phi = \frac{\pi} {3\sqrt{2}}$$ Now as you mentioned Volume of hcp lattice $= 6\sqrt{3} r^2h$. There are 6 atoms in total in hcp. Hence $$\frac{6\left(\frac{4}{3}\right) \pi r^3}{6\sqrt{3} r^2 h} = \frac{\pi} {3\sqrt{2}}$$ Simplifying this we obtain the height of hcp lattice $$h=4r\left(\sqrt{\frac{2}{3}}\right)$$

| improve this answer | |
  • $\begingroup$ We get that their packing fraction is equal after evaluating volume from height,etc. Your answer is working backwards. $\endgroup$ – user600016 Mar 10 at 10:01

To calculate the height of a unit cell, consider a tetrahedral void in an hexagonal closed packing arrangement. It can be imagined as a 3 solid spheres touching each other and at the center-point, you have another sphere stacked over them. An interactive version can be viewed on this site. The situation looks like this:

four blue spheres with a tetrahedral void

If you join the centers of these four spheres, you'll get a tetrahedron. That's basically a pyramid with a triangular base. I'm assuming each edge of our tetrahedron to be equal to $a$.

Now, you have a pyramid ($ABCD$), with an equilateral base ($\Delta BCD$), I would like you to drop a perpendicular from the highest point ($A$) to the center ($G$) triangular base. If you're following me correctly, you'll have a figure like this:

enter image description here

All we have to do now is to calculate the length $AG$. For this, simply use the Pythagorean theorem in $\Delta AGD$.

$$ \begin{align*} AD^2 &= AG^2 + GD^2 \tag{1} \end{align*} $$

Although we know that $AD=a$, the side $GD$ remains unknown. But that's easy to calculate. The point $G$ is the centroid of $\Delta BCD$. Thus, the length $GD$ equals $a/\sqrt{3}$. Pluggging in the values in our first equation, we get $AG=a \sqrt{\frac{2}{3}}$. But note, this is half the height of our unit cell. Thus, the required height is $2a \sqrt{\frac{2}{3}}$.

| improve this answer | |


In the hexagonal closest-packed structure, $a = b = 2r$ and $c = 4 \sqrt{\frac23 }r$, where $r$ is the atomic radius of the atom. The sides of the unit cell are perpendicular to the base, thus $\alpha = \beta = 90^\circ$.

For a closest-packed structure, the atoms at the corners of base of the unit cell are in contact, thus $a = b = 2 r$. The height ($c$) of the unit cell, which is more challenging to calculate, is $c = 2a \sqrt{\frac23} r = 4 \sqrt{\frac23} r$.


Let the edge of hexagonal base equal $a$

And the height of hexagon equal $h$

And radius of sphere equal $r$

The centre sphere of the first layer lies exactly over the void of 2nd layer B.

The centre sphere and the spheres of 2nd layer B are in touch

So, In $\Delta PQR$ (an equilateral triangle):

$\overline{PR} = 2r$, Draw $QS$ tangent at points

$$∴ \text{In } \Delta QRS\text{: } \angle QRS = 30^\circ, \overline{SR} = r$$

$$\cos30^\circ = \frac{\overline{SR}}{\overline{QR}}$$

$$\overline{QR} = \frac{r}{\frac{\sqrt{3}}{2}} = \frac{2r}{\sqrt 3}$$

$$∴ \overline{PQ} = \sqrt{\overline{PR}^2 - \overline{QR}^2} = \sqrt{4r^2 - \frac{4r^2}{3}}$$

$$h_1 = \sqrt{\frac{8r^2}{3}} = 2 \sqrt\frac{2}{3} r$$

$$∴ h = 2h_1 = 4 \sqrt{\frac23} r$$

Hence, in the calculation of packing efficiency of hcp arrangement,the height of the unit cell is taken as $4r\sqrt{\frac{2}{3}}$.


| improve this answer | |
  • $\begingroup$ What does the triangle of dots mean? $\endgroup$ – Jan Sep 25 '19 at 10:00
  • $\begingroup$ How come angle QRS is 30 deg ? $\endgroup$ – Princy Rawat Sep 27 '19 at 4:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.