On the Theory of Crystal Growth and the Formation of Crystalline Nuclei¹

I. N. Stranski and R. Kaishev, Bulgaria

I. Mechanism of Equilibrium of Small Crystals¹

Introduction

After the fundamental works of Gibbs², the physical theory of crystal growth and of the formation of nuclei has only quite recently made a great step forward. This is explained by the fact that the mechanism of these processes could receive a satisfactory interpretation only after the introduction into the theory of precise physical concepts, which was accomplished chiefly by the works of Volmer, Kossel, and Stranski.

The study of the elementary processes of detachment and attachment of building particles during crystal growth (Kossel, Stranski) made it possible to give a clear qualitative interpretation of all phenomena connected with crystal growth³, and led to important conclusions concerning the structure of various crystallographic faces⁴. Until quite recently, a quantitative treatment of the processes of growth and dissolution of crystals, based on elementary processes, was considered impossible. The only attempt to construct a quantitative theory of growth and of the formation of nuclei was Volmer’s theory⁵, in which the probabilities of formation of two- and three-dimensional nuclei were introduced into the theory of crystallization. A treatment such as Volmer’s proves not always possible; for example, in cases where crystal growth does not occur by the building up of separate layers of the lattice⁶. It is essential, however, that in those cases where new lattice layers arise through the formation of two-dimensional nuclei, Volmer’s theory gave the only correct approach to solving the problem posed.

With the aid of the elementary processes of attachment and detachment one can show that, in most cases, crystal growth proceeds through the formation of individual layers, and that for the nucleation of each new layer it is necessary to overcome a certain energy barrier. In exactly the same way one can show that attachment

¹ Yearbook of Sofia University, Faculty of Physics and Mathematics, vol. XXXII, book 2, Chemistry, 1935/36. Translated from the Bulgarian by I. V. Radchenko. Some additions were made by the translator on the basis of Kaishev’s works published in Z. Physik, 102, 684, 1936, and Ann. d. Phys. (5), 30, 184, 1937.

new atomic rows to the growing layer is connected with the overcoming of energy thresholds. However, the internal connection between the theory of crystallization on the basis of elementary processes (Kossel, Stranski) and Volmer’s theory is not obvious, and the opinion has often been expressed that there are fundamental contradictions between these two methods of treatment[^7].

Proceeding from the fact that the essential results of these two theories are identical, we believe that in reality there must exist a close internal connection between them, and that the differences are purely external. Moreover, only with a critical combination of these two methods is it possible to construct a unified quantitative theory of crystal growth and of the formation of nuclei.

Indeed, in a number of works[^8] we have been able to show that the problem of crystal growth and of the formation of nuclei can be solved quantitatively on the basis of elementary processes by the consistent application of the statistical method. The starting point of our treatment was a comparison of the conditions of equilibrium of two- and three-dimensional crystals. In doing so we were able to determine such a statistical quantity by means of which the posed problem receives a complete solution. This quantity is the mean work of detachment of one element from the surface layer of the lattice.

Following the chronological order in the development of the problem, we shall first dwell on its thermodynamic treatment, and then pass to the statistical treatment, which constitutes the principal result of our investigations.

A. Thermodynamic part

1. Equilibrium: three-dimensional nucleus—two-dimensional nucleus—gas phase

We start from the Thomson–Gibbs equation, which for a simple crystallographic form has the form

\[ \ln \frac{p_{r_3}}{p_{\infty}}=\frac{2\sigma M}{r_3 \rho R T}, \tag{1} \]

where \(r_3\) is the radius of the sphere inscribed in the crystal (the central distance), \(p_{\infty}\) is the vapor pressure over an infinitely large crystal, \(\sigma\) is the specific surface energy on a face of the crystal, \(M\) is the molecular weight, \(\rho\) is the density, \(R\) is the gas constant, \(T\) is the absolute temperature. In this case we shall assume that the simple crystallographic form is at the same time also the equilibrium form, determined from the Gibbs condition

\[ \sum F_i \sigma_i = \min \]

at constant volume (\(F_i\) is the area of the \(i\)-th face, \(\sigma_i\) is the surface energy per unit surface). Let us note that the correction to the Thomson–Gibbs equation for the specific peripheral energy

edges is superfluous, since it attains a noticeable magnitude for such small crystallites that this equation in general loses physical meaning.

According to the above equation, to each crystallite with radius \(r_3\), satisfying Gibbs’ condition for an equilibrium form, there corresponds a definite vapor pressure \(p_{r_3}\), at which the crystallite is in (labile) equilibrium with a sufficiently large gas phase. We set ourselves the task of clarifying what is the mechanism of growth and dissolution of such a crystallite, and what in general is the mechanism of its equilibrium with the gas phase.

Let us assume, in accordance with Volmer’s theory, that the growth and dissolution of individual layers of the lattice occur by the formation of two-dimensional nuclei. Consider the case in which a three-dimensional nucleus and a two-dimensional nucleus formed on one of its faces are simultaneously in equilibrium with one and the same gas phase at pressure \(p_r\). (By a crystalline nucleus we shall understand such a crystal as is in equilibrium only with the medium surrounding it.)

The relation between the sizes of the three-dimensional and two-dimensional nuclei for one and the same gas phase can be determined if the specific surface energy \(\sigma\) and the specific peripheral energy \(\chi\) are known, since for a two-dimensional nucleus there also exists a relation between the magnitude of the nucleus and the equilibrium pressure \(p_{r_2}\), expressed by an equation quite analogous to the Thomson–Gibbs equation \({}^{3}\),

\[ \ln \frac{p_{r_2}}{p_\infty} = \frac{\chi M}{r_2 \rho \delta RT}, \tag{2} \]

where \(r_2\) is the radius of the circle inscribed in the two-dimensional crystal, \(\delta\) is the thickness of the two-dimensional nucleus. In this case the necessary condition for an equilibrium form will be a minimum of the peripheral energy

\[ \sum L_i \chi_i = \min \]

at constant area (\(L_i\) is the length of the \(i\)-th side, \(\chi_i\) the peripheral energy per unit length).

The formation of a new lattice layer of a three-dimensional crystallite occurs through the formation of a two-dimensional nucleus, by condensation of the gas phase. In exactly the same way, upon dissolution of a surface layer a two-dimensional nucleus is formed by the separation of a certain number of structural particles from the upper layer of the lattice. The probability of these two processes generally determines the probability of growth and, correspondingly, of dissolution of the crystal. When a crystallite is in (labile) equilibrium with the medium, i.e. in the case when it is a nucleus, the probabilities of growth and dissolution must be the same. Analogously, in the case of a two-dimensional nucleus, the probabilities of attachment and detachment of one peripheral row must be the same. These quite general conditions are absolutely necessary for establishing equilibrium. The elementary

the necessity that the probabilities of these two mutually opposite processes (dissolution and growth) be exactly identical for the nucleus leads, as we shall see below, not only to the conclusion that an energy threshold exists in the separation of one surface layer of the lattice, but also to the consequence that this energy threshold must be equal to the energy threshold in the formation of a new lattice layer by condensation1.

Let us note that, while remaining within thermodynamics, nothing can be said regarding the attachment and detachment of rows at the periphery of a two-dimensional crystal, since it is impossible to give a thermodynamic definition of a one-dimensional nucleus[^9]. We shall return to this question in the statistical part of this work.

Let us first dwell on the thermodynamic consideration of two simple examples: a homeopolar cubic crystal and a heteropolar crystal of the NaCl type.

2. Simple homeopolar cubic crystal

For simplicity we shall restrict ourselves to forces acting between nearest neighbors in the crystal lattice. In this case the equilibrium form will be a cube (see below), and deformation at the surface of the crystal is completely absent.

One may ask whether there is any sense at all in considering such schematized models. To this question one should answer that what is fundamentally essential in the processes of growth and dissolution of crystals is to a considerable extent independent of the special properties of crystals, and it can be discerned in the simplest model, quite independently of whether such a model is realized in nature or not, provided, of course, that this model is devoid of internal contradictions. The model chosen by us fully satisfies this condition.

If we denote by $\varphi$ the work that must be performed in order to separate from one another two nearest neighboring particles, and by $r_0$ the distance between them, then the specific surface energy on the face of the cube $(100)$ is $\sigma=\dfrac{\varphi}{2r_0^2}$, while the specific peripheral energy on the side $(10)$ of a two-dimensional crystal consisting of $n$ atomic layers is $\chi_n=\dfrac{n\varphi}{2r_0}$, whence it follows that $\sigma n r_0=\chi_n$.

First of all we shall prove that the equilibrium form of our model is a cube for a three-dimensional crystal and a square for a two-dimensional crystal. To find the equilibrium form one usually uses the Gibbs–Wulff method, which consists in the following: if, from some point lying inside the crystal, normals are drawn to all possible crystallographic faces and on them segments proportional to the corresponding specific surface

energies, and through the ends of these segments draw planes perpendicular to them, then the part of the crystal bounded by the planes nearest to the central point will be the required equilibrium form.

In a completely analogous way the equilibrium form of a two-dimensional crystal is determined. In this case, segments proportional to the corresponding specific peripheral energies are laid off.

If we denote by \(d_{100}, d_{110}, d_{111}\) the central distances\(^1\) from the corresponding faces, and by \(\sigma_{100}, \sigma_{110}, \sigma_{111}\) the corresponding specific surface energies, then, according to the Gibbs–Wulff rule,

\[ d_{100}:d_{110}:d_{111}=\sigma_{100}:\sigma_{110}:\sigma_{111}. \]

Since for our model\(^ {10}\)

\[ \sigma_{100}=\frac{\varphi}{2r_0^2};\qquad \sigma_{110}=\frac{\varphi}{\sqrt{2}\,r_0^2};\qquad \sigma_{111}=\frac{\sqrt{3}\,\varphi}{2r_0^2}, \]

then

\[ d_{110}:d_{100}=\sqrt{2} \]

and

\[ d_{111}:d_{100}=\sqrt{3}, \]

i.e. the faces \((110)\) and \((111)\) do not belong to the equilibrium form and only touch it along edges (respectively, for the two-dimensional crystal, at vertices). It can be shown that all other conceivable crystallographic planes likewise do not belong to the equilibrium form, since they do not even touch it.

Similarly, if for the two-dimensional crystal we denote by \(d_{10}\) and \(d_{11}\) the central distances to the faces \((10)\) and \((11)\), and by \(x_{10}\) and \(x_{11}\) the corresponding specific peripheral energies, then we obtain the following equation:

\[ d_{10}:d_{11}=x_{10}:x_{11}. \]

Since

\[ x_{10}=\frac{\varphi}{2r_0},\qquad x_{11}=\frac{\varphi}{\sqrt{2}\,r_0}, \]

then, consequently,

\[ d_{10}:d_{11}=\sqrt{2}, \]

i.e. the side \(d_{11}\) touches the vertex of the square formed by the sides \((10)\). All the other conceivable crystallographic sides do not even touch the equilibrium square form.

For a cubic three-dimensional embryo and, correspondingly, a square two-dimensional embryo, the equations hold

\(^1\) Radii of the inscribed spheres.

\[ RT \ln \frac{p_{a_3}}{p_\infty}=\frac{4\sigma M}{\rho a_3^3}, \tag{3} \]

\[ RT \ln \frac{p_{a_{2}}}{p_\infty}=\frac{2x_n M}{nr_0\rho a_{2,n}^2}, \tag{4} \]

where \(a_3\) is the edge length of the cube of a three-dimensional nucleus, and \(a_{2,n}\) is the side of a two-dimensional square nucleus consisting of \(n\)-atomic layers. At equilibrium of these two nuclei with the same gas phase, \(p_{a_3}=p_{a_2}\), we obtain

\[ \frac{2\sigma}{a_3}=\frac{x_n}{nr_0a_{2,n}} \]

and, for \(n=1\),

\[ \frac{a_3}{a_{2,1}}=\frac{2\sigma r_0}{x_1}=2. \]

Before proceeding further, let us prove that, of all conceivable types of two-dimensional nuclei, the most probable is a nucleus consisting of a monoatomic layer, and that, as the most probable one, only it should be taken into account in our subsequent discussion. For the proof it is sufficient to determine the work \(W_{2,n}\) necessary for the formation, at a given supersaturation, of a two-dimensional nucleus consisting of \(n\)-atomic layers. This work can be determined by means of the following cyclic process\(^9\).

1. Let us detach the nucleus from a layer of infinite dimensions, consisting of \(n\)-atomic layers and located on the surface of an infinitely large crystal. We shall denote the vapor pressure over the infinitely large crystal by \(p_\infty\). Let us place the detached nucleus on a crystal located under such a pressure \(p_{a_2}\) of the gas phase that the two-dimensional nucleus will be in equilibrium with the gas. The work of detaching the nucleus is equal to

\[ A_1=4a_{2,n}\cdot x_n. \]

1. Let us evaporate the two-dimensional nucleus back into the gas phase with pressure \(p_{a_2}\), for which we perform the work

\[ A_2=-W_{2,n}-\frac{\rho nr_0a_{2,n}^2}{M}RT. \]

1. Let us separate the amount of vapor \(\rho nr_0a_{2,n}^2\) thus obtained and expand this vapor reversibly from the pressure \(p_{a_2}\) to \(p_\infty\). In doing so, the work performed will be

\[ A_3=-\frac{\rho nr_0a_{2,n}^2}{M}RT\ln\frac{p_{a_2}}{p_\infty}. \]

1. Let us condense this vapor onto the initially taken infinitely large layer of an infinitely large crystal. The work performed is equal to

\[ A_4=-\frac{\rho n r_0 a_{2,n}^{\,2}}{M}RT . \]

Since

\[ A_1+A_2+A_3+A_4=0, \]

we have

\[ W_{2,n}=4a_{2,n}\chi_n-\frac{\rho n r_0 a_{2,n}^{\,2}}{M}RT\ln\frac{p_{a_2}}{p_\infty}, \]

or, taking (4) into account,

\[ W_{2,n}=2a_{2,n}\chi_n . \]

This same work can be calculated in the following way:

\[ W_{2,n}=\int_0^{a_{2,n}} dA, \]

where

\[ dA=\frac{dm}{M}RT\ln\frac{p_a}{p_{a_2}} =\frac{dm}{M}RT\ln\frac{p_a}{p_\infty} -\frac{dm}{M}RT\ln\frac{p_{a_2}}{p_\infty}, \]

\(a\) is the side of the variable two-dimensional crystal, and \(a_{2,n}\) is the side of the two-dimensional nucleus.

From the equation

\[ m=\rho n r_0 a^2 \]

it follows that

\[ dm=2\rho n r_0 a\,da . \]

Taking equation (4) into account, we obtain

\[ dA=4\chi_n\,da-\frac{4\chi_n a\,da}{a_{2,n}} . \]

Hence for \(W_{2,n}\) we have

\[ W_{2,n}=\int_0^{a_{2,n}}4\chi_n\,da -\int_0^{a_{2,n}}\frac{4\chi_n a\,da}{a_{2,n}} =2a_{2,n}\chi_n . \]

Since \(\chi_n=n\chi_1\) and \(a_{2,n}=a_{2,1}\), it follows that

\[ W_{2,n}=2a_{2,1}\chi_1 n, \]

i.e. \(W_{2,n}\) increases with \(n\) and will be smallest for \(n=1\), as was required to prove.

To find the probability of the processes of growth and dissolution, we shall also determine the work \(W^*_{2}\) of formation of a two-dimensional nucleus on one of the faces of a three-dimensional nucleus, by destroying the layer itself. In what follows we shall take into account only those two-dimensional nuclei which consist of a monatomic layer. To determine \(W^*_{2,1}\), we shall carry out the following reversible isothermal process.

Let the two-dimensional nucleus in size coincide exactly with one of the faces of the three-dimensional nucleus \((a'_{2,1}=a_3)\). Let there correspond to it the vapor pressure

\[ p'_{a_{2,1}} < p_{a_3}=p_{a_{2,1}} . \]

1. Let us reversibly evaporate this nucleus under the pressure \(p'_{a_{2,1}}\), whereby the work performed will be

\[ A_1=-W'_{2,1}. \]

1. Let us compress the vapor obtained under the pressure \(p'_{a_{2,1}}\) to \(p_{a_{2,1}}=p_{a_3}\). The work done in this is

\[ A_2=\frac{m'_{a_{2,1}}}{M}RT\ln\frac{p_{a_{2,1}}}{p'_{a_{2,1}}} = \frac{a_{2,1}^{\prime\,2} r_0\rho}{M}\cdot \frac{2\varkappa_1 M}{\rho r_0} \left( \frac{1}{a_{2,1}}-\frac{1}{a'_{2,1}} \right). \]

1. At the pressure \(p_{a_{2,1}}=p_{a_3}\), form a two-dimensional nucleus with side \(a_{2,1}\), for which we perform the work

\[ A_3=W_{2,1}. \]

Consequently,

\[ \begin{aligned} W^*_{2,1} &=A_1+A_2+A_3 =-W'_{2,1}-2a_{2,1}^{\prime\,2}\varkappa_1 +2a_{2,1}^{\prime\,2}\varkappa_1\frac{a'_{2,1}}{a_{2,1}} +W_{2,1} \\ &=-W'_{2,1}-W'_{2,1} +W'_{2,1}\frac{a'_{2,1}}{a_{2,1}} +W_{2,1} \\ &=W_{2,1}+W'_{2,1} \left( \frac{a'_{2,1}}{a_{2,1}}-2 \right). \end{aligned} \]

Since

\[ \frac{a'_{2,1}}{a_{2,1}} = \frac{a_3}{a_{2,1}} =2, \]

then

\[ W^*_{2,1}=W_{2,1}. \]

The work \(W^*_{2,1}\) can also be calculated by another method, analogous to the calculation of the work \(W_{2,1}\), namely:

\[ W^{*}_{2,1} = \int_{a_3}^{a_{2,1}} dA = \int_{a_3}^{a_{2,1}} 4x_1\,da - \int_{a_3}^{a_{2,1}} \frac{4x_1 a\,da}{a_{2,1}} = \]

\[ = 2a_{2,1}x_1 + 2a_3x_1\left(\frac{a_3}{a_{2,1}}-2\right) = 2a_{2,1}x_1 = W_{2,1}. \]

From the foregoing it is clear that the work of formation of a two-dimensional nucleus of a definite kind on a three-dimensional nucleus does not depend on whether the two-dimensional nucleus is obtained by growth or by dissolution. In both cases it is the same. However, we must draw further conclusions as well; namely, on the basis of the general conditions for the possibility of equilibrium one can prove that the probabilities of formation of a two-dimensional nucleus (on one of the faces of a three-dimensional one), determined according to Volmer by the general expression \(Ae^{-\frac{W}{kT}}\) (where \(W\) is the work of formation of the nucleus), must be identical:

\[ Ae^{-\frac{W_{2,1}}{kT}} = A^{*}e^{-\frac{W^{*}_{2,1}}{kT}}. \]

Since \(W_{2,1}=W^{*}_{2,1}\), it follows that \(A=A^{*}\).

If one takes into account that the quantities \(A\) are proportional to the possibilities of attachment of a two-dimensional nucleus to a face of a three-dimensional nucleus, and if one considers that in the two methods of obtaining a nucleus (by condensation from the gas phase and by destruction of the surface layer of the crystal face) one and the same two-dimensional nucleus is obtained on one and the same substrate, then the above statement becomes obvious.

3. Heteropolar Crystal of the NaCl Type

The model we have adopted proved convenient for calculations also in the case of a heteropolar crystal. An energetic consideration of the elementary processes of growth and dissolution led to the conclusion that, in a sufficiently large NaCl crystal, it is precisely the \((100)\) faces that can be realized as fully built-up faces[^11].

All other faces (which are never realized) undergo internal changes. They grow and dissolve without the formation of two-dimensional nuclei in Volmer’s sense. As for the cube faces, their growth indeed proceeds through the formation of two-dimensional nuclei. However, these nuclei are formed almost exclusively at the corners, and not on the edges or in the centers of the faces, as in the case of homeopolar crystals. This conclusion, obtained from a molecular-energy consideration, was subsequently confirmed by Brandes and Volmer[^9], who showed that the differences in the works of formation of a two-dimensional nucleus (at 1% supersaturation) on an edge and in the center of a face are of the order of \(10^{-9}\) erg, and therefore \(10^5\) times greater than \(kT\) at room temperature.

Let us consider the equilibrium: three-dimensional nucleus — two-dimensional nucleus — gas phase. Since the equilibrium form of NaCl is only a cube, there is no need to consider all other possible faces except the faces of the cube.

Suppose that only Coulomb electrostatic and Born forces act between the ions in the crystal lattice. In addition, let us assume that deformations of the lattice and of the ions are absent. Under this condition there is no basis for taking London forces into account.

Proceeding from the results of the investigations of Brandes and Vollmer, namely that an angular nucleus is, as the most probable one, the only nucleus that must be taken into account in considering the processes of growth of NaCl, we shall first find at what thickness of the two-dimensional nucleus the work of its formation \((W_{2,n,E})\) is minimal. For this we define

\[ \frac{\partial W_{2,n,E}}{\partial n}=0. \]

In our case, taking equation (4) into account, we obtain

\[ W_{2,n,E}=2a_{2,n,E}\cdot \bar{\chi}_n \quad \text{and} \quad a_{2,n,E}=\frac{a_{2,1,E}}{\chi_1}\cdot \frac{\bar{\chi}_n}{n}, \]

where

\[ \bar{\chi}_n=\frac{\chi_n+\chi_{n,k}}{2}, \]

\(\chi_n\) is the specific peripheral energy of a two-dimensional crystal consisting of \(n\) ionic layers and situated in the middle of a face of the cube; \(\chi_{n,k}\) is the specific peripheral energy of a two-dimensional crystal in contact with an edge of the cube, for the side coinciding exactly with the edge of the cube.

From the foregoing it follows that

\[ \frac{\partial W_{2,n,E}}{\partial n} = \frac{4a_{2,1,E}}{\chi_1}\cdot \frac{\bar{\chi}_n}{n}\cdot \frac{\partial \bar{\chi}_n}{\partial n} - \frac{2a_{2,1,E}}{\chi_1}\cdot \frac{\bar{\chi}_n^2}{n^2} =0, \]

whence the following condition for the minimum of \(W_{2,n,E}\) is obtained:

\[ \frac{\partial \bar{\chi}_n}{\partial n} = \frac{\bar{\chi}_n}{2n}. \]

In the case of the NaCl lattice, with a good approximation

\[ \bar{\chi}_n = n(A-2B)+2B-C, \]

where \(A=11.68\cdot 10^{-6}\) dyn, \(B=3.77\cdot 10^{-6}\) dyn, \(C=7.36\cdot 10^{-6}\) dyn. Hence for the required minimum

\[ n=\frac{4B-C}{2(A-2B)}=0.67\sim 1. \]

Consequently, the most probable of all angular two-dimensional nuclei is the one that consists of a monatomic layer.

It should be noted that this is not true with respect to a two-dimensional nucleus formed at the middle of a cubic face. In this case the equation is

\[ x_n=n(A-2B)+2B, \]

where \(A\) and \(B\) have the values indicated above.

The minimum work of formation of a two-dimensional nucleus is obtained for

\[ n=\frac{2B}{A-2B}=1.86\sim 2. \]

This result is closely connected with the fact that a single-ion two-dimensional nucleus is larger in magnitude than the face of a three-dimensional nucleus that serves as its substrate,

\[ \frac{a_3}{a_{2,1}}=\frac{2r_0\sigma_{100}}{x_1} =\frac{2\cdot 2.8\cdot 10^{-8}\cdot 152}{11.68\cdot 10^{-6}}=0.73. \]

This conclusion in itself seems paradoxical, since when the two-dimensional nucleus reaches the edge of the cube face serving as its substrate, \(x\) changes discontinuously. However, a two-ion two-dimensional nucleus can fit on the face of the cube, since

\[ \frac{a_3}{a_{2,2}}=\frac{2\cdot 2r_0\sigma_{100}}{x_2} =\frac{2\cdot 2\cdot 2.8\cdot 10^{-8}\cdot 152}{15.80\cdot 10^{-6}}=1.08. \]

Let us also find the ratio of the edge of the cubic three-dimensional nucleus \(a_3\) to the side length of a single-ion two-dimensional angular nucleus \(a_{2,1,E}\):

\[ \frac{a_3}{a_{2,1,E}}=\frac{2r_0\sigma_{110}}{x_1} =\frac{2\cdot 2.8\cdot 10^{-8}\cdot 152}{8\cdot 10^{-6}}=1.06. \]

It is especially important to determine the ratio of the work of formation of a single-ion two-dimensional angular nucleus by condensation from the gas phase \(W_{2,1,E}\) to the work of its formation by destruction of the surface layer \(W^{*}_{2,1,E}\). To determine \(W^{*}_{2,1,E}\), let us carry out the following reverse process.

Let us transform the surface layer on one of the faces of a three-dimensional nucleus into a single-ion two-dimensional angular crystal by separating from the edges of the three-dimensional nucleus two ionic rows, as shown in Fig. 1 (first stage). In doing so, the periphery of length \(2a_3\) and specific peripheral energy \(x_{1,k}\) disappears, and in its place a new periphery is formed, of almost the same length, but with specific peripheral energy \(x_1\). The work performed in this process is equal to

\[ A_1=(x_1-x_{1,k})\,2a_3. \]

Let us reversibly dissolve the obtained single-ion two-dimensional corner crystal (at an equilibrium pressure for a three-dimensional embryo) until a single-ion two-dimensional corner embryo is obtained. This

Fig. 1

process is carried out analogously to the process used in calculating \(W^{*}_{2,1}\). The work performed is

\[ A_2 = W_{2,1,E} + 2a_3 \bar{x}_1\left(\frac{a_3}{a_{2,1,E}} - 2\right). \]

Since, on the other hand, the work of formation of the same two-dimensional embryo by condensation from the gas phase is equal to

\[ W_{2,1,E} = 2\bar{x}_1 \cdot a_{2,1,E}, \]

then, after substituting the corresponding values, we obtain\(^1\)

\[ \frac{W^{*}_{2,1,E}}{W_{2,1,E}} = \frac{a_3}{a_{2,1,E}}\cdot \frac{x_1 - x_{1,k}}{x_1} + 1 + \frac{a_3}{a_{2,1,E}} \left(\frac{a_3}{a_{2,1,E}} - 2\right) = 0.98 \sim 1. \]

In reality this ratio must be exactly equal to unity; the deviation (2%) is due to the inaccuracy in the calculation of the quantities we used.

4. Application of the theory to more complex models and some additional consequences

Finally, let us touch upon more complex cases as well, taking into account, in a simple cubic homeopolar crystal, the forces acting not only between nearest neighbors, but also between particles of the crystal lattice that are more distant from one another. In this case the equilibrium form of the crystal will no longer be the simple crystallographic form. Indeed, as a result of the increase in the range of action of the forces between the individual structural particles of the crystal, faces \((110)\), \((111)\), \((211)\), etc., will successively appear. The two-dimensional embryos will differ in size and form for different crystallographic faces. In the general ca—

\(^1\) Let us note that the greater part of the work \(W^{*}_{2,1,E}\) (about \(3/4\)) is spent on separating the first two ionic rows, which are simultaneously also edges of the cube. We shall dwell on this feature exhaustively below.

their form will no longer be a simple crystallographic form. Moreover, for a complex model the simple equations are no longer valid

\[ nr_0\sigma = \chi_n \quad \text{and} \quad \chi_n = n\chi_1. \]

As a result of the action of attractive forces between particles sufficiently far removed from one another, we obtain the inequalities

\[ nr_0\sigma > \chi_n \quad \text{and} \quad \chi_n > n\chi_1. \]

At the same time, in the most general case, for all possible homeopolar crystals one always has \(W_2 = W_2^*\).

It can also be shown that for crystals of the NaCl type (heteropolar) the reverse inequalities hold, namely:

\[ nr_0\sigma < \chi_n \quad \text{and} \quad \chi_n < n\chi_1. \]

These inequalities are valid for all heteropolar crystals. If one computes the ratio of the work of formation of a two-dimensional nucleus to the work of formation of a three-dimensional nucleus, then in the case of heteropolar crystals we obtain the inequality

\[ \frac{W_{2,1}}{W_3^*} > \frac{r_0}{2a_3}, \]

whereas for homeopolar crystals

\[ \frac{W_{2,1}}{W_3^*} < \frac{r_0}{2a_3}. \]

Consequently, in general the conditions for obtaining regular crystals by growing them from the gaseous phase on three-dimensional nuclei are more favorable for homeopolar crystals than for heteropolar crystals.

B. Statistical Part

1. Formulation of the Statistical Problem

In investigating the conditions for the simultaneous equilibrium of two-dimensional and three-dimensional nuclei in the case of a simple homeopolar crystal, we found by a thermodynamic method that, if we restrict ourselves to the forces acting between nearest neighbors in the crystal lattice, then the ratio of the edge of the cube of a three-dimensional nucleus to the side of the square of a two-dimensional nucleus formed on one of the faces of the three-dimensional nucleus is equal to 2, i.e.

\[ \frac{a_3}{a_{2,1}} = 2. \]

At first glance it seems strange that a face of a three-dimensional crystal is in equilibrium with a layer smaller in size than the face itself. However, it must be taken into account that a three-dimensional crystal grows and dissolves by entire plane layers, whereas a two-dimensional...

dimensional crystal—by whole peripheral rows. It thus becomes clear that the quantity which determines equilibrium, and which at equilibrium must be the same for three-dimensional and two-dimensional crystals, is the average work of detachment per building particle for that part of the structure which directly participates in the equilibrium. In the case of a three-dimensional crystal this will be the average work of detachment per particle of the entire surface layer, and for a two-dimensional crystal—the average work of detachment per particle of an entire peripheral row.

Let us calculate the average works of detachment of a particle for the simple model adopted by us, denoting by \(\varphi\) the work that must be performed in order to separate from one another two nearest building particles of the lattice, situated at a distance \(r_0\) from one another.

Fig. 2

In Fig. 2a one surface layer of a cubic crystal is shown. If the length of the edge of the cube is denoted by \(a_3\), then the number of building particles falling on an edge is

\[ n_3=\frac{a_3}{r_0}. \]

Let us remove the surface layer in three stages. First of all, remove all particles except two rows on adjacent edges (Fig. 2b). For the detachment of each of these particles we perform work \(3\varphi\), and in all \((n_3-1)^2\cdot 3\varphi\). Then remove the remaining two rows as well, leaving only one particle at the corner. For detaching each of the particles of the remaining rows, work \(2\varphi\) is required, and for all of them \(2(n_3-1)\cdot 2\varphi\). Finally, remove the last particle as well, performing for this the work \(\varphi\). Thus the desired average work of detachment will be equal to

\[ \bar{\varphi}_{a_3} = \frac{(n_3-1)^2\cdot 3\varphi + 2(n_3-1)\cdot 2\varphi + \varphi}{n_3^2} = \]

\[ = 3\varphi-\frac{2}{n_3}\varphi =3\varphi-\frac{2r_0}{a_3}\varphi. \tag{5} \]

In an analogous way we shall calculate the average work of detachment of a particle of one peripheral row of a two-dimensional crystal. If the length of the side of the two-dimensional crystal is denoted by \(a_2\), then the number of building particles falling on one peripheral row is equal to

\[ n_2=\frac{a_2}{r_0}. \]

We shall carry out the destruction of the row as follows: first of all we remove \(n_2-1\) building particles (Fig. 3), and in doing so the work \((n_2-1)\cdot 3\varphi\) will be performed. Then we detach the remaining particle as well,

having performed the work \(2\varphi\). Thus for the desired mean work of detachment we obtain

\[ \bar{\varphi}_{a_2}=\frac{(n_2-1)3\varphi+2\varphi}{n_2} =3\varphi-\frac{\varphi}{n_2} =3\varphi-\frac{r_0}{a_2}\varphi . \tag{6} \]

Since it has been found thermodynamically that \(a_3=2a_2\), then

\[ \bar{\varphi}_{a_3}=3\varphi-\frac{2r_0}{a_3}\varphi =\bar{\varphi}_{a_2}=3\varphi-\frac{r_0}{a_2}\varphi . \]

On the basis of the above, the result of the investigations of Brandes and Volmer\(^9\) is explained: that the parallel sides of different similarly oriented two-dimensional nuclei formed on the face of a cube of NaCl are equal in magnitude (Fig. 4). With simultaneous equilibrium of nuclei of all three kinds (in a corner, on an edge, in the middle of a face), it is necessary that the mean work of detachment of a structural particle, calculated for the entire peripheral row, be the same for all nuclei. Note, however, that the peripheral rows lying on the edges of the cube do not directly participate in the equilibrium, since with respect to them the medium surrounding the crystal is, at equilibrium, strongly supersaturated.

Fig. 3

Fig. 4

Making use of the results of our calculations, let us try to solve the whole problem of crystal growth and nucleation, taking elementary processes as the basis.

2. Equilibrium: crystal—vapor

The vapor pressure over a crystal, as is known, can be calculated on the basis of Boltzmann statistics. The Boltzmann distribution function

\[ \frac{N_i}{N} = \frac{\sum_i e^{-\frac{\varepsilon_i}{kT}}} {\sum_Z e^{-\frac{\varepsilon_i}{kT}}} \]

must hold both for the gaseous and for the solid phase. Here \(N_i\) is the number of molecules having energy \(\varepsilon_i\), and \(N\) is the total number of molecules. The summation extends over all possible states \(Z\) of the molecules of the system. If the system consists of \(N\) mole-

if of them \(N_g\) are in the gaseous phase, and \(N_f\) in the solid phase, then

\[ \frac{N_g}{N}= \frac{\displaystyle \sum_g e^{-\frac{\varepsilon_i}{kT}}} {\displaystyle \sum_Z e^{-\frac{\varepsilon_i}{kT}}}; \qquad \frac{N_f}{N}= \frac{\displaystyle \sum_f e^{-\frac{\varepsilon_i}{kT}}} {\displaystyle \sum_Z e^{-\frac{\varepsilon_i}{kT}}}, \]

whence it follows that

\[ \frac{N_g}{N_f}= \frac{\displaystyle \sum_g e^{-\frac{\varepsilon_i}{kT}}} {\displaystyle \sum_f e^{-\frac{\varepsilon_i}{kT}}}. \]

\(\sum_g\) and \(\sum_f\) denote that the summation is carried out over all possible states of the molecules of the gaseous and, respectively, the solid phases.

If the system consists of a crystal of a monatomic substance and its vapor, which is a monatomic gas, then the energy of the molecules will consist only of kinetic energy and will be expressed through the momentum components \(p_x, p_y, p_z\) as follows:

\[ \varepsilon_i=\frac{p_x^2+p_y^2+p_z^2}{2m}. \]

Replacing summation by integration, we obtain

\[ \sum_g e^{-\frac{\varepsilon_i}{kT}} = \frac{1}{h^3} \iiint\!\!\iiint e^{-\frac{\varepsilon_i}{kT}} \,dx\,dy\,dz\,dp_x\,dp_y\,dp_z = \]

\[ = \frac{V_g}{h^3} \int_{-\infty}^{+\infty}\!\!\int\!\!\int e^{-\frac{p_x^2+p_y^2+p_z^2}{2mkT}} \,dp_x\,dp_y\,dp_z = \frac{V_g}{h^3}(2\pi mkT)^{\frac{3}{2}}. \]

For simplicity, let us assume that the temperature of the system is so low that the kinetic energy of the molecules of the solid phase may be neglected. Let, on the average, each molecule of the crystal possess potential energy \(-\varepsilon_p\), so that

\[ \sum_f e^{-\frac{\varepsilon_i}{kT}} = e^{\frac{\varepsilon_p}{kT}}\cdot N_f, \]

whence it follows that

\[ \frac{N_g}{N_f} = \frac{\displaystyle \frac{V_g}{h^3}(2\pi mkT)^{\frac{3}{2}}} {\displaystyle N_f\cdot e^{\frac{\varepsilon_p}{kT}}} \]

or

\[ p=\frac{N_g kT}{V_g} = e^{-\frac{\varepsilon_p}{kT}}\cdot \frac{(2\pi m)^{\frac{3}{2}}\cdot (kT)^{\frac{5}{2}}}{h^3}. \]

After taking logarithms we obtain the equation for the vapor pressure

\[ \ln p=-\frac{\varepsilon_p}{kT}+\frac{5}{2}\ln T+i, \]

where

\[ i=\ln \frac{(2\pi m)^{\frac{3}{2}}\cdot k^{\frac{5}{2}}}{h^3} \]

is the so-called chemical constant.

Usually \(\varepsilon_p\) is identified with the heat of evaporation of the molecule. From the foregoing it becomes evident that, on the average, the difference between the potential energies of a gas molecule and of a molecule on the surface of the crystal \((+\varepsilon_p)\) is equal to the mean work of separation found by us. Consequently, knowing the mean work of separation, one can determine the equilibrium concentration of the medium surrounding the crystal.

Let us first consider an infinitely large crystal of the type of our simple cubic homopolar model. The mean work of separating a particle from an infinitely large face of this crystal \(\bar{\varphi}_\infty\) is determined by the work of separating one particle \(\varphi_{1/2}\) from a position on the half-crystal \(^{12}\) (Fig. 5), which in our case is equal to \(3\varphi\).

Fig. 5

It is obvious that, in the destruction of an infinitely large layer of the lattice, the differences between the work of separating the last structural particle of each row and the work of separation from a position on the half-crystal do not appreciably affect the magnitude of the mean work of separation, since the number of particles in the middle of the face of the crystal is incomparably greater than at the edges. Consequently, for an infinitely large crystal \(\varepsilon_p=\bar{\varphi}_\infty=\varphi_{1/2}\), and

\[ \ln p_\infty=-\frac{\varphi_{1/2}}{kT}+\frac{5}{2}\ln T+i. \tag{7} \]

However, the work of separation \(\varphi_{1/2}\) from the half-crystal cannot serve as a quantity determining the equilibrium of crystals of finite dimensions, since it practically (in our case completely) does not depend on the dimensions of the crystal, whereas the vapor pressure over the crystal must be a function of the crystal dimensions. In fact, as we have already seen, equilibrium is always determined by the quantity \(\bar{\varphi}\).

Identifying \(\varepsilon_p\) with \(\bar{\varphi}\), we obtain an equation determining the vapor pressure over an arbitrarily large crystal as a function of its dimensions, under the condition that it has an equilibrium shape, i.e., under the condition that the mean work of separation from any face of the crystal is one and the same \(^{13}\):

\[ \ln p_{a_s}=-\frac{\bar{\varphi}_{a_s}}{kT}+\frac{5}{2}\ln T+i. \tag{8} \]

From equations (7) and (8) it follows that

\[ \ln \frac{p_{a_3}}{p_{\infty}}=\frac{(\varphi_i-\overline{\varphi}_{a_3})}{kT}. \tag{9} \]

Equation (9) can be derived with the aid of purely kinetic considerations2. Let us first assume that gas molecules do not arrive at the surface of the crystal. Then the number of particles with work of detachment \(\varphi_i\) evaporating in the time interval between \(t\) and \(t+dt\) is equal to

\[ dN=N\beta e^{-\frac{\varphi_i}{kT}}\,dt=NP_i^*dt, \]

where \(N\) is the number of surface particles possessing, at time \(t\), the work of detachment \(\varphi_i\). The mean duration \(\tau_i\) of the stay of a particle on the surface of the crystal in the position to which the work of detachment \(\varphi_i\) corresponds is equal to

\[ \tau_i=\frac{1}{\beta}e^{\frac{\varphi_i}{kT}}=\frac{1}{P_i^*}. \tag{10} \]

On the other hand, the number \(dN\) of molecules settling during the time \(dt\) on the surface of the crystal (in the absence of evaporation) is proportional to the number \(N\) of sites at which, at the moment considered, molecules can settle, and to the number of molecules striking during the time \(dt\) the surface \(\alpha^2\), which corresponds to one site of attachment of the particle,

\[ dN=N\alpha\delta^2\sqrt{\frac{kT}{2\pi m}}\,n_gdt=NP_{n_g}dt, \]

where \(n_g\) is the number of molecules in a cubic centimeter of gas. The mean time \(\tau_{n_g}\) during which the attachment of one particle to a definite site of the crystal surface occurs, at a concentration \(n_g\) of molecules in the gas, is equal to

\[ \tau_{n_g}=\frac{1}{\alpha\delta^2}\sqrt{\frac{2\pi m}{kT}}\cdot\frac{1}{n_g}=\frac{1}{P_{n_g}}. \tag{11} \]

As we have already said, the equilibrium of an infinitely large crystal is determined by the work of detachment from the position on the face of the crystal. The mean duration of the stay of a particle on the face of a crystal at the equilibrium concentration \(n_{g\infty}\) is equal to the mean time during which the attachment of one particle \(\tau_{n_g\infty}\) occurs:

\[ \tau_{\varphi_i}=\frac{1}{\beta}e^{\frac{\varphi_i}{kT}}=\tau_{n_g\infty} =\frac{1}{\alpha\delta^2}\sqrt{\frac{2\pi m}{kT}}\cdot\frac{1}{n_{g\infty}}, \]

whence it follows that

\[ n_{g\infty}=\frac{p_{\infty}}{kT}=\frac{\beta}{\alpha\delta^2}\sqrt{\frac{2\pi m}{kT}}\cdot e^{-\frac{\varphi_1}{kT}}; \tag{12} \]

accordingly

\[ \ln p_{\infty}=-\frac{\varphi_1}{kT}-\frac{1}{2}\ln T+\ln\left[\frac{\beta}{\alpha\delta^2}(2\pi mk)^{\frac12}\right]. \tag{13} \]

Comparing equation (13) with equation (8) (valid for temperatures at which the thermal energy of the crystal may be neglected and the gas is monatomic), we find

\[ \frac{\beta}{\alpha\delta^2}=\frac{2\pi m}{h^3}(kT)^2. \]

For temperatures at which the Dulong–Petit law is valid, as Stern showed \(^{15}\),

\[ \ln p_{\infty}=-\frac{\varphi_1}{kT}-\frac{1}{2}\ln T+\ln\left[\frac{(2\pi m)^{\frac32}}{k^{\frac12}}\,\nu^3\right]. \]

Comparing this expression with (13), we obtain

\[ \frac{\beta}{\alpha\delta^2}=\frac{2\pi m}{kT}\nu^3. \]

The concentration of molecules \(n_g\) of the medium surrounding the crystal may be expressed by means of relations analogous to (12), regardless of whether or not the crystal is in equilibrium with the medium. For a two-dimensional complex consisting of \(n_0\) particles, whose mean residence time on the crystal surface is equal to the mean time of its formation at the given concentration of the gas phase, we obtain

\[ \frac{1}{\beta^{n_0}}e^{\frac{\sum\limits_1^{n_0}\varphi_i}{kT}} = \left(\frac{1}{\alpha\delta^2}\sqrt{\frac{2\pi m}{kT}}\cdot\frac{1}{n_g}\right)^{n_0}, \]

whence it follows that

\[ n_g=\frac{\beta}{\alpha\delta^2}\sqrt{\frac{2\pi m}{kT}}\cdot e^{-\frac{\sum\limits_1^{n_0}\varphi_i}{n_0 kT}} = \frac{\beta}{\alpha\delta^2}e^{-\frac{\varphi n_0}{kT}}. \tag{14} \]

The expression

$$ \frac{\sum_{1}^{n_0}\varphi_i}{n_0}=\varphi_{n_0}=\overline{\varphi}_{n_g} $$

is the average work of detachment of a single particle, calculated for the whole complex.

The question arises: what is this complex? To answer it we shall reason as follows: at a given concentration of the gas phase the crystal can be in equilibrium with the medium under the condition that, for each of its faces, the average duration of residence of the complex on its surface is equal to the average time of formation of the complex. Hence it follows directly that the complex, whose average work of detachment depends on the concentration according to the relation given above, is nothing other than one of the plane nets of the crystal, in equilibrium with the vapor (i.e. a three-dimensional nucleus in equilibrium with the vapor). It is easy to see that, in order to determine $\overline{\varphi}_{n_g}=\overline{\varphi}_{a_3}$, one may take the peripheral row of a two-dimensional nucleus corresponding to the same concentration of the gas phase.

From equations (12) and (14) it follows directly that

$$ \frac{n_g}{n_{g\infty}}=\frac{p}{p_\infty}=e^{\frac{\varphi_{\frac12}-\overline{\varphi}_{a_3}}{kT}} $$

and, correspondingly,

$$ kT\ln\frac{p}{p_\infty}=\varphi_{\frac12}-\overline{\varphi}_{a_3}. $$

It can be shown that this equation (for a simple crystallographic form) is identical with the Gibbs–Thomson equation. If, for example, one takes into account that for our model

$$ \sigma=\frac{\varphi}{2r_0^2};\qquad \varphi_{\frac12}=3\varphi;\qquad \overline{\varphi}_{a_3}=\varphi_{\frac12}-\frac{2r_0}{a_3}\varphi, $$

then the above equation assumes the following form:

$$ RT\ln\frac{p_{a_3}}{p_\infty} = \frac{4r_0^3N}{a_3}\cdot\frac{\varphi}{2r_0^2} = \frac{4M\sigma}{a_3\rho}, $$

i.e. we obtain equation (3).

To each crystal (in equilibrium form) there corresponds one precisely determined equilibrium pressure, but not conversely. Since the crystal grows by the formation of plane layers, the vapor pressure must assume definite discrete values lying on the curve determined by the Thomson–Gibbs equation. The equilibrium pressure is determined from the condition of equal probability of the processes of growth and dissolution of the given crystal.

The equation (9) derived by us is much more general than the Thomson–Gibbs equation, since it is applicable to any face of a crystal, irrespective of whether the given face belongs to the equilibrium form or not.

Putting \(\bar{\varphi}_{a_3}=\bar{\varphi}_{a_2}\), we obtain \(a_3=2a_2\). Since, at equilibrium, the medium surrounding the crystal is supersaturated with respect to the peripheral rows of any square layer larger than a two-dimensional nucleus (because the mean work of separation for a longer row is greater than the mean work of separation \(\bar{\varphi}_{a_2}\) calculated for the peripheral row of a two-dimensional nucleus), it is evident that the equilibrium medium is supersaturated with respect to the atomic rows on the edges of a cubic three-dimensional nucleus1. This conclusion is of great importance for the following reasons. From it, for example, it follows directly that, in the formation of a two-dimensional nucleus on a three-dimensional nucleus by the detachment of building elements from the surface layer of the three-dimensional nucleus, a work is required exactly equal to the work of formation of the same two-dimensional nucleus by condensation from the gas phase.

The magnitudes of these works can be calculated on the basis of the elementary processes of attachment and detachment, since the change in free energy necessary for the formation of a two-dimensional nucleus is equal to \(\pm\sum(\varphi_i-\bar{\varphi}_{a_2})\), irrespective of whether its formation occurs by condensation or by dissolution. Here \(\varphi_i\) is the work of separation of the \(i\)-th particle, and \(\bar{\varphi}_{a_2}\) is the quantity determining equilibrium. In this case the summation must be extended over all the building particles that must be attached or, respectively, detached in order to obtain a two-dimensional nucleus. In condensation the sign before \(\sum\) is negative, and in dissolution it is positive.

The validity of the above can be verified by the following simple arguments. Suppose that we have succeeded in finding such a place on the crystal where the attachment of a building particle to the crystal lattice is accompanied by the release of energy \(\bar{\varphi}_{a_2}\). Then this place will be an “equilibrium position,” since in the medium it will be occupied as often as it is free[^16] (in reality no such position exists on a finite crystal, whereas on an infinitely large crystal such a position will be a position at the middle of the crystal). For dissolution of the surface layer of the lattice we proceed as follows. First of all we move the building particle into the equilibrium position, for which the work \(\varphi_i-\bar{\varphi}_{a_2}\) is required. Then we transfer this particle into vapor, which is not accompanied by a change in free energy. In exactly the same way, but in the reverse order, we proceed in constructing a new layer of the crystal lattice by condensation of vapor.

Let us first calculate the work of formation of a two-dimensional nucleus by condensation of vapor. The construction of a two-dimensional nucleus

to be carried out in three stages, shown in Fig. 6. First we place one particle on a completely filled face. Since the work of adding this particle is \(\varphi\), the change in free energy necessary for its attachment is equal to

\[ \overline{\varphi}_{a_2}-\varphi = 3\varphi-\frac{\varphi}{n_2}-\varphi = 2\varphi-\frac{\varphi}{n_2}. \]

Then we attach such building particles as form two adjacent peripheral rows of the two-dimensional nucleus. Since

Fig. 6

the work of adding each of these particles is equal to \(2\varphi\), the change in free energy upon addition of one particle is equal to

\[ \overline{\varphi}_{a_2}-2\varphi = \varphi-\frac{\varphi}{n_2}, \]

and the total change in free energy is equal to

\[ 2(n_2-1)\left(\varphi-\frac{\varphi}{n_2}\right). \]

Finally, let us attach all the remaining \((n_2-1)\) particles necessary for the complete construction of the two-dimensional nucleus. Since the work of attaching each of them is equal to \(3\varphi\), and the change in free energy in this case is equal to

\[ \overline{\varphi}_{a_2}-3\varphi = -\frac{\varphi}{n_2}, \]

the total change in free energy in this process is equal to

\[ -(n_2-1)^2\cdot\frac{\varphi}{n_2}. \]

Consequently, the total change in free energy required for the formation of a two-dimensional nucleus by condensation is equal to

\[ W_2 = 2\varphi-\frac{\varphi}{n_2} +2(n_2-1)\left(\varphi-\frac{\varphi}{n_2}\right) -(n_2-1)^2\frac{\varphi}{n_2} = \]

\[ = n_2\varphi=\frac{a_2}{r_0}\varphi. \]

If we take into account that \(\dfrac{\varphi}{2r_0}=\chi\) (the specific peripheral energy), then \(W_2=2a_2\chi\), i.e. it is equal to one half of the peripheral energy of the nucleus, which is also obtained by the thermodynamic method.

In an analogous manner we shall calculate the work of formation of a two-dimensional nucleus by separation of the building elements from the surface—

ness of the three-dimensional nucleus. First of all (Fig. 7), let us separate all the structural elements for which the work of separation is equal to \(3\varphi\). Their number is \(n_3^2 - n_2^2 - 2(n_3 - n_2) = 3n_2^2 - 2n_2\). The required change in free energy upon separation of each of these particles is equal to

Fig. 7

\[ 3\varphi - \overline{\varphi}_{a_2} = \frac{\varphi}{n_2}. \]

Consequently, the total change in free energy in this process is equal to

\[ (3n_2^2 - 2n_2)\frac{\varphi}{n_2}. \]

Next, let us separate the particles adjacent to the two-dimensional nucleus, the number of which is \(2(n_3 - n_2) = 2n_2\). Since the work of detaching each of the particles is equal to \(2\varphi\), and the change in free energy

\[ 2\varphi - \overline{\varphi}_{a_2} = \frac{\varphi}{n_2} - \varphi, \]

the total change in free energy over this stage is equal to

\[ 2n_2\left(\frac{\varphi}{n_2} - \varphi\right). \]

Finally, the formation of the two-dimensional nucleus can be moved from the corner to the middle of the face, which is not associated with a change in free energy. Thus, the work of forming a two-dimensional nucleus by destruction of the surface layer on one of the faces of the three-dimensional nucleus is equal to

\[ W_2^* = (3n_2^2 - 2n_2)\cdot\frac{\varphi}{n_2} + 2n_2\left(\frac{\varphi}{n_2} - \varphi\right) = \]

\[ = n_2\varphi = \frac{a_2}{r_0}\varphi = 2a_2\chi. \]

Thus, the work of formation of a two-dimensional nucleus by condensation of vapor proves to be equal to the work of formation of the same nucleus by dissolution of the surface layer.

Previously we could not examine in detail the processes of attachment or detachment of individual atomic rows of the growing layer of the lattice, precisely because the work of formation of a one-dimensional nucleus cannot be determined thermodynamically. However, with the aid of the average work of detachment introduced by us, it is possible to consider the processes of attachment and detachment of atomic rows even without determining the work of formation of a one-dimensional nucleus.

Let us consider the mechanism of equilibrium of a two-dimensional nucleus. The attachment of the first particle to the periphery of a two-dimensional nucleus (the work of attachment \(2\varphi < \overline{\varphi}_{a_2}\)) requires a change in free energy. With respect to all the remaining building particles, successively attaching to the periphery of the two-dimensional nucleus, the equilibrium medium is supersaturated, since the work of detachment of each of the particles, remaining unchanged \((3\varphi = \overline{\varphi}_{a_1})\), is greater than \(\overline{\varphi}_{a_2}\). At equilibrium, the change in free energy necessary for the attachment of the first particle \((W_1 = \overline{\varphi}_{a_2} - 2\varphi)\) must be equal to the change in free energy necessary for the detachment of all the building particles of the peripheral row except one \([W_1^* = (n_2 - 1)(3\varphi - \overline{\varphi}_{a_2}) = \overline{\varphi}_{a_2} - 2\varphi]\), which in essence coincides with the definition of the average work of detachment of a particle calculated for the entire peripheral row of a two-dimensional nucleus. If, upon a change in free energy near the periphery of a two-dimensional nucleus, a supersaturation is produced sufficient for the attachment to it of one particle, then the completion of the whole row occurs automatically. Conversely, if, upon a change in free energy near a two-dimensional nucleus, there is produced an undersaturation sufficient for the detachment of all the building particles of the row except one, then the detachment of this last particle already occurs automatically.

Since \(W_1 = W_1^*\), and since, by arguments analogous to those set forth in considering a three-dimensional nucleus, the quantities preceding the powers of \(e\) in the expressions for the probabilities of attachment and of the corresponding detachment of entire rows must be identical, one may write the equality

\[ A e^{-\frac{W_1}{kT}} = A^* e^{-\frac{W_1^*}{kT}} . \]

The lability of the equilibrium of a two-dimensional nucleus is due to the following. When a new row is attached to one of the faces of a square two-dimensional nucleus, the probability of attachment (or detachment) of new rows to the same side of the two-dimensional nucleus remains unchanged. However, the probability of attachment of new rows to the neighboring sides of the two-dimensional nucleus becomes greater because of the increase of \(A\), whereas the probability of detachment of rows

on these sides decreases as a result of the increase of \(W_1^*\). As new rows are attached to the neighboring sides of the two-dimensional nucleus, the probability of attachment of new rows to the initially chosen side also increases. Analogously, with the detachment of rows from a given side of a two-dimensional nucleus, the probability of detachment of rows from neighboring sides immediately increases and the probability of attachment of new rows to them decreases.

In quite a similar way one can also explain the lability of the equilibrium of a three-dimensional nucleus. Through the attachment of new lattice layers to one of the faces of a cubic three-dimensional nucleus, the probabilities of formation on them of two-dimensional nuclei by vapor condensation and by detachment of the surface layer of the lattice remain unchanged and equal to one another. However, the probabilities of formation of a two-dimensional nucleus on neighboring faces of the three-dimensional nucleus change, since their areas increase. The probability of formation on them of two-dimensional nuclei by vapor condensation (owing to the increase of \(A\)) increases, whereas the probability of formation of two-dimensional nuclei by detachment of particles decreases (owing to the increase of \(W_1^*\)). As soon as the attachment of new layers on these faces takes place, the probability of growth of the initially chosen face also increases. Analogously, the detachment of one surface layer of the lattice of a three-dimensional nucleus leads to an increase in the probability of dissolution of the whole nucleus.

3. Equilibrium: crystal—melt16

In considering the formation of nuclei from the melt, instead of supersaturation we shall use supercooling. As a measure of supercooling we choose the logarithm of the ratio of the vapor pressure \(p\) of the supercooled melt to the vapor pressure \(p'\) of an infinitely large crystal at the same temperature. According to the Clausius–Clapeyron equation, as is known,

\[ \ln \frac{p}{p'}=\frac{L}{RTT_S}(T_S-T), \]

where \(T_S\) is the melting temperature and \(L\) is the molar heat of fusion.

For a supercooled melt one always has \(p>p'\), i.e., the vapor at pressure \(p\) is supersaturated with respect to an infinitely large crystal (with vapor pressure \(p'\)). According to the definition given above, a nucleus is a crystallite which is in equilibrium with this supersaturated vapor. For such a nucleus, as was shown earlier by us,17 the relation is valid

\[ kT\ln \frac{p}{p'}=\varphi_1^+-\bar{\varphi}_{N_0}', \tag{15} \]

where \(\varphi_1^+\) is the work of evaporation of one particle from a position on the half-crystal (equal to the heat of evaporation), \(\bar{\varphi}_{N_0}'\) is the mean work of detachment of one element, calculated for the entire surface layer.

must be finite. Therefore it should be assumed that, in the superheated melt, crystallites are formed which, owing to the impossibility of spontaneous crystallization, are subject to an equilibrium distribution, the number \(Z_N(\Delta N)\) of crystallites consisting of a number of particles between \(N\) and \(N+\Delta N\) being proportional to \(e^{-\frac{W_N}{kT}}\cdot \Delta N\):

\[ Z_N(\Delta N)=Ae^{-\frac{W_N}{kT}}\cdot \Delta N = Ae^{-\frac{\Phi_N}{kT}+\frac{NL}{N_L kT T_S}(T_S-T)}\cdot \Delta N = \left\{ \begin{array}{l} \\[-1.2em] \displaystyle =Ae^{-\frac{NL}{N_L kT_S}}\cdot e^{\frac{\frac{NL}{N_L}-\Phi_N}{kT}}\cdot \Delta N. \end{array} \right. \tag{17} \]

The difference \(\dfrac{NL}{N_L}-\Phi_N\), equal, according to the above, to the difference

\[ N\varphi_{\frac12}-\sum_1^N(\varphi_{\frac12}-\varphi_i)=\sum_1^N\varphi_i, \]

is always positive.

It follows from the equation given above that, with an increase in temperature, the relative number of crystallites of a given size decreases. With increasing superheating the equilibrium distribution shifts toward smaller crystallites. This shift should affect the course of the specific heat capacity of the melt immediately above the melting point.

The same conclusions can be reached with the aid of kinetic considerations. For this purpose we shall use the model of a simple homeopolar crystal with a simple cubic lattice, taking into account only the forces acting between nearest neighbors1.

For the further calculations it is important that, also in the case of a superheated melt, one can formally define a nucleus as a crystallite obeying equation (15a). However, in this case the nucleus will not be convex, but concave (konkav), since only for concave crystallites can the mean work of detachment of one particle \(\overline{\varphi}_{N_0}\), calculated for the entire surface layer, be greater than the work of detachment \(\varphi_{\frac12}\) of one particle from a position at the half-crystal[^21].

Thus, in expression (16), also at a temperature above the melting point, \(kT\ln \dfrac{p}{p'}\), according to (15a), may be taken equal to the difference \(\varphi_{\frac12}-\overline{\varphi}_{N_0}\), where \(\overline{\varphi}_{N_0}\) refers to the concave nucleus corresponding to the given superheating,

\[ W_N=\Phi_N-N(\varphi_{\frac12}-\overline{\varphi}_{N_0}). \]

The quantity \(\overline{\varphi}_{N_0}\) can be easily calculated for our crystalline model\({}^{22}\). For a cubic concave crystal consisting of \(N\) particles, we have

\[ \overline{\varphi}_{N_0}=3\varphi+\frac{2\varphi}{N_0^{1/3}}=\varphi_s+\frac{2\varphi}{n_0}=\overline{\varphi}_{N_0}, \]

where \(\varphi\) is the work of detaching one particle from its nearest neighbors in the lattice, \(n=N^{1/3}\) is the number of particles lying on an edge of a cubic crystal consisting of \(N\) particles. Therefore

\[ \varphi_s-\overline{\varphi}_{N_0}=kT\ln\frac{p}{p^1}=-\frac{2\varphi}{n_0}. \tag{15b} \]

Further, for our model,

\[ \Phi_N=3\varphi n^2, \]

as a consequence of which

\[ W_N=W_n=3\varphi n^2+\frac{2\varphi}{n_0}n^3. \]

Let us suppose that in the melt, to which at temperature \(T>T_s\) there corresponds a concave cubic nucleus with \(n\) particles on an edge, there is an equilibrium distribution of concave cubic nuclei. Then the rate \(v_{n\to n+1}\) of transition of crystallites with \(n\) particles on an edge into a crystallite with \(n+1\) particles is equal to the rate \(v_{n+1\to n}\) of transition of a crystallite with \(n+1\) particles on its edge into a crystallite with \(n\) particles:

\[ v_{n\to n+1}=v_{n+1\to n}. \]

Further, we put

\[ v_{n\to n+1}=Z_n P_{n\to n+1} \]

and

\[ v_{n+1\to n}=Z_{n+1}P_{n+1\to n}, \]

where \(Z_n\) (respectively, \(Z_{n+1}\)) is the number of crystallites with \(n\) (respectively, \(n+1\)) particles on the edge of the cube, contained in a unit volume, and \(P_{n\to n+1}\) (respectively, \(P_{n+1\to n}\)) are the corresponding transition probabilities. Then

\[ Z_n P_{n\to n+1}=Z_{n+1}P_{n+1\to n}, \]

accordingly

\[ \left(\frac{Z_{n+1}}{Z_n}=\frac{P_{n\to n+1}}{P_{n+1\to n}}\right). \]

Taking the logarithm of this expression, we obtain

\[ \ln Z_{n+1} - \ln Z_n = \ln \frac{P_{n \to n+1}}{P_{n+1 \to n}} . \]

For large \(n\)

\[ \ln Z_{n+1} - \ln Z_n = \frac{d \ln Z_n}{dn}, \]

therefore

\[ \frac{d \ln Z_n}{dn} = \ln \frac{P_{n \to n+1}}{P_{n+1 \to n}} . \]

Putting, approximately, \(P_{n+1 \to n} = P_{n \to n-1}\) and integrating, we obtain

\[ \ln Z_n = \int \ln \frac{P_{n \to n+1}}{P_{n \to n-1}}\, dn + \mathrm{const} \]

and, correspondingly,

\[ Z_n = C e^{\int \ln \frac{P_{n \to n+1}}{P_{n \to n-1}}\, dn}. \tag{18} \]

Let \(W'_n\) denote the energy threshold for the attachment of one new layer on one of the faces of a cubic crystallite with \(n\) particles on an edge, and \(W_n^{\prime *}\) the energy threshold for the dissolution of one surface layer of the same crystallite. Then the transition probabilities

\[ P_{n \to n+1}, \quad \text{respectively } P_{n \to n-1}, \]

are proportional to

\[ e^{-\frac{3W'_n}{kT}} \quad \text{respectively} \quad e^{-\frac{3W_n^{\prime *}}{kT}}, \]

i.e.

\[ P_{n \to n+1} = A_n e^{-\frac{3W'_n}{kT}}, \]

\[ P_{n \to n-1} = A_n e^{-\frac{3W_n^{\prime *}}{kT}} . \]

The factor 3 appears in the exponent as a consequence of the fact that, for the transition of a crystallite with \(n\) particles on an edge into a crystallite with \(n+1\) particles, the simultaneous deposition of three layers on three adjacent faces of the cubic crystal is required.

Since at temperatures above the melting point all concave crystallites must be regarded as pre-nuclei, the energy barrier upon dissolution of the surface layer of a crystallite with \(n\) particles on an edge is equal to \(W_n^* = 0^{23}\) (see below). Conversely, the energy barrier \(W_n'\) upon the imposition of one new layer is finite. The quantity \(W_n'\) can be calculated by a method analogous to that by which we calculated \(W_n\).

1. Let us separate from an infinitely large crystal a square surface layer consisting of \(n^2\) particles, and deposit it on the surface of a crystallite in the melt having \(n\) particles on an edge. In doing so we perform the work

\[ A_1 = 2\varphi n . \]

This will be the free energy of the newly formed boundary surface between the crystal and the melt.

1. Let us dissolve the crystallite reversibly in the melt

\[ A_2 = -W_n' . \]

1. Let us transform into vapor \(\dfrac{n^2}{N_L}\) moles and compress the vapor thus formed from the vapor pressure of the melt \(p\) to the vapor pressure \(p'\) of an infinitely large crystal at the same temperature

\[ A_3 = -\frac{n^2}{N_L}RT - \frac{n^2}{N_L}RT \ln \frac{p}{p'} . \]

1. Let us condense the vapor on the surface of an infinitely large crystal

\[ A_4 = \frac{n^2}{N_L}RT . \]

Since

\[ A_1 + A_2 + A_3 + A_4 = 0, \]

then

\[ W_n' = 2\varphi n + \frac{2\varphi}{n_0} n^2 . \]

Substituting this value of \(W_n'\) into the expression for \(P_{n \to n+1}\) and taking into account that \(W_n^* = 0\), we obtain

\[ \frac{P_{n \to n+1}}{P_{n \to n-1}} = e^{-\frac{3W_n'}{kT}} = e^{-\frac{3}{kT}\left(2\varphi n + \frac{2\varphi}{n_0}n^2\right)}, \]

respectively

\[ \ln \frac{P_{n \to n+1}}{P_{n \to n-1}} = -\frac{3}{kT}\left(2\varphi n + \frac{2\varphi}{n_0}n^2\right). \]

Substituting the value of \(\ln \dfrac{P_{n\to n+1}}{P_{n\to n-1}}\) into equation (18) and finding the integral of the expression in the exponent, we finally obtain

\[ Z_n=Ce^{-\frac{1}{kT}\left(3\varphi n^2+\frac{2\varphi}{n_0}n^3\right)} =Ce^{-\frac{W_n}{kT}}, \]

thus arriving once again at equation (17)\(^{24,25}\).

The distribution of pre-nuclei in a superheated melt, when the temperature is raised, as indicated above, shifts toward smaller crystallites. Since, upon dissolution of the surface layer, for a crystallite unstable at these temperatures the energy barrier is absent or, at least, smaller by an order of magnitude than the work of formation of a two-dimensional nucleus, one may expect that, when the temperature is raised, the equilibrium distribution should be established considerably more rapidly than upon cooling. Therefore, upon rapid lowering of the temperature of a superheated melt below the melting point, the distribution established at the higher temperature may remain almost unchanged, and the crystallites present in the melt will act as nuclei even at the lower temperature. In this case one should expect that, when the melt solidifies at a temperature below the crystallization point, the number of crystallization centers will be the smaller, the higher the temperature at which the equilibrium distribution was established. By studying the dependence of the number of crystallization centers on superheating, it is in principle possible to determine experimentally the distribution of pre-nuclei in a superheated melt.

The conclusions we have drawn apparently agree well with experimental data. For example, our theory makes it possible to interpret the results of Otmer’s experiments\(^{26}\).

II. KINETIC DERIVATION OF THE RATE OF NUCLEUS FORMATION\(^{27}\)

Introduction

To explain phenomena associated with the formation of new phases from a homogeneous medium, Gibbs\(^{28}\) introduced the concept of the nucleus. This concept is closely connected with the well-known Gibbs–Thomson equation, according to which the equilibrium concentration of the medium surrounding a given phase depends on the free surface energy of that phase and, consequently, on its dimensions. A nucleus is precisely that phase which is in exact equilibrium with a medium supersaturated with respect to the given phase of infinitely large dimensions. When the medium is sufficiently extensive, the equilibrium of the nucleus is labile. Gibbs derived thermodynamically the work of formation of a nuc-

...nucleus at a given supersaturation. This work is equal to one third of the free surface energy of the nucleus

\[ W=\frac{F\sigma}{3}. \]

Recently, the development of the theory of the formation of a nucleus from a supersaturated system has been especially greatly advanced by the work of Volmer\({}^{29}\). Volmer gave a theoretically well-founded expression for the rate of formation of nuclei. If by \(I\) we denote the number of nuclei formed per unit time in unit volume, then, according to Volmer,

\[ I=Ae^{-\frac{F\sigma}{3kT}} . \tag{19} \]

This expression is in agreement with known experimental data\({}^{30}\) and gives an explanation of Ostwald’s “rule of stages” (Stufenregel)\({}^{31}\).

The above equation of Volmer was derived in a purely formal manner. The derivation is based on the assertion that the probability of formation of a nucleus must be proportional to

\[ e^{-\frac{W}{kT}} . \]

We consider it extremely important to derive this equation on the basis of a detailed consideration of the mechanism of the processes associated with the formation of nuclei. In this respect, of some significance is the attempt by Farkas\({}^{32}\) to calculate the rate of formation of nuclei from a vapor. Starting from Szilard’s ideas, Farkas calculated the equilibrium distribution of particles by size, and from it the magnitude \(I\), although not by an entirely irreproachable method. The quantity \(A\) calculated by him, entering into Volmer’s equation, proved equal to\({}^{33}\)

\[ A=\frac{Cp_r}{F\rho RT}\sqrt{\frac{2\sigma NM}{\pi}} . \]

Before proceeding to the study of the question of the formation of crystalline nuclei from a supersaturated system, let us consider in detail the formation of liquid nuclei from the gaseous phase and of gaseous nuclei from a superheated liquid.

A. Rate of formation of three-dimensional nuclei

1. Rate of formation of liquid nuclei from supersaturated vapor

Let us consider the stationary distribution by size of droplets of liquid in a vapor, which we shall keep supersaturated, removing the nuclei that are formed. To bring about a stationary distribution it is necessary that the number of particles consisting of an arbitrary number \(n\) of molecules not change with time.

If we assume that the transition of a droplet consisting of \(n\) molecules into a droplet of \(n+1\) molecules, or the reverse transition, occurs through the attachment to the particle of one molecule of vapor or, correspondingly, through the evaporation of one molecule of the droplet, i.e. if the scheme

\[ n-1 \underset{v_{n\to n-1}}{\stackrel{v_{n-1\to n}}{\rightleftarrows}} n \underset{v_{n+1\to n}}{\stackrel{v_{n\to n+1}}{\rightleftarrows}} n+1, \]

is valid (where \(v_{n\to n+1}\) is the rate of formation, from a droplet with \(n\) molecules, of a droplet consisting of \(n+1\) molecules), then the general equation holds

\[ v_{n-1\to n}+v_{n+1\to n}=v_{n\to n+1}+v_{n\to n-1} \]

or

\[ I=v_{n-1\to n}-v_{n\to n-1}=v_{n\to n+1}-v_{n+1\to n}. \]

Denoting by \(Z'_n\) the number of droplets composed of \(n\) molecules, by \(p_n\) the vapor pressure over one such droplet, by \(F_n\) its surface area, by

\[ \alpha=\frac{N}{\sqrt{2\pi rTM}} \]

the number of molecules that will fall on \(1\ \mathrm{cm}^2\) at \(p=1\), and by \(p_{r_0}\) the pressure of saturated vapor, we obtain the following equation:

\[ \alpha p_{r_0} Z'_{n-1} F_{n-1}+\alpha p_{n+1} Z'_{n+1} F_{n+1} = \alpha p_n Z'_n F_n+\alpha p_{r_0} Z'_n F_n . \]

After replacing the products \(Z'_n F_n\) (i.e. the total surface area of droplets composed of \(n\) molecules) by \(Z_r\), and if we put\(^1\) \(p_{n+1}=p_n\), this equation takes the form

\[ \frac{p_{r_0}}{p_n}=\frac{Z_{n+1}-Z_n}{Z_n-Z_{n-1}} . \]

For sufficiently large \(n\) one may put

\[ Z_{n+1}-Z_n=\frac{dZ_n}{dn}; \]

then the above equation becomes

\[ \frac{p_{r_0}}{p_n} = \frac{\left(\dfrac{dZ_n}{dn}\right)_{n+1}} {\left(\dfrac{dZ_n}{dn}\right)_n}. \]

\(^1\) It is quite permissible to neglect \(Z_{n+1}(p_{n+1}-p_n)\) in comparison with \(p_n(Z_{n+1}-Z_n)\), as Farkas had already done; however, this still requires a more rigorous justification.

or

\[ \ln \frac{p_{r_0}}{p_n} = \ln \left(\frac{dZ_n}{dn}\right)_{n+1} - \ln \left(\frac{dZ_n}{dn}\right)_n . \]

For sufficiently large \(n\),

\[ \ln \frac{p_{r_0}}{p_n} = \frac{d \ln \dfrac{dZ_n}{dn}}{dn}. \tag{20} \]

This differential equation determines the stationary distribution. To integrate it we shall use the Gibbs–Thomson equation

\[ \ln \frac{p_r}{p_\infty} = \ln \frac{p_n}{p_\infty} = \frac{2\sigma M}{r\rho RT}, \]

where \(r\) is the radius of a droplet consisting of \(n\) molecules, \(p_r\), respectively \(p_n\), is the pressure of its vapor, and the remaining letters have their usual meaning.

If in equation (20) one introduces, instead of \(n\) (the number of molecules in one droplet), the droplet radius \(r\), using for this purpose the equation

\[ n=\frac{4\pi\rho N r^3}{3M} \quad \text{or} \quad dn=\frac{4\pi\rho N r^2 dr}{M}, \]

then we obtain the following equation:

\[ \frac{d \ln \dfrac{dZ_n}{dn}}{dr}\cdot \frac{M}{4\pi\rho N r^2} = \ln \frac{p_{r_0}}{p_r} = \ln \frac{p_{r_0}}{p_\infty} - \ln \frac{p_r}{p_\infty} = \ln \frac{p_{r_0}}{p_\infty} - \frac{2\sigma M}{r\rho RT}. \]

Integration of this equation gives

\[ \frac{dZ_n}{dn} = Ce^{ \frac{4\pi\rho N \ln \dfrac{p_{r_0}}{p_\infty}}{3M} - \frac{4\pi N\sigma}{RT} r^2 }, \]

which, for

\[ r=r_0=\frac{2\sigma M}{\rho RT \ln \dfrac{p_{r_0}}{p_\infty}}, \]

becomes

\[ \left(\frac{dZ_n}{dn}\right)_{r=r_0} = Ce^{ -\frac{16\pi N\sigma^3 M^2}{ 3R^3T^3\rho^2 \ln^2 \dfrac{p_{r_0}}{p_\infty} } } = Ce^{-\frac{F\sigma}{3kT}} . \]

The number of nuclei formed in unit volume per unit time will be calculated from the equation

\[ I=v_{n\to n+1}-v_{n+1\to n}=\alpha p_{r_0}Z_n-\alpha p_{n+1}Z_{n+1}. \]

If we again put \(p_{n+1}=p_n\), add and subtract \(\alpha p_n Z_n\) on the right-hand side of the equation, we obtain

\[ -\frac{I}{\alpha p_n}=Z_n\frac{p_{r_0}-p_n}{p_n}-\frac{dZ_n}{dn}. \]

For \(r=r_0\) we obtain \(p_n=p_{r_0}\), and consequently,

\[ I=-\alpha p_{r_0}\left(\frac{dZ_n}{dn}\right)_{r=r_0} \]

If one takes into account that \(I\) is always an essentially positive quantity, then \(\dfrac{dZ_n}{dn}\) must be negative, which is possible provided that the constant of integration is also negative.

If in the above equation one substitutes the value \(\left(\dfrac{dZ_n}{dn}\right)_{r=r_0}\), replaces \(\alpha\) by \(\dfrac{N}{\sqrt{2\pi RTM}}\), and, in addition, puts \(C=-C'\), then we obtain the final expression for \(I\):

\[ I=\frac{C'p_{r_0}N}{\sqrt{2\pi MRT}}\cdot e^{-\frac{16\pi M\sigma^3N^2}{3R^3T^3\rho^2\ln^2\frac{p_{r_0}}{p_\infty}}} = \frac{C'p_{r_0}}{\sqrt{2\pi MRT}}-e^{-\frac{F\sigma}{3kT}}. \tag{21} \]

The value of \(A\) in the expression derived by us for the rate of formation of liquid nuclei from supersaturated vapor differs from the expression given by Farkas by the factor \(\dfrac{2M}{F\rho}\cdot\sqrt{\dfrac{\sigma}{NRT}}\). The value of \(A\) found by us is directly proportional to the number of impacts of gas molecules per unit surface at the given pressure \((p_{r_0}, \alpha)\).

2. Rate of formation of gaseous nuclei from a superheated liquid

The formation of gaseous nuclei from a superheated liquid has hitherto not been considered theoretically at all. Since this case of nucleus formation is no less important than the preceding one, we shall analyze it here in detail.

Let the pressure at the surface of a column of liquid be equal to \(p_a\). The pressure \(p_{r_0}\) of the gas contained in a bubble of radius \(r_0\) at depth \(h\), as is known, is equal to

\[ p_{r_0}=p_a+\frac{2\sigma}{r_0}+h\rho g, \]

i.e., the sum of the external, capillary, and hydrostatic pressures. In this case the bubble is in equilibrium, i.e., it neither contracts nor expands.

If the bubble contains the vapor of the liquid in which it itself is located, then according to the Thomson–Gibbs equation

\[ p_{r_0}=p_\infty e^{-\frac{2\sigma M}{r_0\rho RT}} . \tag{22} \]

It follows from this that equilibrium is possible when \(p_a+h\rho g=p_a'<p_\infty\). In what follows we shall call the pressure \(p_a'=p_a+h\rho g\) the external pressure.

The Thomson–Gibbs equation (22) is valid for a bubble only under the condition that the bubble is in equilibrium. But this is possible only when the liquid is under the pressure \(p_{r_0}-\dfrac{2\sigma}{r_0}\). Since the external pressure

\[ p_a'=p_{r_0}-\frac{2\sigma}{r_0}<p_\infty \]

is specified in advance, this uniquely determines the vapor pressure on the liquid: it must be equal to \(p_{r_0}\), i.e., to the vapor pressure in a bubble under pressure \(p_a'\) in (labile) equilibrium.

Fig. 8

For greater clarity, in Fig. 8 are given the graphs

\[ p_r=p_\infty e^{-\frac{2\sigma M}{r\rho RT}} \]

(curve \(I\)), \(\dfrac{2\sigma}{r}\) (curve \(II\)), \(p_a'+\dfrac{2\sigma}{r}\) (curve \(III\)) and \(p_r-\dfrac{2\sigma}{r}\) (curve \(IV\)). The radius of the gaseous nucleus at the given external pressure \(p_a'\) is determined by the point at which curves \(I\) and \(III\) intersect, respectively, by the point at which the straight line \(p_a'\) intersects curve \(IV\).

It is obvious that a bubble with radius \(r_0\) is in labile equilibrium. The pressure exerted by the liquid on bubbles of smaller radius \((r<r_0)\) is greater than the vapor pressure on the liquid (determined exclusively by the pressure under which the liquid is located), as a consequence of which these bubbles disappear spontaneously. Bubbles with \(r>r_0\) must expand spontaneously.

The number of gaseous nuclei \(I\) formed per unit time in a unit volume at a given external pressure \(p_a'<p_\infty\), is calcu-

as before, for the case of a stationary distribution of gas bubbles, when the external pressure is maintained constant and the forming embryos are removed.

Let us assume that a bubble consisting of \(n\) gas molecules is obtained from a bubble with \(n+1\) molecules by transfer of one molecule to the liquid, or from a bubble with \(n-1\) molecules when one molecule passes from the liquid into the bubble. In other words, let us assume that here too the scheme

\[ n-1 \;\underset{v_{n\to n-1}}{\overset{v_{n-1\to n}}{\rightleftarrows}}\; n \;\underset{v_{n+1\to n}}{\overset{v_{n\to n+1}}{\rightleftarrows}}\; n+1 . \]

is valid. Then

\[ v_{n-1\to n}+v_{n\to n+1} = v_{n\to n-1}+v_{n\to n+1} \]

and, correspondingly,

\[ I=v_{n-1\to n}-v_{n\to n-1} = v_{n\to n+1}-v_{n+1\to n}. \]

Let us denote by \(Z_n\) the product of the number of bubbles containing \(n\) molecules by the surface area of one bubble, by \(p_a'\) the external pressure, and by \(p_{r_0}\) the vapor pressure over the liquid at the external pressure \(p_a'\). Then, for the stationary distribution, the equation holds

\[ \alpha p_{r_0} Z_{n-1} + \alpha\left(p_a' + \frac{2\sigma}{r_{n+1}}\right) Z_{n+1} = \alpha p_{r_0} Z_n + \alpha\left(p_a' + \frac{2\sigma}{r_n}\right) Z_n, \]

or, if we put approximately

\[ p_a' + \frac{2\sigma}{r_{n+1}} = p_a' + \frac{2\sigma}{r_n}, \]

then

\[ \frac{Z_{n+1}-Z_n}{Z_n-Z_{n-1}} = -\frac{p_{r_0}}{p_a' + \dfrac{2\sigma}{r_n}} . \]

Applying the same method as in calculating \(I\) for liquid embryos, we obtain the equation

\[ \frac{ d\ln \dfrac{dZ_n}{dn} }{ dn } = \ln \frac{p_{r_0}}{p_a' + \dfrac{2\sigma}{r_n}} . \tag{23} \]

This equation gives the stationary distribution.

If equation (22) is taken into account and it is noted that

\[ n= \frac{ 4\pi N\left(p_a' + \dfrac{2\sigma}{r_n}\right) }{ 3RT } \, r_n^3, \]

respectively,

\[ dn=\frac{4\pi Np_a'}{RT}\,r_n^2\,dr_n+\frac{16\pi N\sigma}{3RT}\,r_n\,dr_n, \]

then integration of (23) gives

\[ \ln\frac{dZ_n}{d_n}=\ln C+\frac{4\pi Npa'r_n^3}{3RT} \left( \ln\frac{p_{r_0}}{p_\infty} -\ln\frac{p_a'+\dfrac{2\sigma}{r_n}}{p_\infty} \right)+ \]

\[ +\frac{8\pi N\sigma r_n^2}{3RT} \left( \ln\frac{p_{r_0}}{p_\infty} -\ln\frac{p_a'+\dfrac{2\sigma}{r_n}}{p_\infty} \right) -\frac{4\pi N\sigma}{3RTp_a'^2} \left(p_a'+\frac{2\sigma}{r_n}\right)^2 r_n^2+ \]

\[ +\frac{16\pi N\sigma^2}{3RTp_a'^2} \left(p_a'+\frac{2\sigma}{r_n}\right)r_n . \]

For \(r=r_0\)

\[ p_a'+\frac{2\sigma}{r_0}=p_{r_0} \]

and

\[ \frac{p_{r_0}}{p_a'}=1+\frac{2\sigma}{p_a'r_0} =1-\frac{\rho RT}{Mp_a'}\ln\frac{p_{r_0}}{p_\infty}, \]

whence

\[ \left(\frac{dZ_n}{dn}\right)_{r=r_0} = C e^{ \frac{16\pi N\sigma^3}{3RTp_a'^2} -\frac{16\pi N\sigma^3M^2}{3R^3T^3\rho^2\ln^2\dfrac{p_{r_0}}{p_\infty}} }. \]

The number \(I\) of nuclei formed per unit time in a unit volume we compute from the equation

\[ I=v_{n\to n+1}-v_{n+1\to n} =\alpha p_{r_0}Z_n-\alpha\left(p_a'+\frac{2\sigma}{r_{n+1}}\right)Z_{n+1}. \]

Let us put, approximately,

\[ p_a'+\frac{2\sigma}{r_{n+1}}=p_a'+\frac{2\sigma}{r_n}; \]

add to and subtract from the right-hand side of the above equation

\[ \alpha\left(p_a'+\frac{2\sigma}{r_n}\right)Z_n, \]

and thereby obtain

\[ \frac{I}{\alpha} = Z_n\left[ p_{r_0}-\left(p_a'+\frac{2\sigma}{r_n}\right) \right] -\left(p_a'+\frac{2\sigma}{r_n}\right)dZ_n. \]

For \(r=r_0\) we obtain

\[ I=-\alpha p_{r_0}\left(\frac{dZ_n}{dn}\right)_{r=r_0}, \]

Substituting the values of $\alpha$ and $\left(\dfrac{dZ_n}{dn}\right)_{r=r_0}$ into the above equation, we obtain for $I$ the expression

\[ I= -\frac{C'p_{r_0}N}{\sqrt{2\pi kTM}}\cdot e^{\frac{16\pi N\sigma^3}{3RTp_a r^2}}\cdot e^{-\frac{16\pi N\sigma^3M^2}{3R^3T^2\rho^2\ln^2 \frac{p_{r_0}}{p_\infty}}} = \]

\[ = \frac{C'p_{r_0}N}{\sqrt{2\pi kTM}}\cdot e^{\frac{15\pi N\sigma^3}{3RTp_a r^2}}\cdot e^{-\frac{F\sigma}{3kT}} \tag{24} \]

where $C'=-C$.

Thus the rate of formation of gaseous nuclei in a superheated liquid is determined by an expression analogous to the corresponding expression for the rate of formation of liquid droplets from a supersaturated vapor. However, in this case the Volmer quantity $A$ contains one more factor favorable to the formation of nuclei.

3. Rate of Formation of Crystalline Nuclei from a Supersaturated Vapor

The formation of crystalline nuclei from a supersaturated vapor is a process much more complicated than the formation of liquid and gaseous nuclei. This complexity follows from the fact that the forming phase possesses a definite structure. Owing to the existence of such a structure, the processes of growth and dissolution of crystals also prove to be more complicated than the same processes for a liquid. Until recently it seemed impossible to give a kinetic derivation of the quantity $I$ for crystalline nuclei. However, after we succeeded in clarifying the mechanism of growth and dissolution, and in general the equilibrium of small crystallites,^34 we were able to set ourselves this problem as well.

For simplicity we shall use here too the model of a simple cubic homeopolar crystal and restrict ourselves to forces acting between nearest neighbors in the crystal lattice. The equilibrium form of such a crystalline model is a cube, and the processes of growth and dissolution proceed through the formation of two-dimensional nuclei on the faces of the cube.

To solve the problem posed, it is first of all necessary to extend the mechanism of growth and dissolution of crystals proposed by us also to crystallites of sizes smaller than the size of the nucleus. In doing so, it seems expedient to divide such crystallites into two classes: 1) crystallites whose edge is smaller than the edge of the three-dimensional nucleus, but larger than the side of the two-dimensional nucleus $a_2$; 2) crystallites whose edge is smaller than the side of the two-dimensional nucleus.

The mean work of detachment of one row from the edge of a cube of a crystallite of the first class is greater than the mean work of detachment $\tilde{\varphi}_{a}$ of the surface layer of the lattice of a three-dimensional nucleus (respectively

average work of detachment \(\overline{\varphi}_{a_2}\) of the peripheral row of a two-dimensional nucleus). Growth of such a crystallite is much less probable than its dissolution. However, since the gas phase is supersaturated with respect to the edges of the crystallite, there must exist an energy threshold for the formation of a two-dimensional nucleus by separation of the surface layer of the lattice. For crystallites of the second class such an energy threshold is absent.

Having calculated the change in the free energy \(W_a\) necessary for the formation of a new lattice layer at a given supersaturation as a function of the edge length of the cube \(a\), we find that, as \(a\) increases, \(W_a\) grows until the edge of the crystallite becomes equal to the side \(a_2\) of the two-dimensional nucleus. With further increase of \(a\), the quantity \(W_a\) remains constant and equal to the work of formation of a two-dimensional nucleus at the given supersaturation. For \(a \leq a_2\)

\[ W_a = \frac{(a-r_0)^2}{r_0^2} \left(3\varphi-\frac{r_0}{a_2}\varphi-3\varphi\right) +2\,\frac{a-r_0}{r_0} \left(3\varphi-\frac{r_0}{a_2}\varphi-2\varphi\right) + \]

\[ +3\varphi-\frac{r_0}{a_2}\varphi-\varphi = \frac{a}{r_0}\varphi\left(2-\frac{a}{a_2}\right). \]

Here \(\varphi\) is the work necessary to separate two neighboring particles situated in the crystal lattice at a distance \(r_0\) from one another. For \(a=a_2\)

\[ W_{a_2}=W_2=\frac{a_2}{r_0}\varphi=2a_2\chi, \]

i.e. \(W_a\) is equal to the work of formation of a two-dimensional nucleus.

Having calculated the change in the free energy \(W_a^*\) necessary for dissolution of one surface layer of the lattice as a function of the edge length of the cube \(a\), we find that for all crystallites with an edge smaller than the side \(a_2\) of the two-dimensional nucleus, \(W_a^*=0\). With further increase of \(a\), the quantity \(W_a^*\) grows. As was already shown, the work of detachment of the surface layer of a three-dimensional nucleus \(W_{a_3}\) is exactly equal to the work \(W_{a_2}=W_2=W_2^*\), i.e. equal to the work of formation of a two-dimensional nucleus by condensation of the gas phase. For

\[ W_a^* = \left( \frac{a^2-a_2^2}{r_0^2} -2\,\frac{a-a_2}{r_0} \right) \left(3\varphi-3\varphi+\frac{r_0}{a_2}\varphi\right) + \]

\[ +2\,\frac{a-a_2}{r_0} \left(2\varphi-3\varphi+\frac{r_0}{a_2}\varphi\right) = \frac{a^2}{a_2 r_0}\varphi -\frac{2a}{r_0}\varphi +\frac{a_2}{r_0}\varphi. \]

Since for a crystallite with edge \(a<a_2\), \(W_a^*=0\), while for a crystallite with \(a>a_2\), \(W_a=W_2=\dfrac{a_2}{r_0}\varphi\), for our crystalline model in general

\[ W_a^*-W_a=\frac{a}{r_0}\varphi\left(\frac{a}{a_2}-2\right). \tag{25} \]

Having calculated the number of crystalline nuclei formed per unit volume per unit time under a stationary distribu-

tion. Let a crystalline cube with edge \(a\) be able to be formed from a cube with edge \(a+r_0\) by separation of surface layers from three adjacent faces, or from a cube with edge \(a-r_0\) by attachment of layers to the indicated three faces. In other words, let the formation of a cube with edge \(a\) proceed according to the scheme

\[ a-r_0 \underset{v_{a\to a-r_0}}{\stackrel{v_{a-r_0\to a}}{\rightleftarrows}} a \underset{v_{a+r_0\to a}}{\stackrel{v_{a\to a+r_0}}{\rightleftarrows}} a+r_0, \]

where \(v_{a\to a+r_0}\) is the rate of formation of a crystallite with edge \(a+r_0\) from a crystallite with edge \(a\), etc.

For a stationary distribution, the number of crystallites of any size must not change with time, and therefore

\[ v_{a-r_0\to a}+v_{a+r_0\to a}=v_{a\to a-r_0}+v_{a\to a+r_0} \]

and, correspondingly,

\[ I=v_{a-r_0\to a}-v_{a\to a-r_0}=v_{a\to a+r_0}-v_{a+r_0\to a}. \]

In addition, the following equation is also valid:

\[ Z_{a-r_0}P_{a-r_0\to a}+Z_{a+r_0}P_{a+r_0\to a} = Z_aP_{a\to a+r_0}+Z_aP_{a\to a-r_0}, \]

where \(Z_a\) is the number of crystallites with edge \(a\), \(P_{a\to a+r_0}\) is the probability of transformation of a crystallite with edge \(a\) into a crystallite with edge \(a+r_0\), etc.

Let, approximately,

\[ P_{a-r_0\to a}=P_{a\to a+r_0} \quad\text{and}\quad P_{a+r_0\to a}=P_{a\to a-r_0}. \]

Then the above equation can be put in the form

\[ \frac{Z_{a+r_0}-Z_a}{Z_a-Z_{a-r_0}} = \frac{P_{a\to a+r_0}}{P_{a\to a-r_0}}. \]

For sufficiently large \(a\),

\[ Z_{a+r_0}-Z_a = r_0\left(\frac{dZ_a}{da}\right)_{a+r_0} \]

and

\[ \ln\left(\frac{dZ_a}{da}\right)_{a+r_0} - \ln\left(\frac{dZ_a}{da}\right)_a = r_0 \frac{d\ln\frac{dZ_a}{da}}{da}. \]

Whence it follows that

\[ r_0 \frac{d\ln\frac{dZ_a}{da}}{da} = \ln \frac{P_{a\to a+r_0}}{P_{a\to a-r_0}}. \tag{26} \]

The ratio of the probability of transition of a crystallite with edge \(a\) into a crystallite with edge \(a+r_0\), respectively, to the probability of transition of the same crystallite into a crystallite with edge \(a-r_0\), according to the foregoing, is equal to

\[ \frac{P_{a\to a+r_0}}{P_{a\to a-r_0}} = \left( \frac{ e^{-\frac{W_a}{kT}} }{ e^{-\frac{W_a^*}{kT}} } \right)^3 . \]

The occurrence of 3 in the exponent of this expression is due to the fact that, for the transition of a crystallite with edge \(a\) into a crystallite with edge \(a+r_0\), the simultaneous attachment of three layers to three adjacent faces is necessary. In exactly the same way, for the transformation of a crystallite with edge \(a\) into a crystallite with edge \(a-r_0\), the simultaneous detachment of three lattice layers from three neighboring crystal faces is necessary.

If we use equation (25) and replace \(a_2\) by \(\dfrac{a_3}{2}\), then integration of equation (26) gives

\[ \frac{dZ_a}{da} = C\cdot e^{\frac{\varphi}{2r_0^2 kT}\left(\frac{4a^3}{a_3}-6a^2\right)} . \tag{27} \]

The function \(\dfrac{dZ_a}{da}\) determines the slope of the distribution function \(Z_a\). Since \(C\) must be negative (see below), the function \(\dfrac{dZ_a}{da}\) must always be negative and has a maximum at \(a=a_3\). The function \(Z_a\), with increasing \(a\), decreases monotonically and at \(a=a_3\) has an inflection point. In Fig. 9 the graphs of \(Z_a\) and \(\dfrac{dZ_a}{da}\) in the region near \(a_3\) are schematically represented. Putting \(a=a_3\) and taking into account that \(\dfrac{\varphi}{2r_0^2}=\sigma\), we obtain from equation (27)

\[ \left(\frac{dZ_a}{da}\right)_{a=a_3} = C e^{-\frac{F\sigma}{3kT}} \]

where \(\dfrac{F\sigma}{3}=2a_3^2\sigma\) is the work of formation of a three-dimensional nucleus.

Fig. 9

In addition to the derivations of \(W_2\) and \(W_2^*\), given earlier \(^{35}\), let us calculate the work of formation of a three-dimensional nucleus \(W_3\), using for this purpose the average work of detachment \(\overline{\varphi}_{a_3}=3\varphi-\dfrac{2\varphi}{n_3}\) (where \(n_3=\dfrac{a_3}{r_0}\) is the number of particles fitting along the edge of the cube).

Let us compute for this purpose the change in free energy

$$ \sum (\bar{\varphi}_{a_3}-\varphi_i), $$

necessary for the formation of a nucleus at the given supersaturation. Here the summation extends over all particles of the three-dimensional nucleus.

Fig. 10

For simplicity of calculation, let us construct a three-dimensional nucleus in the following way. First of all, we construct a three-dimensional nucleus by the method shown in Fig. 10 \((a, b, c)\). For this the necessary change in free energy is

$$ 3\varphi-\frac{2\varphi}{n_3} +2(n_3-1)\left(3\varphi-\frac{2\varphi}{n_3}-\varphi\right) +(n_3-1)^2\left(3\varphi-\frac{2\varphi}{n_3}-2\varphi\right). $$

On the layer of the lattice thus formed we then construct successively \(n_3-1\) lattice layers by the method \((d, e, f)\) of Fig. 10; for the construction of each of them the necessary change in free energy is

$$ 3\varphi-\frac{2\varphi}{n_3}-\varphi +2(n_3-1)\left(3\varphi-\frac{2\varphi}{n_3}-2\varphi\right) + $$

$$ +(n_3-1)^2\left(3\varphi-\frac{2\varphi}{n_3}-3\varphi\right), $$

which is equal to zero, as also follows from the definition of \(\bar{\varphi}_{a_3}\). Consequently, the work of formation of a three-dimensional nucleus is equal to

$$ W_3=3\varphi-\frac{2\varphi}{n_3} +2(n_3-1)\left(2\varphi-\frac{2\varphi}{n_3}\right) +(n_3-1)^2\left(\varphi-\frac{2\varphi}{n_3}\right)= $$

$$ =n_3^2\varphi=2a_3^2\sigma=\frac{F\sigma}{3}. $$

The number \(I\) of nuclei formed per unit volume per unit time is obtained from the equation

$$ I=v_{a\to a+r_0}-v_{a+r_0\to a} =Z_aP_{a+r_0}-Z_{a+r_0}P_{a+r_0\to a}. $$

If we set, approximately,

\[ P_{a-r_0\to a}=P_{a\to a-r_0} \]

and

\[ P_{a+r_0\to a}=P_{a\to a-r_0}, \]

and add and subtract \(Z_aP_{a\to a-r_0}\) from the right-hand side of the equation, then we obtain

\[ I=Z_a\left(P_{a\to a+r_0}-P_{a\to a-r_0}\right)-P_{a\to a-r_0}\left(Z_{a+r_0}-Z_a\right). \]

For \(a=a_3\), i.e. for a three-dimensional nucleus, the probabilities of growth and dissolution become identical, whence

\[ I=-\left(P_{a\to a-r_0}\right)_{a=a_3}\cdot \left(\frac{dZ_a}{da}\right)_{a=a_3}\cdot r_0 . \]

Since \(I\) is positive, the constant \(C\) in equation (27) is negative.

If in the expression for \(I\) we substitute

\[ \left(P_{a\to a-r_0}\right)_{a=a_3} = \left(\operatorname{const}\, a_3^2\cdot e^{-\frac{W_2}{kT}}\right)^3, \]

\[ \left(\frac{dZ_a}{da}\right)_{a=a_3} = C e^{-\frac{F\sigma}{kT}}, \]

then we obtain

\[ I=C'r_0\left(a_3^2\cdot e^{-\frac{W_2}{kT}}\right)^3\cdot e^{-\frac{W_3}{kT}} = C'r_0\left(a_3^2\cdot e^{-\frac{Lx}{2kT}}\right)^3\cdot e^{-\frac{F\sigma}{3kT}} . \tag{28} \]

The appearance of the factor \(a_3^2\) is connected with the fact that the probability of formation of a two-dimensional nucleus on one of the faces of the three-dimensional one is proportional to the area serving as the substrate for the two-dimensional nucleus.

The exponent of the first exponential factor in (28) contains the work of formation of a two-dimensional nucleus \(W_2=\dfrac{Lx}{2}\) (i.e. half the peripheral energy of the two-dimensional nucleus), whereas the second exponential factor is identical with the factor in Volmer’s equation (19).

It follows from equation (28) that the rate of formation of crystalline nuclei (other conditions being comparable) is many times smaller than the rate of formation of liquid and gaseous nuclei.

The factor \(e^{-\frac{W_2}{kT}}\) indicates that the incipient phase possesses a definite structure; for the growth of a crystallite, a single impact of gas molecules on its surface is insufficient, and the formation from these molecules of a new lattice layer is also necessary.

Folmer’s equations made it possible to interpret the well-known Ostwald “rule of stages.” The equation we have now derived permits an even better interpretation. As Stranski and Totomanow have shown\(^{36}\), Ostwald’s “rule of stages” is often obeyed because, at a temperature close to the transition point, the supersaturation \(\ln \dfrac{p}{p_\infty}\) is practically the same for both modifications, whereas \(\dfrac{b\sigma^3}{\rho^2}\) is larger for the modification stable at the lower temperature. Here \(b\) is a geometrical factor, of the same order of magnitude for both modifications. The specific surface energy \(\sigma\) and the density \(\rho\) are almost always greater for the modification stable at the lower temperature. But since \(\sigma\) enters the expression given above to the third power, and \(\rho\) to the second, this whole expression is also larger for the modification stable at the lower temperature.

Equation (28) contains, in addition, the factor

\[ e^{-\frac{W_2}{kT}} = e^{-\frac{L_x}{kT}} = e^{-\frac{C x^2 M N}{\rho R^2 T^2 \ln \frac{p}{p_\infty}}}. \]

For a temperature close to the transition point, the supersaturation \(\ln \dfrac{p}{p_\infty}\) is practically the same for both modifications, and almost always \(\dfrac{C x^2}{\rho}\) is greater for the modification stable at the lower temperature. \(C\) is a geometrical factor, of the same order of magnitude for both modifications.

B. Rate of Formation of Two-Dimensional Nuclei

1. Linear rate of crystallization

Folmer and Marder reduced the linear rate of crystallization to the rate of formation of two-dimensional nuclei\(^{37}\). They found that, in the construction of a growing face of a crystal, most of the time is required for the formation of a two-dimensional nucleus. They considered the rate of formation of two-dimensional nuclei (as well as the rate of formation of three-dimensional nuclei) to be proportional to \(e^{-\frac{W_2}{kT}}\), where \(W_2\) is the work of formation of a two-dimensional nucleus. Further, on the basis of the assumption that the process of crystallization from the melt is analogous to the transformation of one crystalline modification into another, a factor \(e^{-\frac{Q}{kT}}\) was introduced into the expression for the rate of crystallization. By means of this factor they took into account the activated molecules of the liquid, i.e., those molecules which are able to pass from the liquid onto the surface of the crystal. Thus Folmer and

Marder obtained the following expression for the linear rate of crystallization:

\[ v=K\cdot e^{-\frac{Q}{kT}}\cdot e^{-\frac{W_2}{kT}} . \tag{29} \]

In crystallization from the melt it is convenient to express the work of formation of a two-dimensional nucleus

\[ W_2=\frac{Lx}{2}=\frac{C\varkappa^2 M}{\rho RT\ln \frac{p}{p_\infty}} \]

not through the free energy \(RT\ln \dfrac{p}{p_\infty}\) of the melt supercooled relative to the crystal, but through the supercooling \(\Delta T\). In Fig. 11 the curve of sublimation pressure over the crystal (curve \(I\)) and the curve of vapor pressure over the liquid (curve \(II\)) are shown schematically; they intersect at the melting point \(T_S\), at which the vapor pressure is equal to \(p_S\). At a temperature \(T\), lower than \(T_S\), the vapor pressure over the supercooled melt is equal to \(p'_\infty\), while the sublimation pressure over the crystal is equal to \(p_\infty < p'_\infty\). Applying the Clapeyron–Clausius equation to both curves, we obtain, for a temperature close to \(T_S\):

Fig. 11

\[ \ln \frac{p_S}{p_\infty} = \frac{\lambda}{R}\left(\frac{1}{T}-\frac{1}{T_S}\right) = \frac{\lambda}{RTT_S}\Delta T. \]

and

\[ \ln \frac{p_S}{p'_\infty} = \frac{\lambda'}{R}\left(\frac{1}{T}-\frac{1}{T_S}\right) = \frac{\lambda'}{RTT_S}\Delta T, \]

where \(\lambda\) is the heat of evaporation of the liquid, and \(\lambda'\) is the heat of sublimation of the crystal.

From these equations it follows that

\[ \ln \frac{p'_\infty}{p_\infty} = (\lambda'-\lambda)\frac{\Delta T}{RTT_S} = \frac{\lambda''}{RTT_S}\Delta T, \]

where \(\lambda''=\lambda'-\lambda\) is the heat of melting of the crystal. Substituting this expression into the expression for \(W_2\), we obtain

\[ W_2=\frac{C\varkappa^2 MT_S}{\rho\lambda''\Delta T} = \frac{\mathrm{const}}{\Delta T}. \]

Using this expression, we can write equation (29) in the form

\[ v = K_1 e^{-\frac{K_2}{T}}\cdot e^{-\frac{K_3}{T\Delta T}} . \tag{29a} \]

With the aid of this equation, Volmer and Marder attempted to interpret the results of their measurements of the rate of crystallization of supercooled glycerin and obtained better agreement of the theoretical curves with the experimental ones than had been possible on the basis of earlier theoretical deductions \(^{37}\). The Volmer–Marder equation (29), and, correspondingly, (29a), was derived by them purely formally. The introduction into it of the factor \(e^{-\frac{Q}{kT}}\) is insufficiently justified.

On the basis of the mechanism of growth and dissolution proposed by us, and in general of the equilibrium of two-dimensional crystals, one can derive the rate of formation of two-dimensional nuclei and, consequently, the proportional linear rate of crystallization.

For simplicity, here too, as a model, we shall use the lattice of a simple cubic homeopolar crystal, taking into account only the forces acting between nearest particles.

Let us again assume that the distribution with respect to the magnitude of the two-dimensional crystallites is stationary and that the supersaturation is maintained constant. The nuclei that form will automatically be removed from the system, since, having grown, they will cover the faces of the three-dimensional crystal, displacing the crystal–medium boundary surface parallel to itself. Let us suppose that a crystallite with side \(a\) can be formed either by the attachment of two rows to two adjacent sides of a crystallite with side \(a-r_0\), or by the separation of two adjacent rows from a crystallite with side \(a+r_0\). Denoting by \(v_{a\to a+r_0}\) the rate of formation of a crystallite with side \(a+r_0\) from a crystallite with side \(a\), etc., and assuming that the processes proceed according to the scheme

\[ a-r_0 \;\underset{v_{a\to a-r_0}}{\overset{v_{a-r_0\to a}}{\rightleftarrows}}\; a \;\underset{v_{a+r_0\to r_0}}{\overset{v_{a\to a+r_0}}{\rightleftarrows}}\; a+r_0, \]

we find that

\[ v_{a-r_0\to a}+v_{a+r_0\to a} = v_{a\to a-r_0}+v_{a\to a+r_0} \]

and, correspondingly,

\[ I = v_{a-r_0\Rightarrow a} - v_{a\Rightarrow a-r_0} = v_{a\Rightarrow a+r_0} - v_{a+r_0\Rightarrow a}, \]

where \(I\) is the number of two-dimensional nuclei formed per unit time per unit surface area.

In addition,

\[ Z_{a-r_0}P_{a-r_0\to a} + Z_{a+r_0}P_{a+r_0\to a} = Z_aP_{a\Rightarrow a+r_0} + Z_aP_{a\to a-r_0}. \]

where \(Z_a\) is the number of two-dimensional crystallites with side \(a\), \(P_{a\to a+r_0}\) is the probability of transition of a crystallite with side \(a\) into a crystallite with side \(a+r_0\), etc.

If, approximately, we put

\[ P_{a-r_0\to a}=P_{a\to a+r_0}, \]

\[ P_{a+r_0\to a}=P_{a\to a-r_0}, \]

then the above equation can be brought to the form

\[ \frac{Z_{a+r_0}-Z_a}{Z_a-Z_{a-r_0}} = \frac{P_{a\to a+r_0}}{P_{a\to a-r_0}}. \]

For sufficiently large \(a\),

\[ Z_{a+r_0}-Z_a = r_0\left(\frac{dZ_a}{da}\right)_{a+r_0} \]

and

\[ \ln\left(\frac{dZ_a}{da}\right)_{a+r_0} - \ln\left(\frac{dZ_a}{da}\right)_a = r_0 \frac{ d\ln \dfrac{dZ_a}{da} }{ da }, \]

whence it follows that

\[ r_0 \frac{ d\ln \dfrac{dZ_a}{da} }{ da } = \ln \frac{P_{a\to a+r_0}}{P_{a\to a-r_0}}. \tag{30} \]

The ratio of the probability of transition of a crystallite with side \(a\) into a crystallite with side \(a+r_0\) to the probability of transition of a crystallite with side \(a\) into a crystallite with side \(a-r_0\) is equal to

\[ \frac{P_{a\to a+r_0}}{P_{a\to a-r_0}} = \left( \frac{ e^{-\frac{W_a}{kT}} }{ e^{-\frac{W_a^*}{kT}} } \right)^2, \]

where \(W_a\) is the change in free energy necessary for the formation of a peripheral row of length \(a\), and \(W_a^*\) is the change in free energy upon dissolution of one such row.

The exponent in the expression for the ratio of probabilities is equal to two because, in order to transform a crystallite with side \(a\) into a crystallite with side \(a+r_0\) (respectively, \(a-r_0\)), the simultaneous attachment (respectively, detachment) of two whole rows to two neighboring sides of the crystallite is necessary.

As was shown earlier, in the case of our crystalline model, when each row is attached to a two-dimensional crystal there is an energy threshold determined by the attachment of the first particle of the row (the work of its attachment is equal to \(2\varphi\)). With respect to all the other particles subsequently attaching themselves

of the row, the medium surrounding the crystal is supersaturated. Since the concentration of the medium near the crystal is determined by the value \(\bar\varphi_{a_2}\), \(W_a\) proves to be equal to

\[ W_a=\bar\varphi_{a_2}-2\varphi=3\varphi-\frac{r_0}{a_2}\varphi-2\varphi =\varphi-\frac{r_0}{a_2}\varphi . \]

The energy threshold for the separation of one peripheral row of a two-dimensional crystal with side \(a\) is determined by the separation of all particles of the row except one. Since the number of these particles is \(\dfrac{a-r_0}{r_0}\), and since the work of detachment of each of them is \(\varphi_i=3\varphi\), \(W_a^*\) is equal to

\[ W_a^*=\frac{a-r_0}{r_0}\left(3\varphi-\bar\varphi_{a_2}\right) =\frac{a-r_0}{r_0}\cdot\frac{r_0}{a}\cdot\varphi =\frac{a-r_0}{a}\cdot\varphi . \]

Consequently,

\[ W_a^*-W_a=\varphi\left(\frac{a}{a_2}-1\right). \]

With the aid of this equation we integrate equation (30)

\[ \frac{dZ_a}{da}=C e^{\frac{\varphi}{2r_0 kT}\left(\frac{2a^2}{a_2}-4a\right)} . \]

For \(a=a_2\), taking into account that \(\dfrac{\varphi}{2r_0}=\chi\), we obtain

\[ \left(\frac{dZ_a}{da}\right)_{a=a_2} = C e^{-\frac{2a_2\chi}{kT}} = C e^{-\frac{l\chi}{2kT}} . \]

The number of nuclei formed per unit time per unit surface is calculated from the equation

\[ I=v_{a\to a+r_0}-v_{a+r_0\to a} =Z_a P_{a\to a+r_0}-Z_{a+r_0}P_{a+r_0\to a}, \]

or, if we put approximately

\[ P_{a-r_0\to a}=P_{a\to a+r_0} \quad\text{and}\quad P_{a+r_0\to a}=P_{a\to a-r_0}, \]

and add and subtract \(Z_aP_{a\to a-r_0}\) to the right-hand side, then

\[ I=Z_a\left(P_{a\to a+r_0}-P_{a\to a-r_0}\right) -P_{a\to a-r_0}\left(Z_{a+r_0}-Z_a\right). \]

For \(a=a_2\), i.e. for a two-dimensional nucleus, the probabilities of growth and dissolution are the same. Therefore

\[ I=\left(P_{a\to a-r_0}\right)_{a=a_2}\cdot \left(\frac{dZ_a}{da}\right)_{a=a_2}\cdot r_0 . \]

If in this expression we substitute the values

and

\[ \left(\frac{dZ_a}{da}\right)_{a=a_2} \quad \text{and} \quad (P_a \to a-r_0)_{a=a_2} = \mathrm{const}\left(a_2 \cdot e^{-\frac{W_a^*}{kT}}\right)^2, \]

then we obtain \(I\), and together with it the linear rate of crystallization proportional to \(I\),

\[ v=C'\left(a_2\cdot e^{-\frac{W_a^*}{kT}}\right)^2\cdot e^{-\frac{L\alpha}{2kT}}, \tag{31} \]

where \(C'=-C\cdot \mathrm{const}\).

If one further takes into account that

\[ W_2=\frac{L\alpha}{2}=\frac{\mathrm{const}}{\Delta T} \]

and

\[ a_2=\frac{\mathrm{const}}{\Delta T}, \]

then equation (31) can be put in the form

\[ v=\frac{K_1}{\Delta T^2}\cdot e^{-\frac{K_2}{T}}\cdot e^{-\frac{K_3}{T\Delta T}}. \tag{31a} \]

This equation differs from the equation of Volmer–Marder (29a) by the factor \(\frac{1}{\Delta T^2}\). From the very derivation of this equation it is clear that the factor \(-\frac{K_2}{T}\) does not depend on whether or not there exists an activation energy of the liquid molecules. In our derivation this factor appears as a consequence of the existence of an energy threshold in the attachment (respectively, detachment) of whole rows to the periphery of a two-dimensional crystal. But if the activation energy of the molecules does in fact exist, this still will not change the form of our equation (31a), since then to the constant \(K_2\) there will be added also \(\left(\frac{Q}{k}\right)\) (where \(Q\) is the activation energy of the molecule, and \(k\) is Boltzmann’s constant).

In Fig. 12 the function (31a) is represented graphically for \(K_1=10^{15,041}\), \(K_2=6993\), and \(K_3=27\,940\). The dotted curve in the same figure represents the experimental results of Volmer and Marder for the linear rate of crystallization of supercooled glycerin. An analogous curve can also be obtained from the Volmer–Marder equation for \(K_1=10^{15,870}\), \(K_2=9507\), and \(K_3=20\,653\). However, the value of \(K_2\) in Volmer–Marder is inadmissibly large.

Volmer and Marder also examined in detail what the causes of the discrepancy between the theoretical and experimental curves may be.^37 The most important cause of the deviations is the assumption that for each layer of the lattice grown on the cleavage there is one single two-dimensional nucleus. At a high rate

...crystallization this assumption is of no essential significance. Therefore the interpretation of the constants based on the fact that the theoretical curves coincide with the experimental ones also becomes unconvincing. At present we can find only the order of magnitude of the constant \(K_2\), determined by the energy barrier in the dissolution of one peripheral row of a two-dimensional nucleus. For our simple crystalline model, for example,

\[ K_2=\frac{2}{k}\left(\varphi-\frac{r_0}{a_2}\varphi\right). \]

Since approximately \(\varphi=\frac{\lambda'}{3N}\) (\(\lambda'\)—the heat of sublimation of the crystal), and in the first approximation \(\frac{r_0}{a_2}\varphi\) may be neglected, it follows that \(K_2 \sim \frac{\lambda'}{3}\). In general one may assert that \(K_2\) must be smaller than \(\lambda'\).

Fig. 12

1. Rate of polymorphic transformations

The rate of transformation into one another of two crystalline polymorphic modifications can be reduced to the rate of formation of two-dimensional nuclei. However, in determining the rate of crystallization from a melt one may neglect the energy barrier associated with the disappearance of the liquid, in comparison with the energy barrier associated with the formation of two-dimensional nuclei on the faces of the growing crystal. Conversely, in polymorphic transformations in some cases it is absolutely necessary to take into account the energy jumps associated with the decomposition of the metastable modification.

Let, at a temperature \(T\), lower than \(T_S\), the sublimation pressure of the modification stable at this temperature be \(p_\infty\), and the sublimation pressure of the modification metastable at this temperature be \(p'_\infty\) (Fig. 11). Then the difference in free energies of the unstable and stable modifications will be equal to \(RT\ln\frac{p'_\infty}{p_\infty}\). Owing to the fact that both phases have a crystalline structure, the transfer of substance from one solid phase to the other occurs not directly, but through the gaseous phase (even if two-dimensional).

Let us first consider the case where both these modifications are represented in the form of infinitely large crystals. In this case the measure of the rate of decomposition of the unstable modification is the rate of formation of “concave” two-dimensional nuclei (according to Volmer–Nohlkeime)\(^{38}\) on the surface of this modification, and the measure...

the rate of growth of the stable modification is the rate of formation of two-dimensional nuclei on the surface of this modification by condensation of the gas phase. Both rates are proportional to

\[ e^{-\frac{W_2}{kT}}, \]

where \(W_2\) is the work of formation of the corresponding two-dimensional nucleus, which is itself inversely proportional to the supersaturation, respectively, to the undersaturation of the gas phase relative to the sublimation pressure of the corresponding modification. Since the gas pressure is \(p\), the measure of decomposition is the undersaturation \(\ln \frac{p}{p'_\infty}\), and that of growth is the supersaturation \(\ln \frac{p}{p_\infty}\). In Fig. 13 the rate of decomposition of the modification unstable at the given temperature (curve \(I\)) and the rate of growth of the stable modification (curve \(II\)) are shown schematically as functions of the gas pressure in the interval from \(p'_\infty\) to \(p_\infty\). If the pressure is equal to \(p'_\infty\), i.e. equal to the sublimation pressure of the unstable modification, then the supersaturation with respect to the stable modification is maximal in the given interval and is equal to \(\ln \frac{p'_\infty}{p_\infty}\), while the undersaturation with respect to the unstable modification is equal to zero, i.e. \(\ln \frac{p'_\infty}{p_\infty}=0\). Conversely, if the gas pressure is equal to the sublimation pressure of the stable modification \(p_\infty\), then the supersaturation with respect to this phase is equal to zero, i.e. \(\ln \frac{p_\infty}{p_\infty}=0\), while the undersaturation with respect to the unstable modification is maximal in the given interval and is equal to \(\ln \frac{p'_\infty}{p_\infty}\). In the pressure interval between \(p'_\infty\)

Fig. 13

and \(p_\infty\), both modification phases are simultaneously present: a decrease in the gas pressure may occur through the growth of the stable modification and, consequently, cannot become less than the sublimation pressure of this modification, equal to \(p_\infty\). On the other hand, an increase in the gas pressure may occur through decomposition of the unstable modification and, consequently, cannot exceed the sublimation pressure of the given modification \(p'_\infty\). Moreover, it may be asserted that at the greatest possible supersaturation (respectively, undersaturation) \(\ln \frac{p'_\infty}{p_\infty}\), the work of formation of a two-dimensional nucleus on the surface of the stable modification by condensation of the gas is always greater than the work of formation of a concave two-dimensional nucleus on the surface of the unstable modification during its decomposition. This follows from the considerations set forth in the interpretation of the remaining

to the “rule of stages.” As a consequence of this, the maximum growth rate of the stable modification in the interval from \(p'_{\infty}\) to \(p_{\infty}\) is represented in Fig. 13 as considerably smaller than the maximum rate of decomposition of the unstable modification in the same pressure interval.

The mechanism of the transformation is, in general outline, as follows. On the surface of the stable modification a two-dimensional nucleus is formed by condensation of the gas enclosed between two sufficiently large layers of the stable and unstable modifications. This nucleus grows at the given supersaturation. At the same time, the concentration in the gas phase rapidly decreases, the undersaturation with respect to the unstable modification increases, and, together with this, the probability of formation on its surface of a concave two-dimensional nucleus also increases. After such a nucleus has formed, the upper lattice layer of the unstable modification is destroyed, as a result of which the concentration in the gas phase again increases. Consequently, the supersaturation with respect to the stable modification, and with it the probability of formation of two-dimensional nuclei on its surface, again increases, and so on. Hence it is evident that the pressure in the gas phase enclosed between the two modifications under consideration will not be constant, but will vary in the pressure interval from \(p'_{\infty}\) to \(p_{\infty}\) about a mean value \(\bar p\), at which the rate of decomposition of the unstable modification becomes equal to the rate of growth of the stable modification. This equilibrium pressure is determined by the point of intersection of the curves \(I\) and \(II\). The fluctuations of pressure about the value \(\bar p\) will be the smaller, the larger the absolute value of \(I\) at the given \(\bar p\). On the other hand, the more rapidly the rate of decomposition of the unstable modification increases with increasing undersaturation (curve \(III\) in Fig. 13), the more \(\bar p\) approaches \(p'_{\infty}\), i.e. the sublimation pressure of the unstable modification, and the more this case approaches crystallization from the melt. To this one may add that, owing to the pressure fluctuations about \(\bar p\), the rate of transformation will be greater than that corresponding to \(\bar p\), and lies between this latter rate and the maximum growth rate of the stable modification1 (when \(\bar p = p'_{\infty}\), we have the case of crystallization from the melt, and the rate of crystallization is equal to the maximum growth rate in the interval from \(p'_{\infty}\) to \(p_{\infty}\)).

The equation (31) derived by us is valid both for the growth rate of the stable modification

\[ I=\frac{C_1}{\ln^2 \frac{p}{p'_{\infty}}}\, e^{-\frac{C_2}{T}}\cdot e^{-\frac{C_3}{T^2 \ln \frac{p}{p_{\infty}}}}, \]

thus also for the rate of decay of the unstable modification

\[ I=-\frac{C'_1}{\ln^{2}\dfrac{p}{p'_{\infty}}}\, e^{-\frac{C'_2}{T}}\cdot e^{-\frac{C'_3}{T^{2}\ln \dfrac{p}{p'_{\infty}}}} . \]

Here the supercooling \(\Delta T\) has been replaced by the undersaturation, respectively the supersaturation, of the gaseous phase with respect to the corresponding modification.

When the absolute values of \(I\) at \(\overline{p}\) are sufficiently large, the fluctuations of \(p\) about the mean value \(\overline{p}\) are small, and in the equation given above one may substitute \(\overline{p}\) for \(p\). The value \(\overline{p}\) is found from the equation

\[ I=\mathrm{const}\, I'. \]

It follows from this equation that the amount of the unstable modification decomposing per unit time, in a stationary process, must be equal to the amount of the stable modification deposited per unit time. Substituting the value of \(\overline{p}\) in one of the equations given above, we can always obtain an expression for the rate of polymorphic transformations.

The case considered by us of infinitely large crystals is not in itself especially important. However, entirely analogous relations will be obtained for such a system of two modifications in which the stable modification is included partly or completely in the unstable modification, i.e., when the crystal of the unstable modification may be regarded as concave.

In the particular case when \(\overline{p}\simeq p'_{\infty}\), the expression for the rate of polymorphic transformation becomes identical with the expression for the rate of crystallization from the melt (31), respectively (31a). Such, in general, is the case when the unstable modification is homeopolar and its crystallites are finite and convex. In this case, on the surface of the unstable modification there are always building particles as strongly bound as in the position at the middle of the crystal \(^{39}\), and at a pressure less than \(p'_{\infty}\) (which, in essence, the gas pressure will always be in the presence of the stable modification), the decay of this modification occurs without any energy thresholds.

When the unstable modification is heteropolar, then even in the case of finite convex crystals its decay is associated with energy thresholds caused by the destruction of rows on the edges of the crystal \(^{40}\).

3. Rate of recrystallization

For completeness we shall say a few words also concerning recrystallization processes.

In principle, recrystallization is one of the cases of polymorphic transformations: differently deformed crystals may be regarded as different modifications. Since the free energy increases as the deformation increases, in the case of differently deformed crystals the principal process will be the growth of weakly deformed crystalline grains at the expense of strongly deformed ones. But this process is determined by the rate of formation of two-dimensional nuclei and is entirely analogous to the process of polymorphic transformations considered above.

It is possible that, in recrystallization, the formation from deformed crystals of such undeformed ones also plays a certain role; their appearance is connected with the formation of three-dimensional nuclei. This case was considered by Becker^41.

LITERATURE

1. I. N. Stranski u. R. Kaischew, Z. physik. Chem. (B) 26, 100, 1934; R. Kaischew u. I. N. Stranski, Z. physik. Chem., 26, 114, 1934; I. N. Stranski u. R. Kaischew, Physik. Z., 36, 393, 1935; abstract by I. N. Stranski at the jubilee Mendeleev Congress, read on 14/IX 1934 in Leningrad.

2. I. W. Gibbs, Thermodynamische Studien, Leipzig, 1892; I. Valeon, Ber. Sächs. Ges., 67, 1, 1915, where the little-accessible works of Gibbs are partially presented. Gibbs was the first to introduce the thermodynamic work of formation of a three-dimensional nucleus and, in addition, pointed out that crystal growth should be divided into two essentially different stages: the formation of a new lattice layer and its growth. The same conclusion was reached later by Volmer, Z. physik. Chem., 102, 270, 1922; K. Spangenberg, Z. Krist., 59, 383, 1923, 1924.

3. W. Kossel, Nachr. Gött. Ges., 1927, 135; Leipziger Vorträge, 1928, 1; P. Debye, Probleme der mod. Physik, Leipzig, 1928, the last article; Naturwiss., 18, 901, 1930; I. N. Stranski, Z. physik. Chem., 136, 259, 1928; (B) 11, 342, 1931; (B) 17, 127, 1932; Naturwiss., 19, 689, 1931; I. N. Stranski u. R. Kaischew, Z. Krist., 78, 373, 1931; I. N. Stranski u. L. Krastanow, Z. Krist., 83, 155, 1932; I. N. Stranski, R. Kaischew, u. L. Krastanow, Z. Krist., 88, 325, 1934; I. N. Stranski, Z. physik. Chem. (A) 142, 453, 1929; I. N. Stranski u. Kuleliew, Z. Physik. Chem. (A) 142, 467, 1929; I. N. Stranski u. Z. C. Mutafitschiew, Z. Elektrochem., 35, 393, 1929; Z. physik. Chem. (A) 150, 135, 1930; I. N. Stranski, Z. Elektrochem., 36, 25, 1930; Z. physik. Chem., Bodenstein-Festband, 230, 1931; R. Kaischew u. L. Krastanow, Z. physik. Chem. (B) 23, 158, 1933; I. N. Stranski, Yearbook of Sofia University, Phys.-Math. Faculty, 24, 2, 297, 1927–1928; 25, 2, 348, 1928–1929; 26, 2, 1929–1930; 29, 2, 1, 1932–1933.

4. I. N. Stranski, Z. physik. Chem. (B) 17, 127, 1932.

5. M. Volmer u. A. Weber, Z. physik. Chem., 119, 277, 1926; L. Farkas, Z. physik. Chem., 125, 236, 1927; M. Volmer, Z. Elektrochem., 35, 555, 1929; H. Brandes, Z. physik Chem., 126, 198, 1927; M. Volmer u. Marder, Z. physik. Chem. (A) 151, 97, 1931; H. Brandes u. M. Volmer, Z. physik. Chem. (A) 155, 466, 1931; M. Volmer u. W. Schultze, Z. physik. Chem. (A) 156, 1, 1931; M. Volmer u. H. Flood, Z. physik. Chem. (A) 170, 273, 1934; H. Flood, Z. physik. Chem. (A) 170, 286, 1934; I. N. Stranski u. D. Totomanow, Naturwiss., 20, 905, 1932; Z. physik. Chem. (A) 163, 399, 1933.

1. I. N. Stranski, Z. physik. Chem. (B) 17, 127, 1932, p. 153.

2. This opinion was expressed especially sharply by Kossel (Kossel, Ann. d. Physik, 21, 475, 1934). See also the discussion on this subject between Volmer (M. Volmer, Ann. d. Physik, 23, 44, 49, 1935) and Kossel (Kossel, ibid., p. 47, 50).

3. I. N. Stranski u. R. Kaischew, Z. physik. Chem. (B) 26, 100, 1934; R. Kaischew u. I. N. Stranski, Z. physik. Chem. (B) 26, 114, 1934; I. N. Stranski u. R. Kaischew, Z. physik. Chem. (B) 26, 312, 1934; R. Kaischew u. I. N. Stranski (B) 26, 317, 1934; (A) 170, 295, 1934; Ann. d. Physik 23, 330, 1935.

4. H. Brandes, Z. physik. Chem., 126, 198, 1927.

5. I. N. Stranski u. R. Kaischew, Z. Krist., 73, 380, 1931.

6. I. N. Stranski, Z. physik. Chem. (B) 17, 127, 1932; there is given an explanation of the experiments of Spangenberg (K. Spangenberger, Neues Jahrb. d. Mineral., Mögge-Festband A 57, 1197, 1928) and Neuhaus (Neuhaus, Z. Krist., 68, 15, 1929) on the growth of spherically polished NaCl crystals.

7. W. Kossel, Nachr. Götting. Ges., 1927, 135; I. N. Stranski and Ezerowich, Sofia Univ. Annual, Phys.-Math. Faculty, 24, 2, 297, 1927—1928; I. N. Stranski, Z. physik. Chem., 136, 259, 1928.

8. I. N. Stranski u. R. Kaischew, Physik. Z., 36, 398, 1935; Ann. d. Physik, 23, 330, 1935.

9. R. Kaischew, Z. Physik, 102, 684, 1936.

10. O. Stern, Z. Elektrochem., 25, 66, 1919.

11. R. Kaischew, Ann. d. Physik, 30, 184, 1937.

12. R. Kaischew u. I. N. Stranski, Z. physik. Chem. (B) 26, 114, 1934; I. N. Stranski u. R. Kaischew, Physik. Z., 36, 395, 1935; Ann. d. Physik (5) 23, 330, 1935; R. Kaischew, Z. Physik, 102, 684, 1936.

13. R. Bloch, Th. Brings u. W. Kuhn, Z. physik. Chem. (B) 12, 415, 1931.

14. M. Volmer u. O. Schmidt, Z. physik. Chem. (B) 35, 467, 1937.

15. I. N. Stranski, Ber. Wien Akad., 145, 840, 1936.

16. I. N. Stranski u. R. Kaischew, Z. physik. Chem. (B) 26, 108, 1934.

17. I. N. Stranski u. R. Kaischew, Z. physik. Chem. (B) 26, 105, 1934; Physik. Z., 36, 395, 1935.

18. R. Kaischew u. I. N. Stranski, Z. physik. Chem. (B) 26, 320, 1934; Physik. Z., 36, 400, 1935.

19. L. Farkas, Z. physik. Chem., 125, 236, 1927; R. Kaischew u. I. N. Stranski, Z. physik. Chem. (B) 25, 317, 1934.

20. The same result can also be obtained by another route; R. Becker u. W. Döring, Ann. d. Physik (5) 24, 719, 1935.

21. P. Othmer, Z. anorg. Chem., 91, 226, 1915.

22. R. Kaischew u. I. N. Stranski, Z. physik. Chem. (B) 26, 317, 1934; (A) 170, 295, 1934; I. N. Stranski u. R. Kaischew, Physik. Z., 36, 393, 1935; see also the abstract at the Mendeleev Congress in Leningrad, 1934.

23. I. W. Gibbs, Thermodynamische Studien, Leipzig, 1892.

24. M. Volmer u. A. Weber, Z. physik. Chem., 119, 277, 1926; M. Volmer, Z. Elektrochem., 35, 555, 1929; in particular, on the formation of two-dimensional nuclei see H. Brandes, Z. physik. Chem., 126, 198, 1927; M. Volmer u. M. Marder, Z. physik. Chem. (A) 154, 97, 1931; H. Brandes u. M. Volmer, Z. physik. Chem. (A) 155, 466, 1931; M. Volmer u. W. Schultze, Z. physik. Chem. (A) 156, 1, 1931.

25. M. Volmer, Z. Elektrochem., 35, 555, 1929; W. Bayerl u. H. Flood, Naturwiss., 21, 273, 1933; M. Volmer u. H. Flood, Z. physik. Chem. (A) 170, 273, 1934; H. Flood, Z. physik. Chem. (A) 170, 286, 1934.

1. I. N. Stranski and D. Totomanow, Naturwiss., 20, 905, 1932; Z. physik. Chem. (A) 163, 399, 1933.

2. L. Farkas, Z. physik. Chem., 125, 236, 1927.

3. Volmer, Z. Elektrochem., 35, 557, 1929.

4. I. N. Stranski and R. Kaischew, Z. physik. Chem. (B) 26, 100, 1934; R. Kaischew and I. N. Stranski, Z. physik. Chem. (B) 26, 114, 1934; I. N. Stranski and R. Kaischew, Z. physik. Chem. (B) 26, 312, 1934; Physik. Z., 36, 393, 1934.

5. R. Kaischew and I. N. Stranski, Z. physik. Chem. (B) 26, 114, 1934; I. N. Stranski and Kaischew, Physik. Z., 36, 395, 1935.

6. I. N. Stranski and D. Totomanow, Naturwiss., 20, 905, 1932; Z. physik. Chem. (A) 163, 399, 1933.

7. Volmer and Marder, Z. physik. Chem. (A), 154, 97, 1931.

8. H. Brandes, Z. physik. Chem., 125, 198, 1927.

9. I. N. Stranski, Z. physik. Chem. (B) 11, 342, 1931; I. N. Stranski and R. Kaischew, Z. Krist., 78, 373, 1931.

10. I. N. Stranski and K. Kaischew, Z. physik. Chem. (B) 26, 108, 1934.

11. R. Becker, Z. Techn. Phys., 7, 547, 1926.

1. We could eliminate the pressure fluctuations about \(\bar p\) by taking, for example, a sufficiently large amount of the corresponding phase. ↩↩↩↩

2. Advances in Physical Sciences, vol. XXI, issue 4. ↩