Chapter 11, Problem 11.157QP

Nanotechnology, or technology utilizing 1–100 nm sized particles, has rapidly expanded in the past few decades, with potential applications ranging across far-reaching fields such as electronics, medicine, biomaterials, and consumer products, to name a few. One of the primary advantages of nanoparticles is the presence of large surface/mass ratios, resulting in enhanced surface activities compared to bulk materials.

1. a Use the density of silver (10.49 g/cm³) to determine the number of Ag atoms in a spherical 20.-nm silver particle.
2. b In the crystalline metallic environment, the measured radii of silver atoms has been measured to be 144 pm. Use this to calculate the atomic packing fraction of a 20.-nm silver particle. In other words, calculate the ratio of the volume taken up by Ag atoms to the volume of the entire nanoparticle.
3. c Based on the result of part (b), silver conforms to which type of cubic crystal lattice?
   1. A simple cubic
   2. B body-centered cubic
   3. C face-centered cubic
4. d A cubic Ag ingot having a mass of 5.0-g is processed to form a batch of 20.-nm Ag nanoparticles. Calculate the ratio of the surface area provided by the batch of nanoparticles to the surface area of the initial cube of Ag.

Interpretation Introduction

Interpretation:

Number of $\text{Ag}$ atom in spherical Silver particle ( $\text{20 nm}$) has to be calculated by using density of silver.

Answer to Problem 11.157QP

Number of $\text{Ag}$ atom in spherical Silver particle is $\text{2}{\text{.5×10}}^{\text{5}}\text{Ag atom}$

Explanation of Solution

Density of silver atom is $\text{10}{\text{.49g/cm}}^{\text{3}}$

$\begin{array}{l}\begin{array}{l}\text{It is convenient to complete this problem using cm instead }\\ \text{of nm units:}\end{array}\hfill \\ \hfill \\ \begin{array}{l}\text{20}\text{. nm}\times \left({\text{10}}^{\text{2}}{\text{cm/10}}^{\text{9}}\text{nm}\right)\text{ = 2}\text{.0}\times {\text{10}}^{\text{-6}}\text{cm diameter }\\ \text{= 1}\text{.0}\times {\text{10}}^{\text{-6}}\text{cm radius}\end{array}\hfill \\ \hfill \\ \hfill \end{array}$

Using the formula for volume of the sphere, $\text{V=}\frac{\text{4}}{\text{3}}\text{πr3}$, the volume of 20-nm spherical

Silver particle is

$\frac{4}{3}\pi {(1.0{\text{ x 10}}^{-6}\text{ cm)}}^3=4.19{\text{ x 10}}^{-18}{\text{ cm}}^3$

This value is multiplied by density of Silver it gives the mass of Silver nanoparticle:

$\text{volume x density = }(4.19{\text{ x 10}}^{-18}{\text{ cm}}^3)(\frac{10.49\text{ g}}{{\text{cm}}^3})=4.39{\text{ x 10}}^{-17}\text{ g}$

On average, each Silver atom has a mass of $\text{107}\text{.9 amu}$.  This is equal to $\text{1}{\text{.79×10}}^{\text{-22}}\text{g}$. By using this values the total number of Silver atom in nanoparticle is calculated.

$(\frac{4.39{\text{ x 10}}^{-17}\text{g}}{1\text{ nanoparticle}})(\frac{1\text{ Ag atom}}{1.79{\text{ x 10}}^{-22}\text{g}})=\frac{2.5{\text{ x 10}}^5\text{ Ag atoms}}{\text{nanoparticle}}$

Interpretation Introduction

Interpretation:

Ratio of the volume occupied by Silver atom to the volume of total nanoparticle has to be calculated by using density of silver.

Answer to Problem 11.157QP

Ratio of the volume occupied by Silver atom to the volume of total nanoparticle is $\text{3}{\text{.13×10}}^{\text{-18}}{\text{cm}}^{\text{3}}$

Explanation of Solution

The volume of Silver atoms in the nanoparticle is determined by multiplying the number of Ag atoms in the nanoparticle by its volume of respective Ag atom:

$2.5{\text{ x 10}}^5\text{ Ag atoms (}(\frac{4}{3}\pi {(144\text{ pm x }\frac{{10}^2\text{ cm}}{{10}^{12}\text{ pm}}\text{)}}^3)/\text{atom)}=3.13{\text{ x 10}}^{-18}{\text{ cm}}^3$

Relating this answer to the volume of total nanoparticle determined in part (a), the packing of atomic fraction for the Silver (Ag) nanoparticle is,

$\left(\text{3}{\text{.13 x 10}}^{\text{-18}}{\text{cm}}^{\text{3}}\text{/ 4}{\text{.19 x 10}}^{\text{-18}}{\text{cm}}^{\text{3}}\right)\text{= 0}\text{.75}\text{.}$

Thus, the volume of 75% of the nanoparticle is occupied by Ag atoms and the volume of residual nanoparticle (25%) of the interstitial space present in between Ag atoms that are touching one another.

Interpretation Introduction

Interpretation:

Based on answer of part (b) type of cubic crystal lattice of Ag has to be predicted.

Answer to Problem 11.157QP

Relating the results to the value of 0.75 calculated in part (b), silver would conform to the face-centered cubic type of crystal lattice.

Explanation of Solution

Atomic packing fraction are determined by using simple space filling geometrical argument with a single unit of each cubic lattice type. In a simple cubic cell one atom is located at each of 8 lattice point of the unit cell.  That is one atom can shared by 8 adjacent unit cells.  If edge length (s) and corners are touch along each edge s is twice that of radius of the atom.  For simple cubic cell, the packing fraction of the atom is determined by dividing the volume of one atom by the cell volume.

$\frac{\text{volume of 1 atom}}{\text{volume of unit cell}}\text{=}\frac{\frac{\text{4}}{\text{3}}\text{π(}\frac{\text{s}}{\text{2}}{\text{)}}^{\text{3}}}{{\text{s}}^{\text{3}}}\text{= 0}\text{.52}$

In body centered cubic cell, arrangement of atoms are similar as simple cubic cell but addition of one atom being located at centre of the unit cell.  So, $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$8$}\right.$ of each corner atoms given to the unit cell and one atom located in the centre, there are 2 atoms per unit cell.  If 3 atoms are assumed to touch along the body diagonal, the length of the body is 4 times the radius of an atom.

Atomic radius in terms of the cell edge length is $\frac{\text{s}\sqrt{3}}{4}$.  For a body-centered cubic cell the atomic packing fraction is determined by dividing the volume of 2 atoms by the cell volume.

$\frac{\text{volume of 2 atoms}}{\text{volume of unit cell}}=\frac{(2)(\frac{4}{3}\pi {(\frac{\text{s}\sqrt{3}}{4}\text{)}}^3}{{\text{s}}^3}=0.68$

In a face-centered cubic cell, in addition to one atom being located at each of the 8 lattice points of the cell.  There is 1 atom on each of the 6 faces of the unit cell.  So, with 1/8 of each atom at corner given to the unit cell and $\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.$ of each face-centered atom in the unit cell, there are 4 atoms per unit cell.  If 3 atoms are expected to touch along the body diagonal, the body length is 4 times the radius of an atom.

Atomic radius in terms of the cell edge length is $\frac{\text{s}\sqrt{2}}{4}$.  For a face-centered cubic cell the atomic packing fraction is determined by dividing the volume of 4 atoms by the cell volume.

$\frac{\text{volume of 4 atoms}}{\text{volume of unit cell}}=\frac{(4)(\frac{4}{3}\pi {(\frac{\text{s}\sqrt{2}}{4}\text{)}}^3}{{\text{s}}^3}=0.74$

Relating the results to the value of 0.75 calculated in part (b), silver would conform to the face-centered cubic type of crystal lattice.

Interpretation Introduction

Interpretation:

The ratio of the surface area given by the batch of nanoparticle to the surface area of the initial cube of Silver has to be calculated.

Answer to Problem 11.157QP

The surface area provided by the nanoparticles is $3.9\times {10}^5$

Explanation of Solution

First calculate the volume of initial cube of Silver by dividing the mass by its density of Silver given in part (b).

$\text{volume of Ag cube = 5}\text{.0 g Ag (}\frac{1{\text{ cm}}^3}{10.49\text{ g}}\text{)= 0}{\text{.477 cm}}^3$

Then, the volume of the cube length of side edge is s, we calculate the length of single side by taking cubed root of the volume.

$\text{s = }\sqrt[3]{{\text{V}}_{\text{cube}}}=\sqrt[3]{0.477{\text{ cm}}^3}=0.781\text{ cm}$

To calculate the surface area of a cube, note that there are six faces showing.  Each having an area of $\text{0}{\text{.781}}^{\text{2}}{\text{cm}}^{\text{2}}$.  Surface area of initial cubic ingot is,

$\text{6 faces x }\frac{{(0.781\text{ cm})}^2}{\text{face}}=3.66{\text{ cm}}^2$

Now, calculate the number of 20-nm Silver nanoparticle formed from $\text{5}\text{.0g}$ of Silver by utilizing the mass of every nanoparticle found in part (a).

$(5.0\text{ g)(}\frac{1\text{ nanoparticle}}{4.39{\text{ x 10}}^{-17}\text{ g}}\text{) = 1}{\text{.14 x 10}}^{17}\text{ nanoparticles formed}$

Assume that every nanoparticle has spherical shape, then use the formula for surface area of the sphere to calculate the surface area given by nanoparticles.

$(1.14{\text{ x 10}}^{17}\text{ nanoparticles)(}\frac{4\pi {(1.0{\text{ x 10}}^{-6}\text{ cm)}}^2}{\text{1 nanoparticle}}\text{) = 1}{\text{.43 x 10}}^6{\text{ cm}}^2$

Finally calculate the ratio of the surface area of the nanoparticles to the surface area of the initial cube.

$\frac{\text{surface area of nanoparticles}}{\text{surface area of initial cube}}=\frac{1.43{\text{ x 10}}^6{\text{ cm}}^2}{3.66\text{ cm}}=3.9{\text{ x 10}}^5$

Thus, the surface area given by the nanoparticles is 390,000 times more than the surface area of the initial ingot.