Deducing Bragg angle for G_220 Bragg peak for LiF

[ChrisJ]

Homework Statement

Calculate the reciprocal lattice vectors for LiF.
Given the lattice parameter $a = 0.402 nm$ calculate the magnitude of the $\textbf{G_{220}}$
reciprocal lattice vector for LiF and thus deduce the Bragg angle for the (2,2,0)
Bragg peak for incoming wavelength of $0.9 nm.$

Homework Equations

$d=\frac{2 \pi}{|\textbf{G}_{220}|}$
$|\textbf{G}_{220}| = \frac{2 \pi}{a}\sqrt{h^2+k^2+l^2}$
$2d \sin{\theta} = n \lambda$

The Attempt at a Solution

This is not coursework, it is a question from a past exam paper I'm trying to do in preparation for my exam.

In a previous bit of the question we are given the positions of the Lithium and Flourine atoms and I deduced it was a FCC with 2 basis atoms, I have never calculated the reciprocal lattice vectors for FCC with basis of 2 atoms before, so just went ahead as I have always done hoping it was the same, but I assume that is not correct and is why I am getting it wrong.

But I don't know why they asked us to calculate them anyway, as in the notes we were given are included the equations in the relevant equations section above, so I just did.

$$|\textbf{G}_{220}| = |\textbf{G}_{220}| = \frac{2 \pi}{a}\sqrt{2^2+2^2}= \frac{4\sqrt{2}\pi}{4.02 \times 10^{-10}}=4.42\times 10^{10} \textrm{m}^{-1} \\ \sin{\theta}=\frac{\lambda |\textbf{G}_{220}| }{4 \pi} = \frac{9 \cdot 4.42}{4 \pi}$$

Which I obviously can't take the inverse sin of since its not between 0-1 .

 

[TSny]

A wavelength of 0.9 nm does seem large. Could it actually be 0.9 A^° = .09 nm?

 

[ChrisJ]

A wavelength of 0.9 nm does seem large. Could it actually be 0.9 A^° = .09 nm?

It 100% says 0.9nm on the past paper, after tearing my hair out for ages I too was starting to think that it could possibly be an error on the paper and be 0.9 angstroms instead, but didn't feel confident in emailing the lecturer to ask. But I wrote the text from the question word for word on this post, so at least its not just me.

Thanks :)

 

[TSny]

OK. Contacting your lecturer is a good idea in my opinion, just to make sure.