Calculating Bragg Angle from Unit Cell in Cubic Crystal

[VulpineNinja]

Homework Statement

The unit cell in a cubic crystal is 2.74Å. Determine the Bragg angle for reflections from the planes (200), (212) and (-112). [sorry i don't know how to put a bar on top of the numbers] if the x-ray wavelength used is 1.54Å

Homework Equations

Given Bragg's condition as 2d sinθ = nλ

and d = 2π /|G|
where G = ha* + kb* + lc* according to Laue condition.

there's also d = a /√(h^2 + k^2 + l^2 )

where a is the edge length.

The Attempt at a Solution

Since d is the distance between planes in atomic lattice, I guess it meant the distance of (200), (212) and (-112) to the origin. So I calculated is by finding the magnitude of G. Then i got confused by the two d equations. Should I use 2π or a to divide with |G| to find the Bragg angle? I thought a is more logical, but the 2π /|G|[/I] is making me confused. What are the difference between them?

 

[NateD]

For (200): |G| = 2π/a, θ = arcsin(nλ/2π) = arcsin(nλ/a)For (212): |G| = √8π^2/a , θ = arcsin(nλ/√8π^2) = arcsin(nλ/a)For (-112): |G| = √8π^2/a , θ = arcsin(nλ/√8π^2) = arcsin(nλ/a)Are these correct?