The X-ray diffraction experiment is carried out so that the wavelength and the
direction for the incident X-ray beam are known. This information may be put into the
reciprocal lattice as follows (see the figure).

**1.** Chose a point according to the orientation of the specimen with respect to the incident beam.

** 2.** Draw a vector **AO** in the incident direction of length **2p/l** terminating at the origin.

**3.** Construct a circle of radius **2p/l** with center at A. Note whether this circle passes through any point of the reciprocal lattice; if it does:

**4. **Draw a vector **AB **to the point of the intersection.

**5. **Draw a vector **OB **to the point of the intersection.

6**. **Draw a line AE perpendicular to **OB.**

**7. **Complete** **the construction to all the intersection points in the same fashion.

To see this construction step by step just click on the Figure above!!!.

We have the following facts:

(1) Since **OB **ends on a point in the reciprocal lattice, it is normal to some set of
lattice planes and is of length **2p/**d, the interplanar spacing for the set. By definition,** OB** =
**2p/**d.

(2) Since **AO** = **2p/l**, then **OE** = **2p/l** sinq and **OB** = 2 ^{. }(**2p/l****)** sinq.

(3) Since **OB **is normal to a lattice plane, AE is the lattice plane, and q is the angle
between the incident wave **AO** and the lattice plane. It is equal to the angle between the
lattice plane AE and reflected wave **AB.**

(4) Since **2p/**d = 2 ^{. }(**2p/l****)** sinq, we obtain l = 2d sinq; the **Bragg** law is satisfied.

Thus, knowing the direction and wavelength of the incident wave, we have been able to determine which plane will diffract it.

If the circle had passed through no points, that would indicate that the particular
wavelength in question would not be diffracted by that crystal in that orientation. Further, if
the magnitude of the vector |AO| < 1/2(**2p/**a), the circle could not pass through any point,
showing that x-ray diffraction cannot occur if the wavelength exceeds 2a. We note also that
the longer the vector **AO** (the shorter the wavelength), the greater is the likelihood of the
circle`s intersecting a point, and hence of diffraction.

There might be more than one point of intersection of the circle with the reciprocal lattice indicating that there is more than one plane which can reflect the incident beam according to the Bragg law.

It is important to realize that the construction has been drawn in two dimensions for
simplicity of demonstration. In three dimensions the Ewald construction is represented by a
sphere of corresponding radius, and so the number of intersecting points (and therefore
diffracting planes ) is accordingly increased.

**Bragg Law in Vector Form**.

By means of the Ewald construction we can write the Bragg law in vector form:
Let **G** = **OB** and **k** = **AO**. For diffraction, it is necessary that the vector **k** + **G**, that is, the
vector **AB**, be equal *in magnitude* to the vector **k** or

(**k** + **G**)^{2} = **k**^{2}

or

2**k ^{.} G** +

If we call the scattered wave vector **k***, than

** k*** =** k** + **G **(2)

so we can write

** k***^{2} = **k**^{2} (3)

and **k*** - **k** = **G**, showing (2) that the scattering changes only the direction on **k**, and (3)
that the scattered wave differs from the incident wave by a reciprocal lattice vector **G**.

Equations (2) and (3) are the momentum and energy conservation law for x-ray diffraction, which is an example of elastic scattering.

Whether or not the **k** circle intersects a lattice vector, and hence reflects, depends
on its magnitude and orientation. Using (1) we can construct in the reciprocal lattice the
locus of all those waves that can produce **Bragg** reflection. This locus represents a set of
planes in three dimensions. The volume terminated by those planes is called Brillouin
zone.