Diffusion dynamics and concentration of toxic materials from quantum dots-based nanotechnologies: an agent-based modeling simulation framework

Abstract

Due to their favorable electrical and optical properties, quantum dots (QDs) nanostructures have found numerous applications including nanomedicine and photovoltaic cells. However, increased future production, use, and disposal of engineered QD products also raise concerns about their potential environmental impacts. The objective of this work is to establish a modeling framework for predicting the diffusion dynamics and concentration of toxic materials released from Trioctylphosphine oxide-capped CdSe. To this end, an agent-based model simulation with reaction kinetics and Brownian motion dynamics was developed. Reaction kinetics is used to model the stability of surface capping agent particularly due to oxidation process. The diffusion of toxic Cd²⁺ ions in aquatic environment was simulated using an adapted Brownian motion algorithm. A calibrated parameter to reflect sensitivity to reaction rate is proposed. The model output demonstrates the stochastic spatial distribution of toxic Cd²⁺ ions under different values of proxy environmental factor parameters. With the only chemistry considered was oxidation, the simulation was able to replicate Cd²⁺ ion release from Thiol-capped QDs in aerated water. The agent-based method is the first to be developed in the QDs application domain. It adds both simplicity of the solubility and rate of release of Cd²⁺ ions and complexity of tracking of individual atoms of Cd at the same time.

Appendix: Parameter specifications

Total mass of CdSe

For a quantum dot with a diameter of 6.5 nm (whole diameter of 8 nm), the volume should be = 143.72 nm³.

The weight of the each CdSe quantum dot is m = V × ρ = 143.72 nm³ × 5.816 g/cm³ = 835.88 × 10⁻²¹ g.

For a 1 cm² area with a quantum layer of 150 nm, the quantity of the close-packed quantum dots should be (1 cm/8 nm) × (1 cm/8 nm) × (150 nm/8 nm) = 2.81 × 10¹³.

So, the whole weight of the quantum dots needed for a 1 cm² area with a quantum layer of 150 nm is

$$m \, = { 2}. 8 1 \times 10^{ 1 3} \times { 835}. 8 8 \times 10^{ - 2 1} {\text{g }} = 2.35 \times 10^{ - 5} {\text{g}} .$$

No. of ions released by a fully oxidized CdSe quantum dot particle

For a CdSe quantum dot with a diameter of 6.5 nm

$$m = V \times \rho = 143.72 \, {\rm{nm}}^{3} \times 5.816 \, {\rm{g}}/\!{\rm{cm}}^{3} = 835.88 \times 10^{-21} \, {\rm{g}}$$

$${\rm{So}}, \; M = 835.88 \times 10^{-21} \, {\rm{g}}/191.37 \, {\rm{g}}/{\rm{mol}} = 4.37 \times 10^{-21} \, {\rm{mol}}$$

So, each CdSe quantum dot contains the quantity of the CdSe molecules

$$4. 3 7\times 10^{ - 2 1} \times { 6}.0 2 2\times 10^{ 2 3} = { 2632}.$$

That is, each CdSe quantum dot with a diameter of 6.5 nm finally will produce 2632 Cd2 + by the free oxidation.

Diffusion coefficients

The diffusion coefficients for particles Cd²⁺ and O₂ will be derived under the following conditions:

T = 293.15 K, b = 6, η_B = 0.001003 kg/ms,

r ₀ (CdSe capped with TOPO) = 4.523 × 10⁻⁹ m,

r ₀ (CdSe) = 3.25 × 10⁻⁹ m,

r ₀ (Cd²⁺) = 4.85 × 10⁻¹¹ m,

r ₀ (O₂) = 1.20  × 10⁻¹⁰ m $\Rightarrow$ D _CdSe = 6.5841 × 10⁻¹¹ m² s⁻¹,

D _{CdSe with TOPO} = 4.731 × 10⁻¹¹ m² s⁻¹,

D _Cd2+ = 4.412 × 10⁻⁹ m² s⁻¹,

D _O2 = 1.814 × 10⁻⁹ m² s⁻¹.

Assuming that the only diffusing particles in our systems are oxygen and Cd²⁺, we do not need to calculate diffusion coefficient and movement probabilities for other particles. However, since CdSe with capping agent and CdSe particles are considered in the model, their sizes are studied for determining the lattice discretization length $(\Delta L)$, which is the size of the largest particle in the system.

Based on Eq. 2, in order to derive movement probability for each particle, first movement probability value of one (T = 1) is assigned to the fastest particle, since no particle will be diffusing faster than this particle. Second, $\Delta L$ will be length of the cubic lattice’s side, which is assumed to be 9.046 × 10⁻⁹ m since it is the size of the largest particle in our system (i.e., CdSe particle capped with TOPO). Third, by inputting the diffusion coefficient of the fastest particle (i.e., Cd²⁺ ion), the equation will be solved in terms of $\Delta t$, which is going to be the fixed time-step in the model. Finally, by knowing $\Delta L$, $\Delta t$, and the diffusion coefficient (D) for each particle, movement probability will be derived for each individual particle (i.e., the agent in the agent-based model). In the following model, particles of the same type are considered to have the same size and same diffusion coefficient, which denotes to have same movement probability in that state.

$\Delta L$ = Size of the Largest Particle = Size of CdSe with TOPO = 9.046 × 10⁻⁹ m

T = 1.0 for Cd²⁺ which is our fastest particle (due to having the largest diffusion coefficient)

D = 4.4120 × 10⁻⁹ m²s⁻¹ for Cd²⁺

$$=> \Delta t = \frac{T}{1} \frac{{\left( {\Delta L} \right)^{2} }}{D} = \frac{{1 \times (9.046 \times 10^{ - 9} )^{2} }}{{4.4120 \times 10^{ - 9} }} = 1.8547 \times 10^{ - 8} {\text{s,}}$$

$$=> T_{{O_{2} }} = \frac{D \times \Delta t}{{(\Delta L)^{2} }} = \frac{{1.814 \times 10^{ - 9} \times 1.8547 \times 10^{ - 8} }}{{(9.046 \times 10^{ - 9} )^{2} }} = 0.404.$$

The volume of quasi 3-D cubic lattice

The simulated quasi 3-D cubic lattice structure has a dimension with length, width, and height of ∆X × ∆L, ∆Y × ∆L, and 1 × ∆L, respectively. We know the fixed value for discretization lattice = ∆L = ($9.046 \times 10^{ - 9} m)$. We assume same length and width ($\Delta X = \Delta Y$), which is determined by matching the concentration of the Cd²⁺ resulting from the simulation with that of Derfus et al. research data (Derfus et al. 2004). It can be established that

$$\begin{aligned} {\text{Cd}}^{ 2+ } {\text{Concentration }}\left( {\text{ppm}} \right) \, & = \, {{\left( {{\text{Cd}}^{ 2+ } {\text{total mass}}} \right)} \mathord{\left/ {\vphantom {{\left( {{\text{Cd}}^{ 2+ } {\text{total mass}}} \right)} {\left( {\text{solution volume}} \right)}}} \right. \kern-0pt} {\left( {\text{solution volume}} \right)}} \\ 0.000056 & = \frac{{1 \times 1.866 \times 10^{ - 19} }}{{1000 \times (\Delta X \times 9.046 \times 10^{ - 9} ) \times (\Delta X \times 9.046 \times 10^{ - 9} ) \times (1 \times 9.046 \times 10^{ - 9} )}} \\ \end{aligned}$$

$$=> \Delta X = 108848.$$

The volume of quasi 3-D cubic lattice structure = (∆X × ∆L). (∆Y × ∆L).(1 × ∆L)

= (108,848 × 9.046.10⁻⁹) (108,848 × 9.046.10⁻⁹) (1 × 9.046.10⁻⁹) = 8.77 × 10⁻¹⁵ m³.

The number of dissolved O₂

The number of dissolved O₂ is calculated based on the number of O₂ in water at 20 C, 1 atm, which is about 1.7 × 10²⁰/L, or about 1.35 × 10⁹ O₂ in the simulated value of 8.77 × 10⁻¹⁵ m³.