The Fundamental Element of the Crystal

Debye U. Scherer. Atombau. Phys. Zeitschr. 19, p. 474, 1918.

From the numerous measurements on crystals carried out by Bragg, it follows that the intensity of the X-rays scattered by the atoms of some element will be proportional to the square of the number of electrons in the atom.

From the data of Barkla it follows that the same quantity will be proportional to the first power of the atomic number \(N\). The two laws can be reconciled by the fact that, in the case of a wavelength large in comparison with the magnitude of the radius of the atom, the scattered waves will coincide in phase, and then the amplitudes of the individual oscillations will be added, and we shall have Bragg’s law. For the case of short waves, the oscillations from the various electrons will differ in phase, and then the intensities will be added, which will correspond to Barkla’s law. Finally, for the case of very small angles of scattering, irrespective of the wavelength, we obtain Bragg’s law, since the phase difference of the oscillations from two electrons whose relative position is determined by the coordinates \(\xi,\eta\) and \(\zeta\) will be

\[ \frac{(\alpha-\alpha_0)\xi+(\beta-\beta_0)\eta+(\gamma-\gamma_0)\zeta}{c}, \]

where \(\alpha,\beta\), and \(\gamma\) are the cosines of the angles of the scattered ray, and \(\alpha_0,\beta_0\), and \(\gamma_0\) are the cosines for the incident ray, \(c\) is the velocity of light. For very small angles ...

the scattering, the expression given for the phase will be close to zero, and consequently the scattering will occur according to Bragg’s law.

Using this, one can establish whether the element of the crystal lattice is an ion or a neutral atom.

Let us take KCl, which consists of two face-centered cubic lattices. From consideration of the structure factor it follows that the intensity of reflection from faces with even indices must be proportional to \((\mathrm{K}+\mathrm{Cl})^2\), where by K and Cl we understand the atomic numbers corresponding to potassium and chlorine, and for planes with odd indices the intensity must be proportional to \((\mathrm{K}-\mathrm{Cl})^2\). If the lattice is built of ions, then the potassium atom must have \(19-1\) electrons, and chlorine \(17+1\).

Consequently, planes with odd indices should not give reflection. In the photograph of KCl obtained by the Debye–Scherer method, lines corresponding also to odd indices were found. The result is negative, as if speaking in favor of the existence of neutral atoms. The null method could give a negative result owing to the existence of thermal motion of the atoms, which causes some, albeit weak, intensity of the reflected rays from faces with odd indices. To resolve the question, another crystal was taken, namely FLi, whose atomic numbers are respectively 9 and 3.

From the photograph the intensity of the scattered rays was determined, corresponding both to planes with even indices and to those with odd indices. For the former, the intensity for small angles must be proportional to \((\mathrm{F}+\mathrm{Li})^2\), and for the latter to \((\mathrm{F}-\mathrm{Li})^2\).

Then curves were constructed expressing the dependence of the scattering intensity on \(H^2=h_1^2+h_2^2+h_3^2\) (where \(h_1, h_2, h_3\) are the indices of the faces), separately for planes with even indices and with odd indices (the sum of the squares of the indices will be proportional to the sines of one-half the angles of deviation).

By comparing the two curves it was possible to construct a curve expressing the dependence of the ratio

\[ \frac{\mathrm{F}+\mathrm{Li}}{\mathrm{F}-\mathrm{Li}} \]

on \(H^2\). The analytical expression of this curve will be

\[ \frac{\mathrm{F}+\mathrm{Li}}{\mathrm{F}-\mathrm{Li}}=1.52. \]

Here

\[ \frac{\mathrm{F}+\mathrm{Li}}{\mathrm{F}-\mathrm{Li}} \]

is a quantity proportional to the ratio of the intensities of reflection from planes with even and odd indices.

At \(H^2=0\) the curve intersects the ordinate axis at the point where

\[ \frac{\mathrm{F}+\mathrm{Li}}{\mathrm{F}-\mathrm{Li}}=1.523+0.0728H^2-0.00234H^4. \]

Three cases are possible: 1) when the atoms are neutral, then

\[ \frac{\mathrm{F}+\mathrm{Li}}{\mathrm{F}-\mathrm{Li}}=\frac{9-3}{9-3}=2; \]

2) when fluorine has acquired one electron from lithium, then

\[ \frac{\mathrm{F}+\mathrm{Li}}{\mathrm{F}-\mathrm{Li}}=\frac{10+2}{10-2}=1.5, \]

and 3) when fluorine has taken two electrons from lithium, then

\[ \frac{\mathrm{F}+\mathrm{Li}}{\mathrm{F}-\mathrm{Li}}=\frac{11+1}{11-1}=1.2. \]

The experimentally found ratio 1.52 corresponds to the second case, i.e. to the case where the lattice consists of ions.

N. Selyakov.