Deducing Bragg angle for G_220 Bragg peak for LiF

In summary, the lattice parameter for LiF is 0.402 nm, and the magnitude of the reciprocal lattice vector is 4.42 × 10^10 m.
  • #1
ChrisJ
70
3

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.

[tex]
|\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}
[/tex]

Which I obviously can't take the inverse sin of since its not between 0-1 .
 
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  • #2
A wavelength of 0.9 nm does seem large. Could it actually be 0.9 A° = .09 nm?
 
Last edited:
  • #3
TSny said:
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 :)
 
  • #4
OK. Contacting your lecturer is a good idea in my opinion, just to make sure.
 

Related to Deducing Bragg angle for G_220 Bragg peak for LiF

1. How is the Bragg angle for G220 Bragg peak for LiF deduced?

The Bragg angle for G220 Bragg peak for LiF can be deduced using the Bragg's Law equation: nλ = 2dsinθ, where n is the order of the diffraction, λ is the wavelength of the incident X-ray, d is the lattice spacing of the crystal, and θ is the Bragg angle.

2. What is the significance of the G220 Bragg peak for LiF?

The G220 Bragg peak for LiF is significant because it is the strongest peak in the X-ray diffraction pattern of LiF. This peak corresponds to the most densely packed planes of atoms in the crystal lattice, providing important information about the crystal structure.

3. How does the Bragg angle for G220 Bragg peak change with different incident X-ray wavelengths?

The Bragg angle for G220 Bragg peak will change with different incident X-ray wavelengths because the wavelength affects the value of λ in the Bragg's Law equation. As the wavelength increases, the Bragg angle will decrease and vice versa.

4. Can the Bragg angle for G220 Bragg peak be calculated for other crystals besides LiF?

Yes, the Bragg angle for G220 Bragg peak can be calculated for any crystal that has a defined lattice spacing. However, the specific value of the angle will vary depending on the crystal's lattice structure and the incident X-ray wavelength.

5. How is the Bragg angle for G220 Bragg peak measured experimentally?

The Bragg angle for G220 Bragg peak can be measured experimentally by setting up an X-ray diffraction experiment and varying the angle of the incident X-rays until the G220 peak is observed. The angle at which the peak is observed is the Bragg angle.

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