Subfield: Modern Physics

X-ray Diffraction from a crystal.

We wish to study X-ray diffraction by a cubic crystal lattice. To do this we start with the diffraction of a plane, monochromatic wave that falls perpendicularly on a 2-dimensional grid that consists of $\mathrm{N}_{1} \times \mathrm{N}_{2}$ slits with separations $\mathrm{d}_{1}$ and $\mathrm{d}_{2}$. The diffraction pattern is observed on a screen at a distance $L$ from the grid. The screen is parallel to the grid and $\mathrm{L}$ is much larger than $\mathrm{d}_{1}$ and $\mathrm{d}_{2}$.
Context question:
a - Determine the positions and widths of the principal maximum on the screen. The width is defined as the distance between the minima on either side of the maxima.
Context answer:
$(x_{n_{1},} y_{n_{2}})=(\frac{n_{1} \cdot \lambda \cdot L}{d_{1}}, \frac{n_{2} \cdot \lambda \cdot L}{d_{2}})$;
$2 \cdot \Delta x=2 \cdot \frac{\lambda \cdot L}{N_{1} \cdot d_{1}} ; \quad 2 \cdot \Delta y=2 \cdot \frac{\lambda \cdot L}{N_{2} \cdot d_{2}}$


Extra Supplementary Reading Materials:

We consider now a cubic crystal, with lattice spacing a and size $\mathrm{N}_{0} \cdot a \times \mathrm{N}_{0} \cdot a \times \mathrm{N}_{1} \cdot a \cdot \mathrm{N}_{1}$ is much smaller than $\mathrm{N}_{0}$. The crystal is placed in a parallel X-ray beam along the z-axis at an angle $\Theta$ (see Fig. 1). The diffraction pattern is again observed on a screen at a great distance from the crystal.

<img_4346>

Figure 1 Diffraction of a parallel X-ray beam along the z-axis. The angle between the crystal and the $y$-axis is $\Theta$.
Context question:
b - Calculate the position and width of the maxima as a function of the angle $\Theta$ (for small $\Theta)$.
Context answer:
position is $(x_{n_{1}},y_{n_{2}},y_{n_{3}}^{\prime})=(\frac{n_{1} \lambda L}{a},\frac{n_{2} \lambda L}{a \cos (\theta)},\frac{n_{3} \lambda L}{a \sin (\theta)})$, and the width $\Delta x=2 \frac{\lambda L}{N_{0} a}$, $\Delta y=2 \frac{\lambda L}{N_{0} a \cos (\theta)}$, $\Delta y^{\prime}=2 \frac{\lambda L}{N_{1} a \sin (\theta)}$


Extra Supplementary Reading Materials:

The diffraction pattern can also be derived by means of Bragg's theory, in which it is assumed that the X-rays are reflected from atomic planes in the lattice. The diffraction pattern then arises from interference of these reflected rays with each other.
Context question:
c - Show that this so-called Bragg reflection yields the same conditions for the maxima as those that you found in $b$.
Context answer:
\boxed{证明题}


Extra Supplementary Reading Materials:

In some measurements the so-called powder method is employed. A beam of X-rays is scattered by a powder of very many, small crystals. (Of course the sizes of the crystals are much larger than the lattice spacing, a).

Scattering of X-rays of wavelength $0.15 \mathrm{~nm}$ by Potassium Chloride [KCl] (which has a cubic lattice, see Fig.2) results in the production of concentric dark circles on a photographic plate. The distance between the crystals and the plate is $0.10 \mathrm{~m}$, and the radius of the smallest circle is $0.053 \mathrm{~m}$ (see Fig. 3). $\mathrm{K}^{+}$and $\mathrm{Cl}^{-}$ions have almost the same size, and they may be treated as identical scattering centres.

<img_4571>

Figure 2. The cubic latice of Potassium Chloride in which the $\mathrm{K}^{+}$and $\mathrm{Cl}^{-}$wons have almost the same size.

<img_4572>

Figure 3. Scattering of X-rays by a powder of $\mathrm{KCl}$ crystals results in the production of concentric dark circles on a photographic plate.

Question: d - Calculate the distance between two neighbouring $\mathrm{K}$ ions in the crystal.

Solution:
d - For the distance, $\sqrt{ } 2$.a, between neighbouring $\mathrm{K}$ ions we have:

$$
\begin{gathered}
\operatorname{tg}(2 \phi)=\frac{x}{L}=\frac{0,053}{0,1} \approx 0,53 \rightarrow a=\frac{\lambda}{2 \cdot \sin (\phi)} \approx \frac{0,15 \cdot 10^{-9}}{2 \cdot 0,24} \approx 0,31 \mathrm{~nm} \\
K-K \approx \sqrt{2} \cdot 0,31 \approx 0,44 \mathrm{~nm}
\end{gathered}
$$

Final answer: 0.44 nm