*3.3.6. Solid solutions*

*3.3.5. Crystal twinnings*

120 Solar Cells - New Approaches and Reviews

crystal. The number n of all cells of graphite is

the equation written above can be reduced to β=12 / (2+*x*).

symbol h, then the value of x can be calculated from the equation

amount of crystalline substance in the cluster of graphite by the formula

and density of crystalline substance of graphite by the expression

density of graphite is taken to be equal 2090 kg/m<sup>3</sup>

mol/m<sup>3</sup> .

line substance

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‐

νm=ν/ 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

n=NA *x*m /4M0+ (1-*х*)m /6M<sup>0</sup> ,

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 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

h1/ h2=*x*/(1 –*x*)

Having determined the value of x from the spectral experiment, one can calculate the specific

νm=ν/ m = (βM0)-1

σ=ν/ V =ρ(βM0)

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

crystalline substance which is to be found for the graphite cluster will be 19,44 x 2090=40640


, then the desired density of the amount of

Mixtures are homogeneous, uniform, multi-component systems that have no separation surface between any parts. The properties of all the parts of such systems are the same. So we can assume that the structural element of the substance of the solution of given composition is the same throughout the volume of solution. Properties of solutions are a function of composition. Consider one of these functions for two cases of specification of the structural elements: formula unit and their number Z in the unit cell of the crystal.

According to formula (2), for one mole of amount of crystalline substance of solid solutions equality is valid:

$$\mathbb{Q}/\mathbb{M} = 1 \;/\; \left(\text{v N}\_{\text{A}}\right)\_{\text{\textdegree}} \tag{3}$$

where ρ is the density of a substance, M is the mass of one mole of a structural element of substance of solid solution, v is a volume of a structural element, N<sup>A</sup> is the Avogadro constant. The mass M of the mole of structural elements will be expressed through mass M0 of mole of formula units of solid solution, and the value of Z which is the number of formula units in structural cell of crystalline material:

$$\mathbf{M} = \mathbf{Z} \,\mathbf{M}\_0 \tag{4}$$

The mass of mole of formula units we express by the sum

$$\mathbf{M}\_0 = \Sigma \mathbf{M}\_1 \mathbf{N}\_{1'} \tag{5}$$

where Mi is a mass of mole of formula unit of i-component of the solution, and N<sup>i</sup> is its mole part. Multiply the left and right parts of the solution of equations (3)-(5) by the multiplier Mi Ni . We get ratio

$$
\Sigma \,\mathbf{M}\_{\mathrm{i}} \mathbf{N}\_{\mathrm{i}} / \left(\rho \mathbf{v} \,\mathbf{N}\_{\mathrm{A}}\right) = \mathbf{M}\_{\mathrm{i}} \mathbf{N}\_{\mathrm{i}} / \Sigma \mathbf{M}\_{\mathrm{i}} \mathbf{N}\_{\mathrm{i}}.\tag{6}
$$

We denote the left part of this ratio as H/NA and the right part we denote by k symbol. The value of k is a dimensionless and it corresponds to the fraction of total mass by way of expression of the concentration of a solution. Therefore, the value of H has the same dimen‐ sionality [the number of particles/mol] that the Avogadro constant. For i-component of the solution we obtain the expression

$$\mathbf{H}\_{\mathrm{i}} \mathbf{\!=} \mathbf{k}\_{\mathrm{i}} \mathbf{N}\_{\mathrm{A}}.\tag{7}$$

Experimentally determined values of p, v, Z and M0i building the amount Hi , are connected with it by a linear dependence with mass fraction ki of i-component of solid solution. The relation (7) is interesting because empirical rules of L. Vegard and Rutgers are particular cases of demonstration of the function found above according to (7).

Indeed, L. Wegard had found out an additivity of sizes for the elementary cell of solid solution:

$$\mathbf{a} = \mathbf{N}\_1 \mathbf{a}\_1 \mathbf{+} \mathbf{N}\_2 \mathbf{a}\_{2'}$$

where N1 and N2 are the molar fractions, a1 and a2 are any parameters of a cell of pure components or any average interatomic distance. Additivity of the cell volumes is set by the rule of Rutgers, which is written as

$$\mathbf{V} = \mathbf{N}\_1 \mathbf{V}\_1 \mathbf{+} \mathbf{N}\_2 \mathbf{V}\_2 \mathbf{y}$$

where V, V1 and V2 are the molar volumes of solid solution and pure components. However, there are quite common deviations from these rules. Examples are solid solutions of NaCl-KCl [71]. Deviations from the rules of L. Vegard and Rutgers can have alternating nature. An example is solid solution Ni-Al.

It should be noted that the additivity of the parameters of a cell does not mean additivity of the cell volume. The equality (6) allows avoiding this contradiction and using parameters and the volume of the cell of the crystal as an example of simple linear function (7) in systems with deviations from the rules of Vegard and Rutgers.

Dependences of lattice constants and densities of solid solution Ni-Al on the structure are presented in Fig. 4a [38]. Solid solutions Ni-Al crystallize in the structural type of cesium chloride and have region of homogeneity between 45 and 60 at.% nickel. The curve of de‐ pendence of the lattice parameter on the composition has a maximum, and density curve has a sharp bend. Such curves indicate that in the region of low concentrations of Nickel we have solid solutions of subtraction: the structure has defects, and a part of places for Nickel atoms remain empty. These voids are statistically distributed throughout the volume of the crystal. Solutions with a Nickel content of more than 50 at.% are normal solutions of substitution, in which the atoms of Nickel are statistically replaced by atoms of aluminium.

Let us transform the diagram "property-structure" in Fig. 4a in the diagram "H-k" of solid solutions Ni-Al as an illustration of the function (7). To do this, we calculate the values of H and k for Nickel. The symbols characterizing its properties will be denoted with footnote index 1. Solid solutions Ni-Al have a cubic cell with a parameter *a*. Therefore, the volume of their structural element is equal to *a*<sup>3</sup> . Then

$$\begin{aligned} \mathbf{H}\_1 &= \mathbf{Z} \, \mathbf{M}\_1 \mathbf{N}\_1 / (\rho \dot{a}^3), \\ \mathbf{H}\_2 &= \mathbf{Z} \, \mathbf{M}\_2 \mathbf{N}\_2 / (\rho \dot{a}^3), \\ \mathbf{H}\_1 / \, \mathbf{H}\_2 &= \mathbf{M}\_1 \mathbf{N}\_1 / \left(\mathbf{M}\_2 \mathbf{N}\_2\right) \end{aligned} \tag{8}$$

The value of the Z of solid solutions of substitution is equal to 2 (body-centered cube), and for solutions of subtraction Z is less than two, even though the cell structure is also a body-centered cub. Assume Z equal to 4N1, as at N1=0.5 solid solutions of subtraction become normal solid solutions of substitution with Z=2. Values of density of crystals p and lattice parameters *a* of solid solutions remain unchanged.

Experimentally determined values of p, v, Z and M0i building the amount Hi

relation (7) is interesting because empirical rules of L. Vegard and Rutgers are particular cases

Indeed, L. Wegard had found out an additivity of sizes for the elementary cell of solid solution:

a=N1а1+ N2а2,

where N1 and N2 are the molar fractions, a1 and a2 are any parameters of a cell of pure components or any average interatomic distance. Additivity of the cell volumes is set by the

V=N1V1+ N2V2,

where V, V1 and V2 are the molar volumes of solid solution and pure components. However, there are quite common deviations from these rules. Examples are solid solutions of NaCl-KCl [71]. Deviations from the rules of L. Vegard and Rutgers can have alternating nature. An

It should be noted that the additivity of the parameters of a cell does not mean additivity of the cell volume. The equality (6) allows avoiding this contradiction and using parameters and the volume of the cell of the crystal as an example of simple linear function (7) in systems with

Dependences of lattice constants and densities of solid solution Ni-Al on the structure are presented in Fig. 4a [38]. Solid solutions Ni-Al crystallize in the structural type of cesium chloride and have region of homogeneity between 45 and 60 at.% nickel. The curve of de‐ pendence of the lattice parameter on the composition has a maximum, and density curve has a sharp bend. Such curves indicate that in the region of low concentrations of Nickel we have solid solutions of subtraction: the structure has defects, and a part of places for Nickel atoms remain empty. These voids are statistically distributed throughout the volume of the crystal. Solutions with a Nickel content of more than 50 at.% are normal solutions of substitution, in

Let us transform the diagram "property-structure" in Fig. 4a in the diagram "H-k" of solid solutions Ni-Al as an illustration of the function (7). To do this, we calculate the values of H and k for Nickel. The symbols characterizing its properties will be denoted with footnote index 1. Solid solutions Ni-Al have a cubic cell with a parameter *a*. Therefore, the volume of their

3

*à à ρ*

The value of the Z of solid solutions of substitution is equal to 2 (body-centered cube), and for solutions of subtraction Z is less than two, even though the cell structure is also a body-centered

3

1 2 11 22

( )

*ρ* (8)

which the atoms of Nickel are statistically replaced by atoms of aluminium.

. Then

H

1 11

H () H ()

= Z M N / , = Z M N / , / = M NH / M N

2 22

with it by a linear dependence with mass fraction ki

rule of Rutgers, which is written as

122 Solar Cells - New Approaches and Reviews

example is solid solution Ni-Al.

structural element is equal to *a*<sup>3</sup>

deviations from the rules of Vegard and Rutgers.

of demonstration of the function found above according to (7).

, are connected

of i-component of solid solution. The

**Figure 4.** (a)-lattice constant (line 1) and density (line 2) for solid solutions Ni – Al as functions of Ni concentration in at.% [38]; (b)-diagram «Н – k» of solid solutions Ni – Al as dependence of the value Н1 on Ni concentration in mass parts of k. Straight line АО corresponds to solid solutions of subtraction, straight line ОВ corresponds to solid solu‐ tions of substitution.

Values k1 of nickel were calculated according to (6) – (7) by the expression

$$\mathbf{k}\_1 = \mathbf{M}\_1 \mathbf{N}\_1 / \left(\mathbf{M}\_1 \mathbf{N}\_1 + \mathbf{M}\_2 \mathbf{N}\_2\right) \tag{9}$$

where M2 and N2 are the mass of mole of aluminium atoms and its molar fraction in solution. Calculated by formulas (8), (9) the values of k1, N1 for Nickel are presented in Table 7. Fig. 4b shows a diagram "H-k" for solid solution Ni-Al, which presents the relation (7) as line AOB. Part of the line AO corresponds to solid solutions of subtraction, the section OB corresponds to solid solutions of substitution.

Now suppose that the atom as formula unit is chosen as a structural element of crystalline substance of solid solution Ni-Al. Then the amount per structural element of the substance of solid solution equals to a3 /Z=a<sup>3</sup> /2 for solutions of substitution as well as for solutions of subtraction. Symbols of values related to the mole of formula units we denote by lower index 0. To show that the value of H0, referred to mole of formula units of crystal, has no universality of the values of H1 found above, we divide both sides of (3) by a molar part Ni of the component. Taking into account the relation (5)

> M0=ΣM<sup>i</sup> Ni ,


**Table 7.** Structure parameters of solid solution Ni – Al [38] and values of k1, Н1 for Ni calculated according to (8), (9).

we obtain the following expression for mole parts of Ni or Al:

$$\begin{aligned} \mathbf{H}\_{01} &= \mathbf{N}\_1 \left( \mathbf{M}\_1 \mathbf{N}\_1 + \mathbf{M}\_2 \mathbf{N}\_2 \right) / \left( \rho \mathbf{v} \right) = \mathbf{N}\_1 \mathbf{N}\_{\mathbf{A}'}\\ \mathbf{H}\_{02} &= \mathbf{N}\_2 \left( \mathbf{M}\_1 \mathbf{N}\_1 + \mathbf{M}\_2 \mathbf{N}\_2 \right) / \left( \rho \mathbf{v} \right) = \mathbf{N}\_2 \mathbf{N}\_{\mathbf{A}'}\\ \mathbf{H}\_{01} &= \mathbf{H}\_{02} \left( \mathbf{N}\_1 / \mathbf{N}\_2 \right) \end{aligned} \tag{10}$$

Denoting left parts of equations as Н0i, we reduce it to a general form

$$\mathbf{H}\_{\rm 0i} = \mathbf{N}\_{\rm i} \mathbf{N}\_{\rm \lambda'} \tag{11}$$

Which formally coincides with (7), but with different form of expressing the concentration: the mole fraction N instead of the mass part k, i.e.

$$\mathbf{H}\_{\mathsf{i}} \mathsf{j} \mathsf{/} \mathbf{H}\_{0\mathsf{i}} = \mathbf{k}\_{\mathsf{i}} \mathsf{/} \mathsf{N}\_{\mathsf{i}}.$$

Values Н01 for Ni are calculated according to the expression

$$\mathbf{H}\_{\rm 01} = \mathbf{N}\_{\rm 1} \left( \mathbf{M}\_{\rm 01} \mathbf{N}\_{\rm 1} + \mathbf{M}\_{\rm 02} \mathbf{N}\_{\rm 2} \right) / \left( \rho a^3 / \mathcal{D} \right) = \mathbf{N}\_{\rm 1} \mathbf{N}\_{\rm 4} \tag{12}$$

they are presented in table. 7. Fig. 5 shows that diagram "H01-N1" in region of homogeneity of solid solutions Ni-Al represents two curves instead of one curve; these two curves are broken on the border of two types of solid solution. The line has a tangent of slope angle 6х1023 close to the numerical value of Avogadro constant. Thus, equation (11) is valid only for the region of solid solutions of substitution with some deviation, value H01 of solid solutions of subtrac‐ tion does not depend on the concentration.

we obtain the following expression for mole parts of Ni or Al:

H

**Mole part of Ni, N1**

**Mass part of Ni, k1**

124 Solar Cells - New Approaches and Reviews

**Lattice parameter, а/Å** **Density of solution ρ, kg /m<sup>3</sup>**

0.46 0.650 2.864 5430 1.84 3.895 2.998 0.47 0.659 2.868 5550 1.88 3.962 3.008 0.48 0.668 2.873 5700 1.92 4.002 2.998 0.49 0.676 2.880 5900 2 4.081 2.957 0.50 0.685 2.882 5920 2 4.142 3.023 0.51 0.694 2.881 5960 2 4.201 3.089 0.52 0.702 2.879 6000 2 4.264 3.158 0.54 0.719 2.875 6140 2 4.345 3.265 0.56 0.735 2.869 6250 2 4.454 3.395 0.58 0.750 2.864 6400 2 4.529 3.501 0.60 0.765 2.858 6500 2 4.642 3.639

**Number of formula units in the cell, Z**

**Н1 mol / 10<sup>23</sup> Н01 mol / 10<sup>23</sup>**

mole fraction N instead of the mass part k, i.e.

( ) ( )

H ( )

01 1 1 1 2 2 1 A 02 2 1 1 2 2 2 A

= N M N + M N / v = N N , = N M N + M N / v = N N ,

**Table 7.** Structure parameters of solid solution Ni – Al [38] and values of k1, Н1 for Ni calculated according to (8), (9).

Which formally coincides with (7), but with different form of expressing the concentration: the

/ Н0i= ki

/Ni .

01 1 01 1 02 2 1 A H = N M N + M N / /2 = N N , *ρa* (12)

Нi

( ) ( ) <sup>3</sup>

( )

*ρ* (10)

H = N N , 0i i A (11)

*ρ*

( )

01 02 1 2

H H /

Values Н01 for Ni are calculated according to the expression

= N N

Denoting left parts of equations as Н0i, we reduce it to a general form

**Figure 5.** Dependence of the value Н0 on the mole part Ni for solid solutions Ni – Al. Horizontal line corresponds to solid solutions of subtraction, the slope line corresponds to solid solutions of substitution. Structural element of crys‐ talline substance is specified as atom of Ni or Al.

Comparing Figs. 7, 8 and 9 shows the three representations of the experimental data on density and parameters of a crystal cell. Two of them are based on the idea of a structural element of the crystal in the form of formula unit. In this case it is an atom of Nickel or aluminium. Curves have maximum, break and other geometrical features. A third way to view *ρ* and *a* of crystals of solid solutions is based on the specification of the structural element of crystalline substance in the form of elementary cell. In this case, there is a linear function Hi of chemical composition for the whole range of homogeneity of solid solutions regardless of the type of structure and its defectiveness. Compactness, simplicity and universality of diagram presentation of the experimental data on density and the crystal unit cell parameters in dependence on mass fraction show its advantage over other methods of presentation of the concentrations of solid solutions.

The value H has dimension [the number of particles/mol]. When the mass of solid solutions remains constant, the value of H indicates a change in the number of crystalline substance of solid solutions with concentration. The range of homogeneity and cell parameters of solid solutions depend on temperature and pressure. Therefore, the amount of crystalline substance of solid solutions is also dependent on temperature and pressure.

**Figure 6.** A change of density of amount of matter in series of liquid alkanes from methane to eicosane as a function of reciprocal mass of mole of molecules. The lower curve corresponds to the specification of the structural element of a liquid substance in the form of molecule, straight lines and points 1, 2 correspond to the specification in form of groups of hydrogen atoms with one atom of carbon.

### *3.3.7. Liquid alcanes*

We calculate the density of a standard amount of substance in a series of liquid alkanes from methane to eicosane by the formula

$$\sigma\_0 \text{=} \emptyset / \mathbf{M}\_0$$

where ρ is the density of the fluid, M<sup>0</sup> is the mass of the mole of molecules. The dependence of the density of the amount of liquid alkanes on chemical composition is presented in Fig. 6 as a function σ<sup>0</sup> from the reciprocal mass of mole of molecules M0. It is seen that σ0 decreases in a series of alkanes from methane to eicosane, and the nonlinearity of the function is explained by the different densities of alkanes, which depends on the length of the carbon chain of the molecule and hydrogen environment of the carbon atom.

Remember that on the Butlerov's theory properties of organic compounds are sufficiently determined by the number and spatial position of carbon atoms in the chain of the molecule. So we give up on the structural element of liquid alkanes in the form of molecules and choose the structural element of a liquid substance in a series of alkanes in the form of groups of hydrogen atoms with one atom of carbon. In this case, the density of the amount of substance must be calculated by the formula

$$\sigma\_n = \text{nq/M}\_0$$

where n is the number of carbon atoms in the molecule. Function σ n-1/ M0 is also presented in Fig. 6. In this case this function is linear and is explained by features of building a carbon chain of alkanes molecules. The fracture of the line can be explained by the different amount of hydrogen associated with the given carbon atom. Indeed, among methane and butane prevails group CH3 (bottom line), in the rest of alkanes prevails the group CH2 (upper line). As

$$\mathbf{v}\_0/\sigma\_0 = \mathbf{v}\_n/\sigma\_n = \mathbf{V}\_\prime$$

the different volumes of atomic groups explain various slope of straight lines in Fig. 6.

In a series of molecules of isoalkanes one can choose groups S, CH, CH2 and CH3. The value σ of normal pentane С5Н12 or

$$\text{CH}\_3\text{-CH}\_2\text{-CH}\_2\text{-CH}\_2\text{-CH}\_3\text{-}$$

corresponds to the top line in Fig. 6, as in its molecule the group CH2 dominates. The value σ of isopentane (CH3) СН2СН belongs to the bottom line (point 1), as in its molecule the group CH3 is predominant. The value σ of neopentane (CH3)4C does not belong to these lines (point 2), as in the molecule, one carbon atom is not bound with hydrogen. As pentane isomers with equal chemical composition have a different density of amount of substance, the phenomenon of isomerism is characterized not only by different geometry of molecules, but a different amount of substance. Kinetic equations of chemical reactions are determined by the number of reacting substances. Therefore, the characteristics of the reaction of pentane isomers can be described with the values of the density of the amount of substance in the kinetic equation.
