# Find atomic radius from angle, wavelength, (311) plane and FCC structure

I've worked this problem and seem to be off by a factor of $$2$$ somehow. From Callister, 6th edition, problem 3.56W (but I don't have access to the "W" web material that actually explains Bragg diffraction). Using other texts, I found that plane spacing $$d$$ satisfies $$\frac{1}{d^2} = \frac{h^2}{a^2} + \frac{k^2}{b^2} + \frac{l^2}{c^2},$$ which for the $$(311)$$ plane in an FCC lattice (therefore cubic) gives $$a = \sqrt{11}d$$.
The problem gives $$\theta = 36.12^\circ$$ and $$\lambda = \pu{0.0711 mm}$$. Bragg's Law says $$n\lambda = 2 d \sin \theta,$$ so for first order diffraction (the problem says reflection but I assume that is a mistake) $$n=1$$ and I get $$d = \pu{0.0603 mm}$$, which gives $$a = \pu{0.2000 mm}$$ and using FCC, $$r = \pu{0.0707 nm}$$, which is about half the correct answer according to Wikipedia $$(\pu{142\pm 7 pm}$$).

Can anyone tell me where I went wrong?

It turns out that glancing angle is $$\theta$$, the angle the incoming beam makes with the surface (not the normal to the surface, as is usual in optics).
Meanwhile the diffraction angle is $$2\theta$$, the total amount the incoming beam gets "turned" to become the outgoing beam.
The problem stated the diffraction angles. They need to be divided by $$2$$ before being plugged into the Bragg Law.