X-ray diffraction and Bragg's law

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Homework Help Overview

The discussion revolves around X-ray diffraction and Bragg's law, specifically addressing issues related to calculating wavelengths and lattice plane spacing from given data. Participants are exploring the implications of the Bragg condition and the relationship between scattering peaks and Miller indices.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants express confusion over obtaining multiple wavelengths from the Bragg condition, questioning the uniqueness of the solution. There are attempts to clarify how to calculate lattice plane spacing, with some suggesting the use of specific formulas related to cubic lattices. Others are exploring combinations of Miller indices to match observed wavelengths.

Discussion Status

Some guidance has been offered regarding the application of formulas and the exploration of Miller indices, but participants continue to express uncertainty and struggle with the calculations. There is no explicit consensus on the correct approach or understanding of the problem.

Contextual Notes

Participants are working with specific angles of scattering and are referencing lecture material, indicating a reliance on previously learned concepts. There is mention of a PDF resource that may provide additional context, but the effectiveness of this resource is not confirmed.

Martin89
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Homework Statement


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The Attempt at a Solution


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Hi All,

I have two issues with this question. First of all when I put the given values into the Bragg condition for diffraction I get two different wavelengths when the question implies there is only one. Secondly, I don't know how I can calculate the lattice plane spacing, d, from the given information. Any help would be appreciated, thanks!
 

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The correct formula is:
λ = 2 dhkl sinθhkl
In case of a cubic lattice with lattice parameter a one has :
dhkl = a / sqr(h2 + k2 + l2)
You have now to find out what (h k l ) planes give rise to the second and what (h k l ) planes give rise to the third scattering peak in the X-ray diffractogramm of CsI.
 
Last edited:
Lord Jestocost said:
The correct formula is:
λ = 2 dhkl sinθhkl
In case of a cubic lattice with lattice parameter a one has :
dhkl = a / sqr(h2 + k2 + l2)
You have now to find out what (h k l ) planes give rise to the second and what (h k l ) planes give rise to the third scattering peak in the X-ray diffractogramm of CsI.
I have no idea how to work that out from the given information...in lectures we were shown how to calculate d when the lattice plane was known
 
There is no "m" in the equation λ = 2 dhkl sinθhkl. Do you understand that the scattering peaks which are observed at 10.8° and 13.3° are due to scattering from different planes with different Miller indices. This has nothing to do with mth-order reflections.
 

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