Crystal Structure Analysis by X Ray Diffraction
Structures of solids can be determined by various techniques. Out of these techniques, X-rays and microscopic studies have helped immensely in studying crystallography of metals. Therefore these two techniques are very popular. Hence X-rays will be discussed in this article and microscopy is discussed in another one.
X-RAYS
X-rays have substantially contributed to reveal the structure of solids. These are electromagnetic waves of ultrahigh frequency. Their wavelength (0.46 oA — 2.29 oA), is almost the same as the inter-atomic spacing in the solid. This is an advantageous situation for X-rays over optical microscopy.
The method of examination by X-rays is called radiography. Radio-graphic determination of solid structures pertains to following studies:
- Type of unit cell
- Disorder in the unit cell
- Orientation of poly-crystalline crystals
- Choice of preferred orientation of crystals
- Revealing of imperfections
- Arrangement of atoms etc.
Production of X-rays
X-rays are produced by striking a solid target material (preferably tungsten or molybdenum) by a beam of cathode ray electrons. Coolidge vacuum tube is employed for this purpose.
The X-rays may be either hard X-rays, or soft X-rays. The hard X-rays require voltages above 50 kV for their production. They are mainly employed in the study of solid structures and in industrial applications.
Whereas, the soft X-rays are generally used for medical applications. Higher the voltage, higher will be the intensity of X-rays produced. They possess intense power to penetrate into the solids. Observations of solid structure appear on a X-ray plate.
Laue Theory of X Ray Diffraction
X-rays possess wave nature. When X-rays are passed through a symmetrically ordered crystal, a series of dark spots are produced on the photographic plate besides a black spot (Laue spot) at the center.
Dark spots are the result of X-rays diffraction in different directions. Such diffraction patterns are known as Laue patterns, and form the backbone in the study of structures of solids.
For diffraction to occur, the spacing among the array of atoms in a crystal must be of the same order as the radiation. As X-rays have wavelengths in the range of inter-atomic spacing, therefore they are diffracted by the crystals. Application of this characteristic is made in the study of crystal structures.
X Ray Diffraction Bragg’s Law
W.H. Bragg and L. Bragg, devised an X-ray spectrometer, and used a crystal as reflecting grating for its study. The crystal was mounted on a turntable. A beam of X-rays was made incident on this crystal. Beam of incident X-rays, at a glancing angle θ, were scattered from the crystal planes which were rich in atoms.
The X-rays were scattered by each individual atom lying on a parallel plane. This phenomenon is known as X-rays diffraction. The Bragg’s law of X-rays diffraction is explained as follows:
A crystal contains millions of atoms. They produce combined scattering i.e. reflection of X-rays from these planes. The system of incident X-rays, glancing angle and the planes in a crystal is shown in Figure 1.
Here A, B, C, etc., are the atoms. Ray number 1 reflects from atom A, ray number 2 from atom B, and so on. Path difference between the two rays will decide whether reflected rays are in phase or not.
The phase difference may be found by drawing perpendiculars AD and AE as shown in Figure. It is obvious that the extra distance traveled by ray number 2 is
= DB +BE = d sinθ + d sinθ = 2 d sin θ
Where d is the inter-planer spacing.
If the path difference between two reflected rays is equal to an integral multiple of wavelength λ, the rays will be in phase. Hence condition of Bragg’s diffraction gives
2d sin θ = nλ ……….(equation 1)
Where n = 1, 2, 3 … is an integer. Order of reflections will be 1, 2 and 3 depending on the value of n. If n = 1, the reflection will be known as first order; for n = 2, it will be called second order reflection; and so on. Equation 1 is known as Bragg’s equation.
X-rays of different wavelengths can be used to get reflections from various set of planes. As sin cannot be more than unity, 1, hence λ is governed by 2d = λ for n = 1.
So the inter-planer spacing of 3 oA gives upper limit of λ as 6 oA. If λ > 6 oA, there will be no reflection.
Normally Ka radiation of molybdenum target is used in diffraction studies as its wavelength 0.71 oA lies within a reasonable range of inter-atomic spacing in solids.
Structure Determination
Lattice constants of a cubic crystal can be obtained using Bragg’s Equation 1. On considering (100) plane and n = 2 and 3, we get
2λ = 2d100sin θ
λ = d100sin θ ……..(2.1)
3λ = 2d100sin θ …..(2.2)
Now considering (200) plane and n = 1. We get
λ = 2d200sin θ …….(3)
Similarly for plane (300) and n = 1, we have
λ = 2d300sin θ ……..(4)
By comparing equations 2.1 and 3, we find that
d100sin θ = 2d200sin θ ……..(5)
It can be shown that d100 = 2d200. Therefore, its substitution in the equation 5 suggests that the equation 2.1 and 3 are identical.
Infact, sets of above two planes are parallel, therefore their reflections superimpose each other.
Similarly Eqs. 2.2 and 4 are identical as d100 = 3d300. Hence third order reflections from (100) planes and first order reflections from (300) planes superimpose each other.
Thus we do not need to consider variation in n, and n = 1 can be taken for all reflections from parallel planes (100), (200), (300), …, etc.
Also we know that for a cubic unit cell, d[hkl] = a/(h2 + k2 + l2)1/2. With the help of this information and the equation 1 we obtain:
Sin2θ = λ2(h2 + k2 + l2)/4a2
Values of θ and λ, are known from the set-up of X-ray system. If lattice constant α or atomic radius r are known, the value of (h2 k2 + l2) can be determined. If
- h2 + k2 + l2= 1, it may mean that 12 + 02 + 02 = 1, or 02+ 12 + 02 = 1, or 02+ 02+12= 1. Therefore (hkl) will be either (100), (010) or (001). Starting from the lowest index plane, we conclude that {hkl} = {100} . Similarly,
- h2 + k2 + l2 = 2, indicates {hkl} = {110},
- h2 + k2 + l2 = 3, indicates {hkl} = {111},
- h2 + k2 + l2 = 4, indicates {hkl} = {200},
- h2 + k2 + l2 = 5, indicates {hkl} = {120},
- h2 + k2 + l2 = 6, indicates {hkl} = {112} ,
- h2 + k2 + l2 = 7, indicates {hkl} = {NIL}. It means that there are no reflections corresponding to this value, and such planes do not exist.
- h2 + k2 + l2 = 8, indicates {hkl} = {220},
- h2 + k2 + l2 = 9, indicates {hkl} = {122}, and so on.
Determining Type of Cubic Lattice
Now a question arises as to how we distinguish between different lattices of a cubic system i.e. how to know whether a crystal has SC, BCC, FCC or DC structure?
Its answer is simple. All combinations of (h2 + k2 + l2) do not lead to reflections for a given lattices. There is no first order reflection in BCC from (100) planes, although a second order reflection is possible which superimposes on first order reflection from (200) planes.
An extinction rule can be framed for cubic crystals as under:
Here, zero is taken as an even number.
Powder Method of Crystal Structure Analysis
Crystal structures can also be analyzed by the powder crystal method. This experimental technique is more suitable for highly symmetric crystals.
The powder consists of a large number of randomly oriented tiny crystals. A monochromatic X-rays beam of generally Ka radiation is made incident on powder as shown in Figure 3.
The Debye-Scherrer method, after the name of its inventor, requires the powder to be taken in a capillary tube of non-diffracting materials. The powder may also be gummed oh a hair or pasted on a thin wire of non-diffracting material.
The powder is placed at the centre of a cylindrical camera, and is surrounded by a photo film completely so as to receive rays diffracted up to 180°. This arrangement is shown in the Figure 4(a).
The powder technique is based on the principle that Bragg’s reflection will occur as all possible diffraction planes are available. This is due to the fact that powder contains many hundreds of finely ground, randomly oriented tiny crystals.
Reflections occur from sets of parallel planes lying at different angles with respect to the incident beam of X-rays. Path of the reflected rays form a cone whose axis lies along the incident X-rays beam. The cone angle is 4θ which is twice the glancing angle for a certain set of planes.
The cone of X-rays intersects the photo film and forms a series of concentric circular rings on it. When narrow width film strip is spread out, parts of circular rings are visible as shown in Figure 4(b).
When Bragg angle is 45°, the cone opens out into a circle; but when this angle is more than 45°, a reversed reflection is obtained. These situations are marked A and B respectively in Figure (a).
To obtain intense reflected lines, the exposure in Debye-Scherrer camera is done for many hours. The geometry of Figure (b) indicates the following relation:
l = 4 Rθ
Where l is the distance between two corresponding arcs on the film, R is radius of the powder camera, and θ is Bragg’s angle in radians.
As lis an experimentally observed value and R a known constant of the experimental arrangement, hence θ can be calculated. This value θ, on substituting in equation 1 gives inter-planer spacing.