Consider the unit cell of aluminum with aluminum ion on every corner and every face-centered site of the cube.

Question

1. Using the value of ionic radius of aluminum ion r=0.143 nm, the length of each edge of the unit cell can be calculated as what nm?

2. Using atomic weight value of aluminum 27.0, the density of aluminum can be calculated as what g/cm^3

1) 0.4045nm

2) 0.0027096117g/cm³

Explanation:

1) Using the value of ionic radius of aluminum ion r=0.143 nm, the length of each edge of the unit cell can be calculated as what nm?

In the question above, using a face centered cubic structure, we have:

For a face centered site of the cube,

The diagonal = 4r

Where r = Atomic or ionic radius of Aluminum

Let the edge length of the cube be represented by X

Therefore, we have based on Pythagoras theorem,

X² + X² = (4r) ²

2X² = 16r²

Divide both sides by 2

X² = 8r²

Find the square root of both sides

X = √8 * r

Since r = 0.143nm

The length of each edge of the unit cell can be calculated as

X = √8 * 0.143nm

X = 0.4044650788nm

Approximately = 0.4045nm

2) Using atomic weight value of aluminum 27.0, the density of aluminum can be calculated as what g/cm^3

Density of an object = Mass of the Object / Volume of the Object.

The object in this question = Cube

Step 1

Volume of a cube = (Length of the cube) ³

In the question above, side length of the cube = 0.4045nm

When would convert 0.4045nm to centimeters

= 1 nm = 1 * 10^-7 cm

0.4045nm =

Cross multiply

= 4.045 * 10^-7 cm

Volume of the Aluminum cube = (4.045 * 10^-7cm) ³

= 6.618439112 * 10^-20cm³

Step 2

The atomic weight value of aluminum is given as 27.0 in the question

A face centered cubic structure has 4 atoms per unit cell.

1 Atomic mass or weight value = 1.6605 * 10^-24 grams

Hence, the mass of Aluminum is calculated as:

(4 atoms / 1 cell) * 1 unit * (27 / 1 atom of Aluminium) * (1.6605 * 10^-24g / 1 Atomic mass value)

= 1.79334 * 10^-22g

Density = Mass/Volume

= 1.79334 * 10^ - 22g / 6.618439112 * 10^-20cm³

= 0.0027096117g/cm³