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

Question

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?

Answer 1

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.

Answer 2

As mentioned by RobertTheTutor, the angle the incoming beam makes with the surface is $\theta$, therefore the diffraction angle is $2\theta$ (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.

$$2\theta = 36.12^\circ \rightarrow \theta = 18.06^\circ$$

The interplanar spacing for the (311) is calculated,

$$d_{311}= \frac{n \lambda}{2 \sin \theta} = \frac{1 \times \pu{0.0707 nm}}{2 \times \sin(18.06^\circ)} = \pu{0.1147 nm}$$

For a cubic lattice,

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

thus for the (311),

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

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,

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