Echoes from evenly spaced cliffs
Guide 1 gave you a single atom that, struck by an X-ray, sprays a faint secondary wavelet out in every direction. On its own that wavelet is far too weak to notice. The magic starts when you have not one atom but the vast, orderly army of a crystal — and here is the move that a 23-year-old William Lawrence Bragg made in 1913, a move so tidy it won a Nobel Prize. Group the atoms into families of parallel planes (the very (hkl) plane families you learned to index), and pretend each family behaves like a stack of extremely faint, half-silvered mirrors. Send in a wave, and most of the time the reflections from all those planes fall out of step and cancel to nothing. But at a few precise angles the echoes from every plane arrive crest-on-crest, add up, and burst out as a bright beam. That burst is a Bragg peak, and the rule that pins its angle is Bragg's law.
The whole trick lives in one number: the planes in a family are evenly spaced, a distance d apart — the interplanar spacing you computed back in the indexing rung. A wave grazes in at some angle, bounces off each plane, and the wave that dived down to reflect off a deeper plane has to travel a little farther than the one that bounced off the plane above it. If that extra distance happens to be a whole number of wavelengths, the returning crests line up perfectly and reinforce — a loud reflection. If instead the extra distance is half a wavelength off, a crest meets a trough, the interference is destructive, and the reflection dies. It is exactly like shouting toward a row of evenly spaced cliff walls: at most angles the echoes smear into mush, but at just the right angle they boom back together as one.
The path difference: 2 d sin theta
Now let us earn the equation instead of just quoting it, with a scrap of right-triangle geometry. Picture two neighbouring planes of a family, a distance d apart, and a ray grazing in at angle theta to each. Follow the ray that reflects off the lower plane: compared with its partner bouncing off the upper plane, it must dip down to the lower plane and climb back up. Drop a perpendicular from the upper reflection point onto the lower ray's path, and the little right triangle it makes has the plane spacing d as its hypotenuse and the extra leg as d sin theta. There is one such extra leg on the way in and a matching one on the way out. Add them and the total surplus distance — the path difference between the two rays — is 2 d sin theta.
- Draw two parallel planes of one family, separated by the spacing d, and send in a ray at glancing angle theta to each.
- The ray that reflects off the lower plane travels an extra d sin theta on the way down, and another d sin theta on the way back up.
- So the total extra path between the two reflected rays is 2 d sin theta.
- The crests reinforce only when that extra path is a whole number of wavelengths: 2 d sin theta = n lambda.
- That is Bragg's law; the integer n is the order of the reflection.
Stand back and admire how little the law asks of you. Bragg's law — n lambda = 2 d sin theta — contains no atom positions, no charges, no messy physics of scattering; just three quantities, the spacing d, the wavelength lambda, and the angle theta, tied by one clean equation. Feed it a plane family of spacing d and a wavelength, and it hands back the exact angle at which that family (its Bragg planes) flashes. The integer n is the order of reflection: n = 1 is the first-order peak, n = 2 the second, and so on, each a case where the path difference is one, two, three whole wavelengths.
A worked peak, and the ceiling on orders
Numbers make it real. Take a plane family of spacing d = 2 angstrom, a typical value for a metal or oxide, and illuminate it with the lab workhorse, copper K-alpha X-rays of wavelength lambda = 1.54 angstrom. For the first-order peak (n = 1), rearrange Bragg's law to sin theta = n lambda divided by (2 d) = 1.54 divided by (2 times 2) = 1.54 / 4 = 0.385. Take the arcsine: theta = arcsin(0.385) = 22.6 degrees. The detector, sitting at twice the Bragg angle, catches this peak at a 2 theta scattering angle of 45.2 degrees. That single line on a diffractogram is a direct, honest fingerprint of a 2-angstrom spacing in your crystal.
Push to higher orders and something instructive happens. For n = 2, sin theta = 2 times 1.54 / 4 = 0.770, so theta = 50.4 degrees and 2 theta = 100.8 degrees — a genuine second peak, farther out. But try n = 3: sin theta = 3 times 1.54 / 4 = 1.155, and a sine can never exceed 1. There is no such angle. This plane family simply cannot produce a third-order reflection with this wavelength; the orders hit a ceiling. (In the modern crystallographer's convention you met in the reciprocal-lattice rung, that n = 2 reflection off the (hkl) planes is just relabelled as the first-order reflection off the (2h 2k 2l) planes, which have half the spacing — the folding of n into the indices.)
BRAGG REFLECTION - two planes a distance d apart, glancing angle theta
incident \ / diffracted
\ /
\ th th /
------------o--------------------o------------ upper plane
\ : /
\ : d /
\ : /
----------------o-----:------o---------------- lower plane
the LOWER ray runs an extra d sin(theta) on the way in,
and another d sin(theta) on the way out:
extra path = 2 d sin(theta)
crests reinforce (loud) when 2 d sin(theta) = n lambda <- Bragg's law
worked: d = 2 A , Cu K-alpha lambda = 1.54 A
n=1 : sin th = 1.54 / 4 = 0.385 -> th = 22.6 deg -> 2th = 45.2 deg
n=2 : sin th = 3.08 / 4 = 0.770 -> th = 50.4 deg -> 2th = 100.8 deg
n=3 : sin th = 4.62 / 4 = 1.155 -> impossible (sin cannot exceed 1)
Why the wavelength must fit the crystal
That n = 3 ceiling was not a fluke of our numbers; it is a universal fence built into the law. Because sin theta can never top 1, Bragg's law demands n lambda no greater than 2 d, and at the very least (n = 1) that means lambda no greater than 2 d. Read it as a rule: to get any reflection at all from planes spaced d apart, your wavelength must be no longer than twice that spacing. This is the wavelength requirement, and it quietly decides which kinds of waves can even attempt to read a crystal.
Put scales on it and the punchline lands hard. Interatomic spacings are a few angstrom — our example was 2 angstrom — so a usable wavelength must be at most about 4 angstrom, and ideally near d itself. Visible light has a wavelength of roughly 4000 to 7000 angstrom, thousands of times too long: to visible light a crystal's planes are as smooth as glass, which is precisely why no light microscope can ever see individual atoms. X-rays, by contrast, live at 0.5 to 2.5 angstrom, fitting the interatomic spacing like a key in a lock. That is the deep reason X-rays — and, for the same reason, electrons and neutrons — are the tools that read atomic structure, while light cannot. Copper K-alpha at 1.54 angstrom is the classic lab choice, a sharp characteristic line from a copper target that comes ready-made at just the right size.
What Bragg's law really says — and what it doesn't
Now for the honest fine print, because the stack-of-mirrors story is a beautiful lie. Atoms do not reflect like mirrors at all — guide 1 was emphatic that each atom scatters a wavelet in every direction, not just the specular one. Bragg's 'reflecting planes' are a bookkeeping fiction that gives exactly the right answer for a sly reason: the mirror-like direction is precisely where the wavelets from all the atoms lying within one plane happen to stay in step, and the condition 2 d sin theta = n lambda is precisely where successive planes stay in step too. Line up both and you get a real beam. The planes reflect nothing; they are a mnemonic for a double coincidence of interference. The more fundamental telling of the same story — waves, not mirrors — is the scattering-vector and Laue picture waiting for you in guide 3.
Those two languages are not rivals; they are the same equation in different clothes. Divide Bragg's law through by lambda d and it reads 2 sin theta / lambda = 1 / d. The right-hand side, 1 / d, is exactly the length of the reciprocal-lattice vector g you built last rung; the left-hand side is the length of the scattering vector. So 'a Bragg peak occurs' is word-for-word identical to 'the scattering vector equals a reciprocal-lattice vector.' The Ewald sphere of guide 3 is nothing but the picture of when that match happens. One physics, two dialects — Bragg speaks in real-space planes and angles, Laue in reciprocal-space dots, and they never disagree.
And here is the limit you must never forget: Bragg's law fixes only WHERE the peaks are — the angles — and nothing about how BRIGHT each one is. The angle 2 theta decodes the spacing d, and the full set of d-values decodes the unit-cell size and shape. But whether a peak the law permits actually blazes, sits dim, or vanishes completely is a separate question, answered by the structure factor — the sum over the motif — coming in guide 4. Some reflections Bragg cheerfully allows are switched fully off by systematic absences from centering and glides, the subject of guide 5. The slogan to carry: peak POSITIONS give you the lattice (the cell); peak INTENSITIES give you the motif (the atoms inside it).
In practice you run the whole argument backwards. A powder pattern hands you a row of peaks; you read each 2 theta off the diffractogram, invert Bragg's law to recover the d-spacing behind it, and from the collection of d-values rebuild the unit cell. Only then do the intensities — a harder tale of the motif, and one haunted by the lost phase of each reflection — tell you where the atoms actually sit inside that cell. Bragg's law is the doorway to structure; the next three guides are what lies through it.