X-ray waves

Fig. 4.1. Interference of incoming and the reflected X-ray waves inthe triangular region above a flat and thick reflecting substrate. The strength ofthe electromagnetic field is represented on the gray scale by instantaneous crests (white) andtroughs (black). Inthe course of time, the pattern moves from the left to the right [4.21].

Dependence of X-Ray Wave-Lengths on Combination, T. Wetterblad, Zeit. Phys., 42, 603, 1927. 

In Fig. 36 the horizontal lines are supposed to represent the equidistant layers of atoms in a crystal. RA, SB, and the other parallel lines represent the path of the incident X-rays which are reflected as shown along AT, BU, and the other parallel lines. The X-ray waves are perpendicular to the paths and so all the reflected beams must be imagined to overlap each other. Draw AN and AM perpendicular to SB and BU. The path SBU is longer than RAT by the distance NB + BM. 

At present time the X-ray range is of most interest for polymer research. One sees that in the conventional sources a gap appears in the region of long X-ray wave length. Here synchrotron radiation is the only source available. This is of importance for the scattering experiments in the anomalous dispersion region which are discussed in the paper by Stuhrmann (this book). At wave length of about 0.1 nm,... 

For the sake of simplicity we now consider electrons as stationary points and disregard the dependence of the scattered intensity on the scattering angle. Each electron then interacts with the incident x-ray wave producing a spherical elastically scattered wave as shown in Figure 2.20. Thus, the scattering of x-rays by a single electron yields an identical scattered intensity in every direction. 

The absolute intensity of the x-ray wave coherently scattered by a single electron, I, is determined from the Thomson equation ... 

Figure 10 (a) The origin of Bragg s law. The X-ray waves of wavelength A are reflected at an angle 0 from successive planes of equivalent atoms, separated by a lattice repeat d. (b) For constructive interference to take place between the waves reflected the path length difference AB + BC needs to be a multiple of the wavelength. This is true when 2d sin 6=nX. 

The Wave-lengths as Functions of the Atomic Numbers.—Many of the laws connecting X-ray wave-lenghts with each other, or with the atomic numbers of the chemical elements may be expressed most simply in terms... 

The simplest theory of diffraction [11, 16] relates the intensity distribution of the X-ray wave scattered on a perfect crystal to the distribution of the electrons in the unit cell of volume Vc ... 

In Figure 1.6, the atomic scattering factors f(s) for hydrogen, carbon, and oxygen are plotted against s = 2(sin 6)/X. In the forward direction (s = 0) the x-ray waves scattered from different parts of the electron cloud in an atom are all in phase, and the wave amplitudes simply add up, rendering /(0) equal to the atomic number Z. As s increases, the waves from different parts of the atom develop more phase differences, and the overall amplitude begins to decrease. The exact shape of the curve f(s) reflects the shape of the electron density distribution in the atom. The... 

NRVS data are commonly interpreted within a harmonic approximation, which describes molecular vibrations in terms of independent oscillations along a set of normal coordinates Qa = ej rjni p related through a linear transformation to the Cartesian coordinates rj of atoms j weighted by their masses nij. Molecular rotation and translation lead to six modes having zero frequency. Projection of the transformation coefficients eja onto the direction k of the X-ray wave vector k = (Eo/hc)k determines the recoil fraction... 

Figure 316(a) from the Braggs text demonstrates the reinforcement of x-ray waves that obey Braggs Law. In Figure 316(b), we see a schematic of their x-ray apparatus in which single crystals (or powders) were placed on a rotating table so that reflections could be collected from all angles. 

X-ray waves O and O are parallel. The extra distance traveled by wave O in traveling through the crystal is AB -1- BC. For diffraction to occur it is necessary that this distance be a whole number of wavelengths, n that is,... 

---