# Kris's Research Notes

## September 8, 2011

### Nanowires, part 3

Filed under: Nanowire — Kris Reyes @ 6:56 pm

In this note, we describe the role of energy parameters in our model of nanowire growth. We give conditions on the energies to encourage growth (rather than etching) and explain how energies determine how fast atom diffuses through the liquid droplet onto the interface. Simulations are done on a hexagonal lattice.

### Hexagonal Lattice

In these simulations, a hexagonal lattice is used so that each atom has six nearest neighbors. Here is the resulting (ensemble averaged) Wulff shape obtained from annealing a sphere of material with bond strength $\gamma = 0.5$ eV at 600K:

### Nanowire growth example

The simulations grow nanowires by VLS. A droplet of material B is initially on a flat substrate of material A. The domain is 256 atoms wide, the droplet is a hemisphere of radius 16 atom. Material A is then deposited uniformly on the Vapor-Solid and Vapor-Liquid interface. The deposition and desorption rates of material A are set so that the solid and vapor phases are in equilibrium at 600K. The liquid droplet does not evaporate, hence is always at constant volume. Recall the relevant energy parameters:

• $\gamma_{AA}, \gamma_{BB}$ – homogeneous bond strengths;
• $\gamma_{AB}$ – the bond strength between A and B;
• $\gamma_{HH}$ – bond strength between intermediate species H, a penalty energy for atom-atom exchanges;
• $\lambda_D$ – barrier for diffusion of an A atom through liquid B;
• $\lambda_S$ – the barrier for incorporation of an A atom on the surface of a B droplet into the bulk of the droplet.

We consider the following values of the above parameters:

$\gamma_{AA} = 0.50$ eV, $\gamma_{BB} = 0.40$ eV, $\latex \gamma_{AB} = 0.35$ eV;

$\gamma_{HH} = 0.55$ eV $\lambda_{D} = \lambda_S = 0.70$ eV.

Material A was deposited at a rate of 1 monolayer/second for over 400 seconds at 600K. Here is the result (B is red, substrate A is green, deposited A is aqua):

We observe several things. First, there is a large amount of mixing within the A crystal. Indeed, if we color the deposited A atoms differently depending on when they were deposited, we see the fast mixing more explicitly:

It mixes really fast. This may not be desirable.

Second, we note that there are no A atoms within the bulk of the B droplet. Indeed, if we calculate the amount of A atoms within the droplet bulk at periodic intervals, we do not see more than one A atom inside at a time. We can address this problem by further understanding our model.

### The Model

As stated above, material A is deposited uniformly on the interface between the system and vacuum. Because deposition and desorption rate is fixed so that there is an equilibrium between solid and vapor phases, an A atom that is deposited onto the solid phase diffuses for some short amount of time until it desorbs. Hence, we may assume those A atoms that incorporate into the nanowire do so via the droplet rather than diffusing along the substrate surface (although, this may still be a significant mechanism during the beginning of nanowire growth — however as the surface area of the wire increases, we assume that most deposited A atoms desorb before eventually reaching the droplet).

Therefore, we must study how an A atom incorporates into, then diffuses through a liquid B droplet. There are three main areas an A atom may occupy within a liquid B droplet: on the surface, in the bulk and at the solid-liquid interface:

As illustrated above, in the first case, the energy about the A atom is $2 \gamma_{AB}$, in the second case, it is $6\gamma_{AB}$ and in the third case, it is $4\gamma_{AB} + 2\gamma_{AA}$. It is clear that if

$0 < \gamma_{AB} < \gamma_{AA}$,

then an A atom on the surface is less energetically favorable than it in the bulk, which is in turn less favorable than it on the interface.

While the energetics is clear, we must discuss the rates of the relevant events. This is a little more complicated. In order for an A atom to reach the solid-liquid interface via the liquid droplet, the droplet must pass from the surface, into the bulk and the onto the interface. The rates of these events depend on different parameters. In passing from the surface to the bulk, we must consider $\lambda_{S}$ (which dictates directly the barrier of this event) vs. $2\gamma_{AB}$, which is the cost for diffusing on the surface rather than passing into the droplet. If the A atom incorporates into the bulk, then only $\lambda_D$ controls how fast it diffuses within this region. Once it gets near the solid-liquid interface, bond strengths $\gamma_{AA}, \gamma_{BB}, \gamma_{AB}$ and $\gamma_{HH}$ once again come into effect. We examine these processes more carefully.

#### Transition between surface and bulk regions

An A atom on the surface incorporates into the bulk with barrier $\lambda_S$ or diffuses on the surface with barrier $2\gamma_{AB}$. An A atom at the “surface-bulk” interface goes back onto the surface with barrier $\lambda_S$ and enters the bulk-proper with barrier $\lambda_D$. We may write

$\lambda_{S} = 2\gamma_{AB} + \epsilon_S$,

where $\epsilon_S$ determines how readily an atom incorporates into the droplet. If $\epsilon_S > 0$, then an atom is more likely to diffuse on the surface, while if $\epsilon_S < 0$, it will probably diffuse into the droplet.

Similarly, we write

$\lambda_{D} = \lambda_S + \epsilon_D$.

If $\epsilon_D > 0$, then diffusion through the droplet (once in the bulk) is slow, including for atoms at the surface-bulk interface. In the extreme case, large $\epsilon_D$ will prevent atoms from even entering the droplet bulk.

#### Transition between bulk and solid/liquid interface

Consider the nucleation of an A atom from the droplet bulk onto the liquid solid interface (which we denote by the transition of states $X \rightarrow X\wedge Y \rightarrow Y$, and indicate the intermediate state as $X\wedge Y$):

B is colored red, A is green and the intermediate species H is blue.

Here bond strength determine the rate of this transition, and its energy barrier is given by

$E_X - E_{X\wedge Y} = 3\gamma_{AB} - \gamma_{AA} - \gamma_{AB} - \gamma_{HH}$,

whereas the reverse transition has barrier

$E_Y - E_{X\wedge Y} = -\gamma_{AB} + \gamma_{AA} + \gamma_{BB} - \gamma_{HH}$,

We can control how fast the interface grows by comparing these two barriers. If $E_X - E_{X\wedge Y}$ < $E_Y - E_{X\wedge Y}$, then the rate of attachment onto the interface is faster than detachment from the interface. This is a desirable property (and is in fact the opposite of what we want if we want to encourage droplet etching). This inequality yields

$\gamma_{AB} < \frac{1}{2} (\gamma_{AA} + \gamma_{BB})$.

This is another upper bound of $\gamma_{AB}$, and hence suggests further that $\gamma_{AB}$ cannot be too big.

Consider once again the state $X$ above. The A atom could alternatively diffuse away from the surface, with energy barrier $\lambda_D$. Comparing the two barriers, we can write:

$3\gamma_{AB} - \gamma_{AA} - \gamma_{BB} - \gamma_{HH} = \lambda_D + \epsilon_I$.

If $\epsilon_I > 0$, we get slow nucleation as atoms will diffuse close to the interface, but will have to pay an extra $\epsilon_I$ in energy to attach to it.

#### Summary

In short, we have the following relationships between energy parameters:

$\gamma_{AB} < \gamma_{AA}$;

$\gamma_{AB} < \frac{1}{2}(\gamma_{AA} + \gamma_{BB})$;

$\lambda_S = 2\gamma_{AB} + \epsilon_S$;

$\lambda_D = \lambda_S + \epsilon_D$;

$3\gamma_{AB} - \gamma_{AA} - \gamma_{BB} - \gamma_{HH} = \lambda_D + \epsilon_I$.

In later posts, we shall study the effect parameters $\epsilon_I, \epsilon_D, \epsilon_S$ on the simulations.