# Kris's Research Notes

## March 21, 2012

### Nucleation Regime

Filed under: GaAs Simulations — Kris Reyes @ 6:58 pm

In this note we discuss the nucleation process, one of the three key processes involved in droplet epitaxy. We derive an expression for the critical nucleus radius and compare it with the size of a liquid droplet at fixed growth conditions. We shall see that this defines a regime in which nucleation is impossible in equilibrium. This regime agrees well with the morphological phase diagram obtained through simulations.

## Model for Critical Radius

Consider a circular domain of radius $R$ and a circular crystal nucleus of radius $r_0$ centered at the origin:

We consider the concentration $c(r, \theta) = c(r)$ of As in the system as well as the persistence of the nucleus under a quasi-static deposition of As flux at a rate of $F_{As}$ monolayers/second. In the interior of the domain, the system satisfies Laplace’s equation:

$\displaystyle r c_{rr} - c_r = 0,$,

which has solutions of the form $c(r) = A \log r + B$. At the boundary $r = R$, we have

$\displaystyle \ell D c_r(R) = F_{As}$,

for diffusion coefficient $D$ and atomic spacing $\ell$. This implies $A = \frac{F_{As}R}{\ell D}$.

We study the growth of the nucleus by examining the sign of $c_{r} = c^+_r + c^{-}_r$, where $c^+_r$ is the contribution of As atoms nucleating from the liquid domain and $c^-_r$ desribes the detachment of As atoms into the liquid domain as governed by the Gibbs-Thomson equation. Specifically:

$\displaystyle c^+_r(r_0) = \frac{d}{dr}\left[A\log r + B\right] = \frac{F_{As}R}{\ell D} \frac{1}{r_0}$,

$\displaystyle c^-_r(r_0) = \frac{d}{dr}\left[ c_0 \exp \left[\frac{\gamma}{k_B T r_0}\right]\right] = -\frac{c_0\gamma}{k_B} \frac{1}{T}\frac{1}{r^2_0}\exp \left[\frac{\gamma}{k_B T r_0}\right]$,

where $k_B$ is Boltzmann’s constant. Hence, the net change in concentration at the interface is given by:

$\displaystyle c_r(r_0) = c^+_r + c^-_r = \frac{1}{r_0}\left(\frac{F_{As}R}{\ell D} - \frac{c_0\gamma}{k_B} \frac{1}{T} \frac{1}{r_0} \exp \left[\frac{\gamma}{k_B T r_0}\right]\right)$.

From this, we may define

$\displaystyle \phi(F_{As}, T, r) = \frac{F_{As}R}{\ell D} - \frac{c_0\gamma}{k_B} \frac{1}{T} \frac{1}{r} \exp \left[\frac{\gamma}{k_B T r}\right].$

From above it is clear that when $\phi(F_{As}, T, r) > 0$, a nucleus of radius $r$ will grow at the specified growth conditions (i.e. $c_{r}(r_0) > 0)$, whereas $\phi(F_{As}, T, r) < 0$ implies the nucleus will shrink. The radius $r_c = r_c(F_{As}, T)$ for which $\phi(F_{As},T, r_c) = 0$ is called the critical radius, and is an equilibrium solution to the above problem. Moreover, the derivative

$\displaystyle \frac{\partial \phi}{\partial r} = \frac{c_0\gamma}{k_B T}\exp\left[\frac{\gamma}{k_BTr}\right]\left(\frac{\gamma}{k_B T}\frac{1}{r} + 1\right) > 0$,

which implies the critical radius $r_c$ is an unstable equlibrium — once $r > r_c$, the nucleus will grow indefinitely (and conversely). That is,

$\phi(F, T, r) > 0 \Leftrightarrow r > r_c$.

## Effect of $F_{As}, T$

We study the effect of experimental parameters $F_{As}, T$ by considering the size of the liquid droplet as a function of temperatures. If the radius of the droplet at a particular $T$ is smaller than that of the critical radius $r_c(F_{As}, T)$, then nucleation cannot occur.

As a first step, we consider $R, D$ in the above expression for $\phi$. Here, $R$ is the radius of the domain — the radius of the liquid droplet which we derive momentarily. The diffusion coefficient $D$ describes the diffusivity of As through the liquid droplet, and may be written in Arrhenius form

$\displaystyle D = R_0 \exp\left[-\frac{E_A}{k_B T}\right]$,

where $R_0$ is the diffusion prefactor (roughly $R_0 \sim 10^{13}$).

Recall, we had established empirically the relation $r \propto \left(\frac{D_G}{F_G}\right)^{\alpha}$ for some $\alpha \leq 1/2$. Here the diffusion coefficient $D_G = R_0 \exp\left[-E_G/k_B T\right]$, for some activation energy $E_G$. Hence, we may replace $r, R$ with the expression

$\displaystyle r, R = \frac{r_0 R_0^\alpha}{F_G^\alpha} \exp\left[-\alpha E_G/k_B T\right]$,

where $r_0$ is an appropriately chosen constant of proportionality. These remarks, coupled with those in the above paragraph, enables us to write $\phi(F_{As}, t, r(T))$ in terms of $F_{As}, T$ exclusively:

$\displaystyle \phi(F_{As}, T, r(T)) = \frac{r_0 R_0^{\alpha-1}}{\ell F_G^\alpha} F_{As}\exp\left[-(\alpha E_G-E_A)/k_B T\right] - c_0\frac{\gamma F_G^\alpha}{k_B r_0 R_0^\alpha} \frac{1}{T} \exp\left[\alpha E_G/k_B T\right] \exp \left[\frac{\gamma}{k_B T}\frac{F_G^\alpha}{r_0 R_0^\alpha} \exp\left[\alpha E_G/k_B T\right]\right].$

After some bookkeepping, we may write in a slighly simpler form:

$\displaystyle \phi(F_{As}, T, r(T)) = C_1 F_{As}\exp\left[-\frac{C_2}{T}\right] - c_0 \frac{C_3}{T}\exp\left[\frac{C_4}{T}\right] \exp \left[\frac{C_3}{T} \exp\left[\frac{C_4}{T} \right]\right].$

where

$\displaystyle C_1 = \frac{r_0 R_0^{\alpha-1}}{\ell F_G^\alpha}$

$\displaystyle C_2 = (\alpha E_G-E_A)/k_B$

$\displaystyle C_3 = \frac{\gamma F_G^\alpha}{k_B r_0 R_0^\alpha}$

$\displaystyle C_4 = \alpha E_G/k_B$

Nucleation is possible when $\phi(F_{As}, T, r(T)) > 0$, i.e. when

$\displaystyle C_1 F_{As}\exp\left[-\frac{C_2}{T}\right] > c_0 \frac{C_3}{T}\exp\left[\frac{C_4}{T}\right] \exp \left[\frac{C_3}{T} \exp\left[\frac{C_4}{T} \right]\right].$

Taking logs yields a necessary condition for nucleation with respect to $\log F_{As}$ and $\frac{1}{T}$:

$\displaystyle \log F_{As} > \log \frac{1}{T} + (C_2 + C_4)\frac{1}{T} + C_3\frac{1}{T} \exp\left[\frac{C_4}{T} \right] + C,$

for some appropriate choice of constant $C$.

The shape of the resulting critical curve in the $\left(\frac{1}{T}, \log F_{As}\right)$ plane depends heavily on the sign and magnitudes of the coefficients $C_2, C_3$ and $C_4$. For example, if $C_2 + C_4$ is large relative to $C_3$, then the linear term dominates and the resulting curve is a straight line. However, if $C_3$ is large then the curve closely resembles the graph of $y = xe^x$. For the simulations, we set:

$\alpha \sim \frac{1}{4}$,

$E_G \sim 0.9,$ eV

$E_A \sim 0.7,$ eV

$\gamma \sim 0.1,$ eV

$F_G \sim 0.1,$ monolayers/second

$R_0 \sim 10^{13}.$ monolayers/second

Setting $r_0 = 2.7$, we obtain a reasonable maximum droplet radius of 128 atoms over all temperatures.

From the above values we may calculate the coefficients:

$\displaystyle C_2 + C_4 = \frac{2\alpha E_G - E_A}{k_B} \approx -2907$,

$\displaystyle C_3 = \frac{\gamma F_G^\alpha}{k_B r_0 R_0^\alpha} \approx 0.1362$,

That is, $C_2 + C_4$ is four orders of magnitude larger that $C_3$, and so the critical curve is essentially a downward sloping line. Points above this line correspond to parameters where the droplet radius is larger than the critical radius for nucleation, whereas the region below the curve defines a regime where nucleation is impossible.

### Effective As Diffusivity

In the above, we describe the diffusivity of As through liquid Ga by the activation energy $E_A = 0.7$. This means the flux of As atoms impinging on the nucleus is given by

$c_r(r_0) \propto \frac{F_{As} R}{D} e^{\frac{E_A}{kT}} \frac{1}{r_0} = F_{As} e^{-\frac{\alpha E_G - E_A}{kT}}$

In reality, our domain is not as above. Instead As atoms impinge on the nucleus faster than what is described above because nucleation occurs on the surface. As a first past approximation to this, we consider an effective As diffusivity by increasing $E_A$ thereby increasing the rate $c_r$ above.

This results in a more negative coefficient $C_2 + C_4$ to the linear term $\frac{1}{T}$ above, resulting in a steeper curve. For example, if we set $E_A = 1.7$ eV, we get $C_2 + C_4 = -14535$, a much steeper curve than above. Here are the contours of $\phi$ for this case:

Compare this to the phase diagram we obtained from simulations: