How Does Copper's X-ray Wavelength Relate to Aluminum's Bragg Angle Calculation?

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SUMMARY

The discussion centers on the relationship between Copper's X-ray wavelength of 1.54 Angstroms and the calculation of the interplanar distance in Aluminum using Bragg's Law. The Bragg angle for reflection from the (111) planes in Aluminum is given as 19.2 degrees. By applying the equation nλ = 2dsinθ, where n is the order of reflection, λ is the wavelength, d is the interplanar distance, and θ is the Bragg angle, participants clarify how to compute the interplanar distance for Aluminum's fcc structure based on the provided parameters.

PREREQUISITES
  • Understanding of Bragg's Law and its application in X-ray diffraction.
  • Familiarity with the face-centered cubic (fcc) crystal structure.
  • Knowledge of X-ray wavelength measurements in Angstroms.
  • Basic concepts of constructive interference in wave physics.
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  • Study the derivation and applications of Bragg's Law in crystallography.
  • Learn about the properties and calculations related to face-centered cubic (fcc) structures.
  • Explore the significance of X-ray wavelengths in material science.
  • Investigate the relationship between atomic spacing and diffraction patterns in X-ray crystallography.
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Students and professionals in materials science, physicists focusing on crystallography, and anyone interested in the practical applications of X-ray diffraction techniques.

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Homework Statement


I'm a bit confused by this question. It starts off by saying a Cu target emits an x-ray line of wavelength λ=1.54 Angstroms. It then goes and says part A of the problem..
A) Given that the Bragg angle for reflection from the (111) planes in Al is 19.2 degrees, computer the interplanar distance from these planes. Recall that aluminum has an fcc structure.

I don't understand why they start off by telling me about copper and then go onto aluminum in the actual question part... is there a connection that I am missing here?


Homework Equations





The Attempt at a Solution

 
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For Bragg's Law you have nλ = 2dsinθ. This corresponds to when constructive interference of the diffracted x-rays occurs. You can also see from this equation that constructive interference depends on both the x-ray wavelength (λ) and the atomic spacing (d). It's the same for light diffraction or water wave diffraction - constructive interference always depends on the wavelength and the lattice spacing.

Note: The whole idea behind using x-rays to image crystals is that the wavelength of x-rays is on the same order as atomic spacings.
 

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