# How do you calculate a unit cell's dimensions from an XRD spectra?

My inorganic lab had us do an XRD measurement, but I've never been explained how to interpret the data.

# Question:

Calculate the unit cell dimensions $a$, $b$, and $c$, for $\ce{YBa2Cu3O7}$ from the indexed X-ray powder pattern provided in [the textbook]. Explain why the crystals are nearly tetragonal in terms of the atomic structure of the compound. The following formula is useful:

$$\sin^2{\theta} = \frac{\lambda^2}{4}\left(\frac{h^2}{a^2}+\frac{k^2}{b^2}+\frac{l^2}{c^2}\right)$$ The three numbers, $hkl$, called Miller indices, indicate the direction of the scattering plane. The Miller indices can be read from the diffractogram provided. For simple reflections of the type $h00$, $0k0$, and $00l$, the value of $h$, $k$, or $l$ corresponds to n in Bragg's law.

# Bragg's law:

$$n\lambda = 2d\sin{\theta}$$

# The spectrum:

(This is a scan from my textbook, while working on it I circled some items. Just ignore that.)

# My Attempt:

$$\lambda = 0.154~\mathrm{nm}$$

Scouring the internet I found the formula for tetragonal atomic structure: the axes $a = b \neq c$, and the axis angles $\alpha = \beta = \gamma = 90^\circ$.

Solving for $c$: \begin{align} \sin^2{(22.9^\circ)} &= \frac{\lambda^2}{4} \left(\frac{h^2}{a^2}+\frac{k^2}{b^2}+\frac{l^2}{c^2}\right) \\ &=\frac{0.154^2}{4} \left(\frac{0^2}{a^2}+\frac{0^2}{b^2}+\frac{3^2}{c^2}\right) \\ 0.151 &= 0.0237\left(\frac{3^2}{c^2}\right) \\ c &= 1.188 \end{align}

For $a$ and $b$:

\begin{align} \sin^2{(22.9^\circ)} &= \frac{\lambda^2}{4} \left(\frac{h^2}{a^2} + \frac{k^2}{b^2} + \frac{l^2}{c^2}\right) \\ &= \frac{0.154^2}{4} \left(\frac{1^2}{a^2} + \frac{0^2}{b^2} + \frac{0^2}{c^2} \right) \\ 0.151 &= 0.0237 \left(\frac{1^2}{a^2}\right) \\ a &= 0.156 \end{align}

Is this the right approach? If I have the right answer, what should the units be?

• You are doing it right. For the units since you are only using nm in the calculation (via the lambda), your results will be in nm too. Howerer, crystallographs are used to express cell parameters in Angstrœm (0,1nm) although it's not an official unit. – Babounet Feb 26 '15 at 8:27
• your are doing it right! – Yomen Atassi Feb 27 '15 at 17:23
• @john sorry for this late comment but if you still want to check whether your calculated unit cell values are correct i have found this small software useful – Eka Apr 30 '15 at 18:15

You should not really have assumed that the compound was tetragonal. If it was indeed tetragonal, then the peaks corresponding to the $(200)$ and $(020)$ planes would overlap. Instead, you can see that one occurs at $2\theta = 46.7^\circ$ and the other at $2\theta = 47.6^\circ$. The question may be considered a little misleading, but it does hint at this; it says that the compound is "nearly tetragonal", and it gives you the general formula for an orthorhombic crystal, which is one with $a \neq b \neq c \,(\neq a)$ and $\alpha = \beta = \gamma = 90^\circ$.

The next problem is that you are given values of $2\theta$ (which happens to be the convention), but the formula reads $\sin^2{\theta}$. As such, you need to make the appropriate adjustments.

This question is relatively easy because the peaks have already been indexed for you (i.e. the values of $h$, $k$, and $l$ are already found out). Therefore, all that is left is to plug the numbers into the formula you are given. Which peaks you use is entirely up to you, although it is obviously easier to use $(h00)$, $(0k0)$ and $(00l)$ peaks.

The reflections of the $(200)$ plane occur at $2\theta = 47.6^\circ$. So:

\begin{align} \theta &= 23.8^\circ \\ \sin^2{(23.8^\circ)} &= \frac{(1.54~Å)^2}{4} \cdot \frac{2^2}{a^2} \\ a &= 3.81~Å \end{align}

Similarly, you can show using the $(020)$ and $(003)$ peaks that $b = 3.89~Å$ and $c = 11.64~Å$. This is actually pretty good. The reported values of $a$, $b$ and $c$ are $3.8231~Å$, $3.8864~Å$, and $11.6807~Å$ respectively.

Regarding units, if you make sure that you leave your units inside the intermediate workings (as they should be), you should never run into confusion about "what unit should the answer be". In your question, you converted $\lambda$ into units of $\mathrm{nm}$. $\sin^2\theta$, $h$, $k$, and $l$ are all dimensionless. Therefore, your calculated values of $a$, $b$, and $c$ must match the units of $\lambda$ you used, i.e. $\mathrm{nm}$.

In my answer, I used ångströms throughout. It is not wrong to use $\mathrm{nm}$, but as Babounet noted in the comment section, ångströms are more commonly used in crystallography.