A typical paper clip weighs 0.59 g

A typical paper clip weighs 0.59 g. Assume that it is made from BCC iron. Calculate

(a) the number of unit cells; and

(b) the number of iron atoms in the paper clip. (See Appendix A for required data.)

2. Aluminum foil used to package food is approximately 0.001 inch thick. Assume that all of the unit cells of the aluminum are arranged so that a0 is perpendicular to the foil surface. For a 4 in. x 4 in. square of the foil, determine

(a) the total number of unit cells in the foil; and

(b) the thickness of the foil in number of unit cells.

The Correct Answer and Explanation is :

To address the given problems, we need to apply concepts from materials science, specifically related to crystal structures and unit cells. Let’s break down each question:

1. Paper Clip Made of BCC Iron

Given:

• Mass of paper clip: 0.59 g
• Crystal structure: Body-Centered Cubic (BCC) iron
• Lattice parameter for BCC iron: ( a = 2.866 \times 10^{-8} ) cm
• Density of BCC iron: 7.87 g/cm³

(a) Number of Unit Cells

1. Calculate the volume of the paper clip:

• Volume ( V = \frac{\text{Mass}}{\text{Density}} = \frac{0.59\, \text{g}}{7.87\, \text{g/cm}^3} \approx 0.0749\, \text{cm}^3 )

1. Calculate the volume of a single BCC unit cell:

• Volume of unit cell ( V_{\text{cell}} = a^3 = (2.866 \times 10^{-8}\, \text{cm})^3 \approx 2.36 \times 10^{-23}\, \text{cm}^3 )

1. Calculate the number of unit cells:

• Number of unit cells ( N_{\text{cells}} = \frac{V}{V_{\text{cell}}} = \frac{0.0749\, \text{cm}^3}{2.36 \times 10^{-23}\, \text{cm}^3} \approx 3.18 \times 10^{21} )

(b) Number of Iron Atoms

1. Determine the number of atoms per BCC unit cell:

• BCC structure has 2 atoms per unit cell.

1. Calculate the total number of atoms:

• Total atoms ( N_{\text{atoms}} = N_{\text{cells}} \times 2 = 3.18 \times 10^{21} \times 2 \approx 6.36 \times 10^{21} ) atoms

2. Aluminum Foil

Given:

• Thickness of foil: 0.001 inch
• Dimensions of foil: 4 in. x 4 in.
• Crystal structure: Face-Centered Cubic (FCC) aluminum
• Lattice parameter for FCC aluminum: ( a = 4.05 \times 10^{-8} ) cm

(a) Total Number of Unit Cells

1. Convert foil dimensions to centimeters:

• 1 inch = 2.54 cm
• Thickness: ( 0.001\, \text{inch} = 0.00254\, \text{cm} )
• Length and width: ( 4\, \text{in.} = 10.16\, \text{cm} )

1. Calculate the volume of the foil:

• Volume ( V = \text{Length} \times \text{Width} \times \text{Thickness} = 10.16\, \text{cm} \times 10.16\, \text{cm} \times 0.00254\, \text{cm} \approx 0.000261\, \text{cm}^3 )

1. Calculate the volume of a single FCC unit cell:

• Volume of unit cell ( V_{\text{cell}} = a^3 = (4.05 \times 10^{-8}\, \text{cm})^3 \approx 6.63 \times 10^{-23}\, \text{cm}^3 )

1. Calculate the number of unit cells:

• Number of unit cells ( N_{\text{cells}} = \frac{V}{V_{\text{cell}}} = \frac{0.000261\, \text{cm}^3}{6.63 \times 10^{-23}\, \text{cm}^3} \approx 3.94 \times 10^{18} )

(b) Thickness of the Foil in Number of Unit Cells

1. Calculate the number of unit cells along the thickness:

• Number of unit cells along thickness ( N_{\text{thickness}} = \frac{\text{Thickness}}{a} = \frac{0.00254\, \text{cm}}{4.05 \times 10^{-8}\, \text{cm}} \approx 62,740 )

Explanation

In these calculations, we utilized fundamental principles of materials science:

• Volume Calculation: The volume of an object is determined by its mass divided by its density. This provides the total space occupied by the material.
• Unit Cell Volume: The volume of a unit cell is calculated by cubing the lattice parameter (( a )), which represents the edge length of the unit cell.
• Number of Unit Cells: By dividing the total volume of the object by the volume of a single unit cell, we determine how many unit cells are present in the object.
• Atoms per Unit Cell: The number of atoms per unit cell is a characteristic of the crystal structure. For BCC iron, there are 2 atoms per unit cell, and for FCC aluminum, there are 4 atoms per unit cell.
• Total Number of Atoms: Multiplying the number of unit cells by the number of atoms per unit cell gives the total number of atoms in the object.

These calculations are essential for understanding the atomic-scale composition of materials and are fundamental in materials science and engineering.