# Calculation of the structure factor of graphite (Solid State Physics/Crystallography)

1. Mar 4, 2012

### spf

Hi,

this is my first post on this forum and I hope that I am in the right category. My question concerns an independent study and not a homework assignment. However, it's a standard homework question in introductory solid state physics courses which makes me both the more frustrated that I cannot figure it out and the more surprised that I cannot find the answer on the web in spite of extensive googling.

1. The problem statement, all variables and given/known data

The structure factor of graphite needs to be calculated.

The following article might be helpful to answer this question but is not necessary as I give all the values and formulas below: Chung, JMS 2002 (Journal of Materials Science), p.2, bottom right

More precisely, the problem lies in the calculation of the following sum in the structure factor

$\sum_{j} e^{- i \vec{G}_m \vec{\rho}_j}$

with $\vec{G}_m$ being the reciprocal lattice vector and $\vec{\rho}_j$ being the coordinates of the four atoms of the graphite unit cell (assuming the standard ABAB stacking).

The correct result (according to Chung, JMS 2002) should be:

$(1 + e^{- i \frac{2\pi}{3}(2m_1-m_2)})(1 + e^{- i \pi m_3})$

which is the same as

$(1 + e^{i \frac{2\pi}{3}(m_1+m_2)})(1 + e^{- i \pi m_3})$

because

$1 = e^{i 2\pi} = e^{i \frac{2\pi}{3} \cdot 3 m_1}$

My problem is that I get

$(1 + e^{i \frac{2\pi}{3}(m_1+m_2)})(e^{- i \frac{2\pi}{3}(m_1+m_2)} + e^{- i \pi m_3})$

The result from Chung JMS 2002 seems more reasonable as it explains the experimentally confirmed 001 Bragg spot extinctions which my result does not. But I cannot find any error in my calculations nor in the start values. Can anybody help me out?

2. Relevant equations

The direct space basis vectors of the graphite lattice are

$\vec{a}_1 = a (\sqrt{3}/2, - 1/2, 0)$
$\vec{a}_2 = a (\sqrt{3}/2, 1/2, 0)$
$\vec{a}_3 = c (0, 0, 1)$

a = 0.246 nm being the graphite in-plane lattice parameter, c = 0.671nm the graphite out-of-plane lattice constant (distance between two A planes in a ABAB stacking) and $a/\sqrt{3}$ being the next-neighbour carbon distance of 0.142nm.

A possible choice of the coordinates of the four atoms of the graphite unit cell (assuming the standard ABAB stacking) are

$\vec{\rho}_A = (0, 0, 0)$
$\vec{\rho}_B = (\frac{a}{\sqrt{3}}, 0, 0)$
$\vec{\rho}_A' = (0, 0, \frac{c}{2})$
$\vec{\rho}_B' = (- \frac{a}{\sqrt{3}}, 0, \frac{c}{2})$

One gets the reciprocal vectors with the following formula:

$\vec{b}_1 = 2\pi \frac{\vec{a}_2 \times \vec{a}_3}{\vec{a}_1 \cdot (\vec{a}_2 \times \vec{a}_3)}$
$\vec{b}_2 = 2\pi \frac{\vec{a}_3 \times \vec{a}_1}{\vec{a}_2 \cdot (\vec{a}_3 \times \vec{a}_1)}$
$\vec{b}_3 = 2\pi \frac{\vec{a}_1 \times \vec{a}_2}{\vec{a}_3 \cdot (\vec{a}_1 \times \vec{a}_2)}$

The reciprocal lattice vector is then

$\vec{G}_m = m_1 \vec{b}_1 + m_2 \vec{b}_2 + m_3 \vec{b}_3$

The sum in the structure factor is then calculated by

$\sum_{j} e^{- i \vec{G}_m \vec{\rho}_j}$

3. The attempt at a solution

By using the formulas and values given above, one gets

$\vec{b}_1 = \frac{4 \pi}{\sqrt{3} a} (+ 1/2, - \sqrt{3}/2, 0)$
$\vec{b}_2 = \frac{4 \pi}{\sqrt{3} a} (+ 1/2, + \sqrt{3}/2, 0)$
$\vec{b}_3 = \frac{2 \pi}{c} (0, 0, 1)$

and thus

$\vec{G}_m = ( \frac{2 \pi}{\sqrt{3} a} (m_1 + m_2), \frac{2 \pi}{a} (m_2 - m_1), \frac{2 \pi}{c} m_3)$

and thus the following sum in the structure factor

$(1 + e^{i \frac{2\pi}{3}(m_1+m_2)})(e^{- i \frac{2\pi}{3}(m_1+m_2)} + e^{- i \pi m_3})$

$(1 + e^{i \frac{2\pi}{3}(m_1+m_2)})(1 + e^{- i \pi m_3})$

Where is the error? Any help would be appreciated.

Thanks,

spf

2. Mar 5, 2012

### spf

Re: Calculation of the structure factor of graphite (Solid State Physics/Crystallogra

OK, up to now I tried many things to figure it out, have redone this calculation many times, rechecked the values and have tried out three different sets of coordinates of the A, B, A' and B' atoms in the unit cell and I always get the same result. I went the other way round by asking myself which coordinates the A, B, A' and B' atoms should have within the unit cell in order to produce the result from the paper and I got AA graphite and not the standard ABAB graphite.

In conclusion, I start to think that my result is correct and the one in the paper has a typo. That's not impossible as (despite peer-review) I spotted three other typos in the paper (sign error in the direct space basis vectors (a), sign error in the reciprocal lattice vector (G), missing indices in the text. I was wrong by writing that my result does not explain the 001 extinctions (or more generally the 00l extinctions for l=2n+1). By inserting m1=0 and m2=0 into my result, it predicts these extinctions as required. (By the way, you can predict also these extinctions based on the fact that graphite (the standard ABAB one) has $6_3$ screw symmetry (which means that if you shift it by c/2 in $\vec{a}_3$ direction and rotate it by 60°), it looks exactly the same again.)

So, I guess that's problem solved then. However, there must be lots of people out there who have calculated the structure factor of ABAB graphite in their introductory solid state physics or crystallography classes, either as students or as teachers, so if one of them could just confirm me that my result is correct, it would mean a lot to me. If not, that's OK as well, I am fairly sure of my result by now.

Last edited: Mar 5, 2012
3. Nov 12, 2012

### w4m0330

Re: Calculation of the structure factor of graphite (Solid State Physics/Crystallogra

Dear spf,
I got the same results as yours when I teach the exercise course in solid state physics. your results is correct!
this results can also predict that the (101) diffraction really exist, while in that paper, (101) is extinction, which is totally wrong from experimental results!