Wave length for a bragg diffraction from a incident x-ray.

[Shawj02]

A cubic crystal with a lattice constant "a" is mounted with a [100] direction parallel to the incident x-ray beam. what would be the wave length for a bragg diffraction to occur from (110) and (111) planes?

So Bragg's law is:
2d sin (theta) = n wavelength.

So my thoughts are:
This is actually two questions; between (100),(110) and (100),(111).
So my first solution looks like
Angle between (100) and (110) = ArcTan(a/a) = pi/4
and I think "d" equals either Sqrt(2)a or a/Sqrt(2)
And then plug the correct d and theta into Braggs for the wavelength.


And my solution for (100),(111).
Angle between (100) and (111) = ArcTan(Sqrt(3)a/a) = pi/3
and "d" is either Sqrt(3)a or a/(Sqrt(8))
And then plug the correct d and theta into Braggs for the wavelength.

Does this seem like I am on the correct path for the solution? which d is the correct method.
Im worried that I've simplified the question too much and there is more too it.

Also I found that "photons x-rays: wavelength = 12.4/E(keV) * [A]"
But I am not too sure what the "[A]" is.

Thanks!

(Im pretty sure that the d values are a/Sqrt(2) and a/(Sqrt(8), can someone check please?)

 

[BigJDubs]

Yes, your d values are correct - a/Sqrt(2) for (100),(110) and a/(Sqrt(8)) for (100),(111). The "[A]" in the equation you found is the angstrom unit, which is a unit of length equal to 0.1 nanometers.