Bragg's Law

Introduction

X-rays are electromagnetic radiation such as visible light and as such has interference effects. This, however, cannot be detected with a typical single or double slit experiment, because due to the small wavelength, the gap distances would have to be extremely small.

In 1912, Max von Laue has the idea to use crystals for interference experiments. Their internal structure is like a grating, its ions are arranged at regular intervals $$d$$.

The X-rays are diffracted by the electron shell of the irradiated atoms and interfere with each other. The path difference of the diffracted waves for a certain angle $$\alpha$$ depends on the distance between the atoms in the crystal lattice and thus of the element-specific lattice spacing $$d$$.

Bragg's Law

In the sketch, the red, green and purple line together form a right triangle with the hypotenuse $$d$$. Using the sine definition we get the following expression:

$$\delta = d \cdot \sin \alpha$$

As can be seen in the sketch on the left, the path difference is the dual of $$\delta$$:

$$\Delta s = 2 \cdot \delta = 2d \cdot \sin \alpha$$

Constructive interference occurs when the path difference $$\Delta s$$ is a multiple of the wavelength.

$$\Delta s = k \cdot \lambda$$

Thus applies to the $$k$$-th maxima:

$$k \cdot \lambda = 2d \cdot \sin \alpha \qquad k = 1, 2, 3, ...$$

Rotational crystal experiment

In the rotational crystal experiment X-rays hit a crystal. The radiation interferes and then hits a detector. The crystal can be rotated so that you can change the angle of incidence $$\ alpha$$ of the radiation.

In this case, a lithium fluoride crystal and X-rays of an unknown wavelength $$\lambda$$ are used. At the angles $$\alpha_1 = 0°$$ $$\alpha_2 \approx 10°$$ and $$\alpha_3 \approx 21°$$ relative intensity maxima are observed. These angles are the so-called Bragg angle or glancing angles.
If one knows the lattice spacing of lithium fluorid $$d_L = 2.01 \times 10^{-10} \,\, m$$, one can use any any of these angles in the Bragg equation and thereby calculate the wavelength of X-rays used in the experiment. For the first maximum the following holds:
$$\lambda = 2 \cdot d \cdot \sin \alpha = 2 \cdot 2,01 \cdot 10^{-10} \,\, m \cdot \sin 10^\circ = 6,98 \cdot 10^{-11} \,\, m$$