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X-Rays · Lesson 5 of 5

Bragg's Law: X-Rays Meet a Crystal

15 min read3 ideas5 checks1 figure to playBy Asheshanand Acharya
What you will learn 0 / 3
Reflection from crystal planes

A grating made by nature

To measure the wavelength of light, one uses a grating of fine lines ruled close together. Its lines must be about as far apart as the wavelength. X-rays have wavelengths near \(0.1\,\text{nm}\), the size of an atom. No ruling machine can space lines so finely. In 1912 Max von Laue saw that nature had already made such a grating: a crystal.

In a crystal of rock salt, sodium and chlorine atoms stand in a pattern that repeats in every direction. Through such a pattern one can draw many sets of parallel crystal planes, each full of atoms. The planes of one set are a fixed distance \(d\) apart. In rock salt, one important set has \(d = 0.282\,\text{nm}\).

Each plane of atoms acts like a very weak mirror. It sends back a tiny fraction of an X-ray beam, and lets the rest go on to the next plane. The reflected ray leaves at the same angle to the plane as the incoming ray. For X-rays the angle is measured from the plane itself. It is called the glancing angle \(\theta\).

The extra path between two neighbouring planesThree parallel planes of atoms, rows of dots a distance d apart. Two parallel X-rays arrive at the glancing angle theta of 30 degrees. Ray 1 reflects from atom A on the top plane; ray 2 reflects from atom B directly below it on the next plane. Perpendiculars dropped from A onto ray 2, before and after its reflection, cut off two pieces of ray 2, from C to B and from B to D, drawn thick. Each piece is d sin theta, so ray 2 travels 2 d sin theta farther than ray 1. θ θ ray 1 ray 2 A B C D d sin θ d sin θ d
Fig 5.1Two parallel X-rays reflect from two neighbouring planes, \(d\) apart, at the glancing angle \(\theta\). Ray 1 reflects at A on the top plane and ray 2 at B on the plane below. The perpendiculars from A cut off the thick pieces CB and BD of ray 2. Each piece is \(d\sin\theta\), so ray 2 travels \(2d\sin\theta\) farther.

A single plane reflects too little to notice. But a beam enters millions of planes, and every one of them reflects a little. Whether these small reflections add up depends on their paths. If they arrive in step, they reinforce into a strong beam. If not, they cancel, as two waves on a string do, and as light does in the chapter on interference.

Braggs law
Maths you need

The extra path

Look again at Fig 5.1. Ray 2 goes deeper than ray 1, by one plane. Drop a perpendicular from A onto ray 2 before it reflects, and another after it reflects. Up to the first perpendicular the two rays have gone equal distances, and after the second they go on side by side. The difference lies in the two thick pieces, CB and BD.

Deriving \(CB + BD = 2d\sin\theta\) fill the gap
1

\[ AB = d \]

Why. B is on the next plane, straight below A.

2

?

Your turn. in the right triangle ACB the angle at A equals \(\theta\), and AB is the hypotenuse. Which line is it?

3

\[ BD = d\sin\theta \]

Why. the outgoing ray makes the same angle \(\theta\) with the plane, so the triangle ADB is a mirror copy of ACB.

4

\[ CB + BD = 2d\sin\theta \]

Read it. add the two pieces; this is the extra path of ray 2.

Which line fills step 2?Unanswered

The two rays reinforce when this extra path is a whole number of wavelengths. Then the crests of ray 2 fall on the crests of ray 1, and the same holds for every deeper plane. This is the condition found by William Lawrence Bragg in 1912, Bragg's law.

law 4.2
Bragg's law

X-rays of wavelength \(\lambda\) reflected from a set of parallel crystal planes a distance \(d\) apart give a strong beam only when \(2d\sin\theta = n\lambda\). Here \(\theta\) is the glancing angle and \(n = 1, 2, 3, \ldots\) is the order of reflection.

RememberBragg's law
\[ 2d\sin\theta = n\lambda, \qquad n = 1, 2, 3, \ldots \]
Holds whenX-rays (or any waves) reflected from one set of parallel planes of a crystal; \(\theta\) is measured from the planes, not from their normal.
Symbols\(d\) the spacing of the planes, \(\theta\) the glancing angle, \(n\) the order of reflection, \(\lambda\) the wavelength
Units\(d\) and \(\lambda\) in the same unit (nm or Å)
Two reflected waves, in step or out of stepOn the left, two neighbouring planes of atoms of rock salt, 0.282 nanometre apart, and two X-rays of copper K-alpha, 0.154 nanometre, arriving at the glancing angle theta set by a slider from 5 to 85 degrees. Ray 1 reflects from the top plane, ray 2 from the plane below; the extra path of ray 2, 2 d sin theta, is drawn thick. On the right, the two reflected waves are drawn one above the other, ray 2 shifted by its extra path, and below them their sum. At most angles the two waves are out of step and, with many planes, the reflections cancel. At 15.8 degrees the extra path is exactly one wavelength, the waves are in step and the sum is twice as tall: a strong first-order reflection. Whole numbers come again at 33.1 degrees, two wavelengths, and 55.0 degrees, three. θ d ray 1 ray 2 together 2d sin θ = 0.154 nm = 1.00 λ n = 1: in step, a strong reflection
SimCopper Kα X-rays, \(0.154\,\text{nm}\), on rock-salt planes \(0.282\,\text{nm}\) apart. Turn the glancing angle. The two reflected waves are drawn on the right, one above the other, and their sum below. At \(15.8°\) the extra path is exactly one wavelength, and the sum is twice as tall. Whole numbers of wavelengths come again at \(33.1°\) and \(55.0°\).

Turn the angle slowly in the figure above. At most angles the extra path is some odd fraction of a wavelength, and the two waves partly cancel. With millions of planes, each a little further out of step, the reflections cancel almost completely. Only near \(15.8°\), \(33.1°\) and \(55.0°\) do all the planes agree, and the crystal sends back a strong beam.

How many orders?

The sine of an angle is never more than one. So \(n\lambda = 2d\sin\theta\) can be at most \(2d\). The orders that exist are the whole numbers \(n \le 2d/\lambda\). For copper Kα on rock salt, \(2d/\lambda = 0.564/0.154 = 3.66\), so there are three orders. A wave longer than \(2d\) gives no strong reflection at all.

ExampleAll the orders of one crystal

X-rays of wavelength \(1.0\,\text{Å}\) fall on a crystal whose planes are \(2.0\,\text{Å}\) apart. Find the glancing angles of the first-order and second-order maxima. What is the highest order that can occur?

Reveal one step at a time and finish the rest yourself.
Common mistake

The glancing angle is measured from the plane, not from the normal. A beam that meets the planes at \(60°\) to their normal has \(\theta = 30°\). The reflected beam is then turned through \(2\theta\) from its first direction, which is why a Bragg detector sits at \(2\theta\) from the beam.

CheckpointUnanswered

X-rays strike a set of crystal planes at \(70°\) to the normal of the planes. What value of \(\theta\) goes into Bragg's law?

Checkpoint · numericUnanswered

The first-order maximum for X-rays of \(0.154\,\text{nm}\) from a set of planes is at a glancing angle of \(30°\). Find the spacing \(d\) of the planes, in nm.

nm± 2 %
Measuring spacing and wavelength

The Bragg spectrometer

The Braggs, father and son, built an instrument to use the law. A narrow X-ray beam falls on a crystal mounted on a turntable. As the crystal turns through the glancing angle \(\theta\), a detector turns through \(2\theta\). So the detector always stands where a reflection would go. A graph of the count against \(\theta\) shows a sharp peak at every Bragg angle.

The Bragg spectrometerOn the left, a narrow beam of copper K-alpha X-rays, 0.154 nanometre, comes from the left and strikes a rock-salt crystal on a turntable. A slider turns the crystal to the glancing angle theta from 5 to 60 degrees; the detector turns with it to 2 theta from the beam, so it always sits where a reflection would go. On the right, a graph of the detector count against theta draws the whole curve, with a dot at the present angle. The count is small at most angles and shows three sharp peaks, first order at 15.8 degrees, second at 33.1 and third at 55.0, each weaker than the last. The readout gives theta and the detector angle 2 theta. X-rays detector crystal θ 2θ 204060Oθ (°)detector count n = 1 n = 2 n = 3 θ = 15.8°, detector at 2θ = 31.6°
SimCopper Kα X-rays, \(0.154\,\text{nm}\), on a rock-salt crystal. Turn the crystal. The detector follows at \(2\theta\), and the dot on the graph shows its count. The count is small at most angles and peaks at \(15.8°\), \(33.1°\) and \(55.0°\), the orders 1, 2 and 3, each weaker than the last.

Check the three peaks in the figure above against Bragg's law. Their sines are \(0.273\), \(0.546\) and \(0.819\), in the ratio \(1 : 2 : 3\). That ratio tells the experimenter that the peaks are orders 1, 2 and 3 of one wavelength. The first gives \(\lambda = 2d\sin\theta = 0.564 \times 0.273 = 0.154\,\text{nm}\).

The spectrometer works in two directions. With X-rays of a known wavelength, the angles give the spacing \(d\) of a crystal's planes, and so the arrangement of its atoms. With a crystal of known \(d\), the angles give the wavelength of unknown X-rays. Moseley measured his Kα lines this way, with crystals of known spacing.

ExampleRock salt and two wavelengths

(a) Rock-salt planes are \(2.82\,\text{Å}\) apart. Find the first-order glancing angle for X-rays of \(1.41\,\text{Å}\). (b) A crystal with \(d = 1.0\,\text{Å}\) gives a first-order maximum at \(30°\). Find the wavelength, and the least tube voltage that can make it.

Try it on paper first. The solution is one tap away when you are ready.
Checkpoint · numericUnanswered

Molybdenum Kα X-rays, \(0.071\,\text{nm}\), fall on rock-salt planes \(0.282\,\text{nm}\) apart. Find the glancing angle of the first-order reflection, in degrees.

°± 2 %
Checkpoint · in your own wordsUnanswered

Why does visible light, of wavelength \(500\,\text{nm}\), give no Bragg reflection from a rock-salt crystal?

The same law works for any wave. In the chapter on Matter Waves, electrons of \(54\,\text{eV}\) reflected from a nickel crystal by exactly this rule. That is why a crystal proved that electrons are waves, and why it proves the same of X-rays.

In the world
One medicine, two crystals

Many medicines are sold as tiny crystals. The same molecule can often crystallise in more than one pattern, and the patterns dissolve at different rates in the body. Pharmaceutical laboratories in India check each batch with X-rays. A powder of the medicine gives rings at the Bragg angles of its planes. The set of angles tells which crystal form is present.

Story · G. N. Ramachandran, Madras, 1954
Three chains twisted together

In 1954 G. N. Ramachandran and Gopinath Kartha, at the University of Madras, worked on collagen, the protein of skin, tendon and bone. They studied the X-ray pictures of collagen fibres, spot by spot. Each spot came from planes that obey Bragg's law. From the spacings they proposed that collagen is three long chains twisted round one another, a triple helix. The idea proved right. Ramachandran later made a famous chart of the shapes a protein chain can take, still used in every laboratory that studies proteins.

Exam traps
  1. Measuring \(\theta\) from the normal.
    Bragg's \(\theta\) is the glancing angle, between the ray and the plane. Subtract the angle from the normal from \(90°\).
  2. Writing \(d\sin\theta = n\lambda\).
    The extra path has two pieces, one going in and one coming out: \(2d\sin\theta = n\lambda\).
  3. Taking equal steps in angle for the orders.
    The sines go up in equal steps, \(\sin\theta = n\lambda/2d\); the angles do not.
  4. Forgetting the limit on the order.
    \(\sin\theta \le 1\), so \(n \le 2d/\lambda\). There are only a few orders, and none if \(\lambda > 2d\).