Geometrical Structure Factor

Geometrical Structure Factor

The ratio of amplitude of the wave scattered on unit cell to the amplitude of wave scattered by a single electron for the same wavelength of the wave.

Let us consider jth atom on the cell having position vector \(\vec r_j=u_j \vec a+v_j \vec b+w_j \vec c\dotsm(1)\)

Let \(f_j\) be the measures of scattering power by jth atom relative to the single electron located at the same position. Where, \(u_j, v_j , w_j\) in equation (1) are called fractional co-ordinates. whereas, \(\vec a, \vec b , \vec c\) are called primitive translational vector.

Let a be the amplitude of the wave scattered by the atom having scattering centre \(\rho\) is given by

$$a=\sum_\rho e^{-i\phi}$$

$$a=\sum_\rho e^{-i \del k\rho}\dotsm(2)$$

Amplitude of wave scattered from jth ion located at \(r_j\) vector, then, $$a=\sum_j\sum_\rho f_j e^{i\del k^\vec(\rho^\vec+r^\vec_j)}$$

$$=\sum_\rho \sum_j f_j e^{-i\rho \delk}.e^{-i\delk^\vec.r_j^\vec}$$

$$=\sum_\rho e^{-i\rho\delk}\sum_j f_j e^{-i\delk.r^\vec_J}$$

$$\therefore\;\;a=\sum_\rho e^{-i\rho\delk}.S\dotsm(3)$$

Where \(S=\sum_j f_j e^{-i\delk.r^\vec_j}\) is called geometric structure factor.

Equation (3) represents the scattering amplitude in terms of geometrical structure.

If scattered wave satisfy the Bragg’s diffraction condition then, \(\delk^\vec=G^\vec\) = reciprocal lattice vector.

$$S=\sum_j f_j e^{-iG^\vec.r^\vec_j}\dotsm(4)$$

But we have \(G^\vec = hA^\vec + kB^\vec + LC^\vec\)

Therefore equation (4) becomes $$S=\sum_j f_j e^{-i(hA^\vec + kB^\vec + LC^\vec).(u_ja6\vec+v_j b^\vec+w_j c^\vec)}$$

$$=\sum_j f_je^{-i(hu_JA^\vec.a^\vec+kv_j.B^\vec.b^\vec+lw_jc^\vec.C^\vec)}$$

$$\therefore\; S=\sum_j f_j e^{-2\pi i(hu_j+kv_j+lw_j)}\dotsm(5)$$

This is the required expression for the geometrical structure factor.

1. Simple Cubic Crystals

In the simple cubic crystals, identical atoms are located at (0, 0, 0) i.e.\( (u_j, v_j, w_j)\) = (0, 0, 0)

Then, geometrical scattering factor $$S=\sum_j f_je^{-2\pi i(u_j h+v_j k +w_j l)}$$

$$S=fe^0$$

Therefore, S = f

Which shows that the intensity peaks are observed during x- ray diffraction in the simple cubic crystals for all types of plane.

2. Body-Centered Cubic Crystals

In body centered cubic crystals, there are two identical atoms per unit cell and are located at (0, 0, 0) and (\(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\) ) i.e. \( (u_1, v_1, w_1) = (0, 0, 0)

And \( (u_2, v_2, w_2) = (\frac{1}{2}, \frac{1}{2}, \frac{1}{2}) \)

Now, geometrical scattering factor \(S = \sum_j f_j e^{-i2\pi(u_j h + v_j k +w_j l)}\)

$$=f +fe^{-i\pi(h+k+l)}$$

$$S=f[1+e^{-i\pi(h+k+l)}]\dotsm(1)$$

Case-I:

If (h+k+l) is odd then

S = F (1 - 1 )

= 0 ; Which implies that diffraction pattern can’t be observed from the atomic plane like (1, 0, 0), (1, 1, 1), (1, 2, 0) etc

Case-II:

If ( h + k + l ) is even then ,

S = F( 1 + 1 )

\(\therefore\) S = 2f

Which implies that the X- ray diffraction can be observed from the atomic plane like (1, 1, 0), (1, 0, 1), (0, 1, 1) etc.

3. Face-Centered Cubic Crystals

In FCC crystals, there are four identical atoms located at (0, 0, 0), \( (\frac{1}{2}, \frac{1}{2}, 0) \), \( (\frac{1}{2}, 0, \frac{1}{2}) \), \( (0, \frac{1}{2}, \frac{1}{2}) \)

Therefore, geometrical scattering factor is

$$S = f + fe^{-i\pi(h+k)}+fe^{-i\pi(h+l)}+fe^{-i\pi(h+l)}$$

$$=f[1+e^{-i\pi(h+k)}+e^{-i\pi(h+l)}+e^{-i\pi(k+l)}]$$

If we consider a atomic plane like (1, 1, 1) then S = 4f

But if we consider a atomic like (1, 1, 0) then, S = 0

From this we can conclude X-ray diffraction are observed from the plane like (1, 1, 1) and absent from the plane like (1, 1, 0).

References:

Adhikari, P.B, Daya Nidhi Chhatkuli and Iswar Prasad Koirala.A Textbook of Physics. Vol. II. Kathmandu: Sukunda Pustak Bhawan, 2012.

Jenkins, F.A and H.E White.Fundamental of optics. New York (USA): McGraw-Hill Book Co, 1976.

wood, R.W.Physical Optics. New York (USA): Dover Publication , 1934.