Chapter 15, Problem 15B.7P

Interpretation Introduction

Interpretation:

The structure factors and the form of diffraction pattern under the given conditions has to be determined.

Concept Introduction:

The structure factor is given by

  $F_{hkl}={\displaystyle \sum _f{f}_j{e}^{i{\varphi }_{hkl}\left(j\right)}}$

Where,

${\varphi }_{hkl}\left(j\right)=2\pi \left(h{x}_j+k{y}_j+l{z}_j\right)=$ Phase difference

$f_j=$ Scattering factor

$f_j{e}^{i{\varphi }_{hkl}\left(j\right)}=$ phase factor

Explanation of Solution

The location of the atoms in a bcc unit cell can be depicted as follows

Figure 1

Calculation of structure factor for bcc unit cell:

The atoms at the corner will give a weight of $\frac{1}{8}$ and the atom at the body center will give a weight of $\frac{1}{2}$. Consider all the atoms are identical.

Table for calculated values of phase difference:

  $\begin{array}{cccccc}Atom& Weight& x& y& z& {\varphi }_{hkl}\\ 1& 1/8& 0& 0& 0& 0\\ 2& 1/8& 0& 1& 0& 2\pi k\\ 3& 1/8& 1& 1& 0& 2\pi \left(h+k\right)\\ 4& 1/8& 1& 0& 0& 2\pi h\\ 5& 1/8& 1& 0& 1& 2\pi \left(h+l\right)\\ 6& 1/8& 1& 1& 1& 2\pi \left(h+k+l\right)\\ 7& 1/8& 0& 1& 1& 2\pi \left(k+l\right)\\ 8& 1/8& 0& 0& 1& 2\pi l\\ 9& 1/2& 1/2& 1/2& 1/2& 2\pi \left(\frac{1}{2}h+\frac{1}{2}k+\frac{1}{2}l\right)\end{array}$

Calculation of phase factor:

For corner atoms:

  $f_j{e}^{i{\varphi }_{hkl}\left(j\right)}=f_A\left(e^{i.0}+e^{i.2\pi k}+e^{i.2\pi \left(h+k\right)}+e^{i.2\pi h}+e^{i.2\pi \left(h+l\right)}+e^{i.2\pi \left(h+k+l\right)}+e^{i.2\pi \left(k+l\right)}+e^{i.2\pi l}\right)$

The phase factor for the corner atoms is $+1$ and as they have a weight of $\frac{1}{8}$, the total contribution to the structure factor is $f_A$.

For the atom at body center,

  $f_j{e}^{i{\varphi }_{hkl}\left(j\right)}=f_B\left[e^{i.2\pi \left(\frac{1}{2}h+\frac{1}{2}k+\frac{1}{2}l\right)}\right]=f_B\left[e^{i\pi \left(h+k+l\right)}\right]$

Hence, the structure factor is given by

  $F_{hkl}=f_A+f_B{e}^{i\pi \left(h+k+l\right)}=f_A+f_B\left[{\left(-1\right)}^{\left(h+k+l\right)}\right]$

If $\left(h+k+l\right)=$ even, $F_{hkl}=f_A+f_B$

If $\left(h+k+l\right)=$ odd, $F_{hkl}=f_A$

Diffraction pattern:

When,

(a) $f_A=f$, $f_B=0$: $f_B=\frac{1}{2}{f}_A$$F_{hkl}=f_A+f_B{e}^{i\pi \left(h+k+l\right)}=f_A+f_B\left[{\left(-1\right)}^{\left(h+k+l\right)}\right]=f$

This implies that there are no systematic absences.

(b) $f_B=\frac{1}{2}{f}_A$:  $F_{hkl}=f_A+f_B{e}^{i\pi \left(h+k+l\right)}=f_A+\frac{1}{2}{f}_A\left[{\left(-1\right)}^{\left(h+k+l\right)}\right]=f_A\left[1+\frac{1}{2}{\left(-1\right)}^{\left(h+k+l\right)}\right]$

When $h+k+l$ is odd, $F_{hkl}=f_A\left(1-\frac{1}{2}\right)=\frac{1}{2}{f}_A$

When  $h+k+l$ is even, $F_{hkl}=f_A\left(1+\frac{1}{2}\right)=\frac{3}{2}{f}_A$

That is, there is an alternation of intensity as $\left(I\alpha F_{{}_{hkl}}^2\right)$ according to whether $h+k+l$ is odd or even.

(c) $f_A=f_B=f$:  $F_{hkl}=f_A+f_B{e}^{i\pi \left(h+k+l\right)}=f+f\left[{\left(-1\right)}^{\left(h+k+l\right)}\right]=f\left[1+{\left(-1\right)}^{\left(h+k+l\right)}\right]$

If $h+k+lisodd$, $F_{hkl}=0$. Thus, all $h+k+l$ odd lines are missing.