*3.3.5. Crystal twinnings*

Natural graphite is a cluster of two modifications of carbon 2H and 3R: the content of rhom‐ bohedral modifications may reach 30%. Therefore, calculating the specific amount of crystal‐ line substance

$$\mathbf{v}\_{\mathfrak{m}} = \mathbf{v} / \mathfrak{m}$$

of the whole crystal requires knowledge of the part of one of the structures. If the part of twolayer packaging of graphite is equal to x, then its mass will be equal xm, and the weight of three-layer packaging of graphite shall be equal to (1-x)m, where m is the mass of the entire crystal. The number n of all cells of graphite is

$$\mathbf{m} = \mathbf{N\_{A}} \ulcorner \mathbf{x} \mathbf{m} \,\, \mathsf{(4M\_{0} + \ (1 - \lambda)\mathbf{m} \,\, \not\in \mathsf{M\_{0}}\mathbf{J\_{0}})} $$

where M0=0.012 kg/mol. We denote the ratio mNA/M<sup>0</sup> by symbol n0. Then the ratio n0/n=β will express the ratio of the number of carbon atoms to the number of all cells of the crystal, and the equation written above can be reduced to β=12 / (2+*x*).

The value of x for mechanical mixtures of pure graphite 2H and 3R is determined by weighing components. If case of clusterization it is necessary to calculate β with knowledge of x which can be determined by the spectrophotometric method, comparing the absorption intensity of UV radiation at wavelengths 219Å and 418Å, the principle of which is described below in the section 6.1. If intensities of these peaks of absorption of UV-radiation in Fig. 8 identified by the symbol h, then the value of x can be calculated from the equation

$$\mathbf{h\_1/h\_2=x/(1-x)}$$

Having determined the value of x from the spectral experiment, one can calculate the specific amount of crystalline substance in the cluster of graphite by the formula

$$\mathbf{v}\_{\mathbf{m}} = \mathbf{v} / \ \mathbf{m} = (\beta \mathbf{M}\_0)^{-1}$$

and density of crystalline substance of graphite by the expression

$$\sigma = \text{v} / \text{ V = q} (\text{\(\beta M\)}^{-1})$$

where ρ is the density of graphite. Let in Fig. 8 x=0.8. Then the value β of is 12/ (0.8+2)=4.29, and specific quantity of crystal substance of crystal is 1/ (4.29x0.012)=19.44 mol/kg. If the density of graphite is taken to be equal 2090 kg/m<sup>3</sup> , then the desired density of the amount of crystalline substance which is to be found for the graphite cluster will be 19,44 x 2090=40640 mol/m<sup>3</sup> .
