Second Peak in X-Ray Diffraction Pattern of bcc Metal

By Bragg’s law, $$2d\sin\theta=n\lambda$$ and $$d=\frac{a}{\sqrt{h^2+k^2+l^2}}$$

Combining the two gives: $$\sin\theta=\frac{n\lambda\sqrt{h^2+k^2+l^2}}{2a}$$

By the reflection rule for bcc, there will be destructive interference when $\sqrt{h^2+k^2+l^2}=\sqrt{1}$, on the (100) plane. The first peak will thus occur when $n=1$ and $\sqrt{h^2+k^2+l^2}=\sqrt{2}$, on the (110) plane. However, I’m confused about the second peak. It will be impossible to get $\sqrt3$ while satisfying the reflection rule for bcc, so I’ve moved on to $\sqrt4$, which is equal to 2. However, mathematically there is no difference between having $n=2,\sqrt{h^2+k^2+l^2}=1$ ((100) plane) and $n=1,\sqrt{h^2+k^2+l^2}=2$ ((200) plane). Which one does the second peak represent, then? Does it represent both? Why then is this peak so much smaller than the other peaks in most diffraction patterns that I see? For example, for $\alpha$-iron, the diffraction pattern is

• Have a look at the rules for systematic absences in a powder spectrum, for example, for body centred cubic crystals, reflections interfere destructively only when $h+k+l=$ odd and so these are absent. – porphyrin Oct 4 '16 at 15:05
• @porphyrin I have made a mistake in including the (100) plane. Thank you for pointing that out. I have corrected my question. – lightweaver Oct 5 '16 at 7:12

The reflection with $n=2,\;(h,k,l)=(1,0,0)$ is the same as $n=1,(h,k,l)=(2,0,0)$. Yes, I said "same", not "similar"; they are one thing, there are no "both". For this reason, you may just as well abandon using $n$ altogether.

As for why the peak is so small compared to others you've seen, let me respectfully suggest that maybe you haven't seen all that many. Indeed, there is no law that all peaks on the diffraction pattern must be about the same size. On the contrary, they are essentially proportional to the squares of Fourier coefficients of electron density, and as such, can be anything from 0 to the maximum. The early textbook exercises usually employ some mercifully simple cubic or tetragonal structures with few well-resolved and well-developed peaks. Not all compounds are like that. Look at the graphite diffraction pattern, for instance: you'll basically see a lonely (002) peak, large as life, and only after the most careful inspection you'll start noticing the rest of them amidst the background noise.

As for the (100) peak, it couldn't be there, since it fell victim to the systematic absences of the bcc lattice. Ditto for (111) and (210). Only those with even $h+k+l$ survived, and that's precisely what you see on the picture.

• Thank you. However, I'm not quite understanding your first assertion. Could you further explain why they are exactly the same (perhaps by way of a diagram)? – lightweaver Oct 5 '16 at 7:06
• What diagram? $1\cdot\sqrt{2^2+0^2+0^2}=2\cdot\sqrt{1^2+0^2+0^2}$, there is nothing more to it. – Ivan Neretin Oct 5 '16 at 7:35
• Yes, the math shows that. But I don’t see why physically that’s the case. Shouldn’t the diffractometer detect a constructive maxima arising from the n = 2, (100) case, and another constructive maxima arising from the n = 1, (200) case? They are different planes, so would the second diffraction peak reflect both contributions? You say they are exactly the same thing, but I don’t see how physically that could be. – lightweaver Oct 5 '16 at 13:11
• Just how would it detect these supposed two maxima, if they appear at exactly the same angle? As for the planes, they are but a mathematical abstraction. There are no planes in crystals, much like there are no huge walls along all meridians and parallels on Earth, even though one might get that impression from looking at the maps. – Ivan Neretin Oct 5 '16 at 13:21

To answer the last part of your question the amplitude of the scatted x-ray is proportional to what is called the structure factor $F_{hkl}$ and the measured intensity to $|F_{hkl}|^2$.

If the unit cell contains i atoms with a scattering factor of $f_i$ each of the atoms are at the coordinate $ax_i+by_i+cz_i$ within the unit cell of where a, b, c are the unit cell dimension, and $x, y, z$ are the fractions of a, b, c. For example an atom at the mid-point of a has $x=1/2$.
The scattering factors $f_i$ have a complicated form depending on scattering angle but for most use can be considered to be a constant proportional the the number of electrons in each atom.

The structure factor is defined as $$F_{hkl}=\Sigma _i f_i e^{i\phi_i(h,k,l)}$$ where the phase is defined as $\phi_i(h,k,l) = 2\pi(hx_i+jy_i+lz_i)$.
(If you are unfamiliar with this notation you can convert to sine and cos using $e^{i\phi} = cos(\phi) + i~sin(\phi)$) where $i=\sqrt(-1)$

The phase arises because every atom in the unit cell contributes to every reflection and so to every {hkl} but because most atoms are not on an {hkl} they contribute by scattering in proportion to their distance between the parallel planes that make up each {hkl}. As these distances are different this results is a slightly different path length and this appears as a scattered wave with a different peak position (a phase delay) relative to a scattered wave from an atom on the {hkl} and this alters the total wave amplitude when waves from all atoms are added up.

The calculation is easy with cubic lattices, but very tedious, and starts by drawing out the unit cell (in reciprocal space) and labelling all the planes {hkl} and working out $x, y, z$ for each atom and doing the sum. The scattering factors for atoms the edge and corners of unit cells have to be taken as fractions of their values, 1/2 for sides and 1/8 for corners for example. Because $cos(\phi(hkl))$ can vary between $\pm1$ it is easy to see how a variation in intensity occurs.
The rules for cubic lattices are that all {hkl} are present for the primitave cubic, $h+k+l=$ odd are absent in bcc and for face centred lattices for a reflection to be present h, k, l are all odd or all even.