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

[PsychonautQQ]

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

 

[Laura]

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.