# Introduction to Inorganic Chemistry/Basic Science of Nanomaterials

## Contents

- 1 Basic Science of Nanomaterials
- 2 11.1 Physics and length scales: cavity laser, Coulomb blockade, nanoscale magnets
- 3 11.2 Semiconductor quantum dots
- 4 11.3 Synthesis of semiconductor nanocrystals
- 5 11.4 Surface energy
- 6 11.5 Nanoscale metal particles
- 7 11.6 Discussion questions
- 8 11.7 Problems
- 9 11.8 References

## Basic Science of Nanomaterials[edit]

## 11.1 Physics and length scales: cavity laser, Coulomb blockade, nanoscale magnets[edit]

## 11.2 Semiconductor quantum dots[edit]

## 11.3 Synthesis of semiconductor nanocrystals[edit]

## 11.4 Surface energy[edit]

## 11.5 Nanoscale metal particles[edit]

## 11.6 Discussion questions[edit]

## 11.7 Problems[edit]

1. Consider a spherical gold nanoparticle that is 3 nm in diameter. If the diameter of an atom is approximately 3 Å, how many atoms are on the surface of the particle? What fraction of the gold atoms in the particle are on the surface?

2. Now consider a 3 nm diameter droplet of mercury. Mercury atoms are also about 3 Å in diameter. The heat of vaporization of bulk mercury is 64.0 kJ/mol, and the vapor pressure of mercury is 0.00185 torr = 2.43 x 10^{-6} atm. The surface tension of mercury (γ_{Hg}) is 0.518 N/m, and the surface excess energy can be calculated as γ_{Hg}A, where A is the surface area. Using this information and the Clausius-Clapyron equation (P = const•exp(-ΔH_{vap}/RT)), calculate the vapor pressure of 3 nm droplets of mercury.

3. James Heath and coworkers (Phys. Rev. Lett. 1995, 75, 3466) have observed Ostwald ripening in thin films of gold nanoparticles at room temperature. Starting with an uneven distribution of particle sizes, they find that the large particles grow at the expense of smaller ones. Can you explain this observation, based on your answers to problems (1) and (2)?

4. The bandgap of bulk silicon is 1.1 eV. What bandgap would you expect for a 3 nm diameter Si nanocrystal? Use the Brus formula,

where d is the particle diameter, ε_{Si} = 12, h = 6.6 10^{-34} J s, 1 eV = 1.6 10^{-19} J, 1 J = 1 kg m^{2}/s^{2}. and e^{2}/4πε_{0} = 1.44 10^{-9} eV m. Assume that the electron-hole reduced mass μ is approximately equivalent to the electron mass, m_{e} = 9.1 10^{-31} kg.