# 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?

Nobody answered this and I figured it out, so I'll explain here.

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.

As mentioned by RobertTheTutor, the angle the incoming beam makes with the surface is θ, therefore the diffraction angle is 2θ (the total angle between the incoming and outgoing beam).

So the angle given of 36.12° needs to be divided by 2 before being plugged into the Bragg's Law.

$$2\theta = 36.12° \rightarrow \theta = 18.06°$$

The interplanar spacing for the (311) is calculated,

$$d_{311}= \frac{n \lambda}{2 \sin \theta} = \frac{1 \times 0.0707\;nm}{2 \times \sin(18.06º)} = 0.1147 \;nm$$.

For a cubic lattice,

$$a = d \sqrt{h^2+k^2+l^2})$$

thus for the (311),

$$a = 0.1147 \sqrt{3^2+1^2+1^2} = 0.1147 \sqrt{11} = 0.3804 \; nm$$

Finally, using Pythagorean theorem to relate the atomic radius, r, to lattice parameter, a, for the face centered cubic structure:

For the front face, $$a^2 + a^2 = (4r)^2 \rightarrow \sqrt{2} a = 4 r$$

So,

$$r = \frac{\sqrt{2}}{4 a} = \frac{\sqrt{2}}{4 \times 0.3804\;nm} = 0.1345\;nm$$