Platinum (at. wt. = 195.1) crystallizes in the fcc form and has the density of \(21.4 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\). Calculate the side of the unit cell. \(\quad\) Ans. \(3.93 \mathrm{~A}\).

Platinum crystallizes in a face-centered cubic (fcc) structure. Understanding the arrangement in an fcc is crucial for solving problems related to unit cells. In an fcc structure, atoms are positioned at each corner of the cube and at the center of each face. Each corner atom is shared among eight different unit cells, and each face-centered atom is shared between two unit cells. This means that in total, there are 4 atoms per unit cell.

Here’s how we break it down:

• Each corner contributes 1/8th of an atom (since it's shared among 8 unit cells), and there are 8 corners, resulting in 1 whole atom from the corners.
• Each face-center contributes 1/2 of an atom (since it's shared among 2 unit cells), and there are 6 faces, resulting in 3 atoms from the faces.
• Adding these, 1 (from corners) + 3 (from faces) = 4 atoms per unit cell.

Thus, when you see 'fcc', always remember it means 4 atoms per unit cell.

Converting the atomic weight of platinum from atomic mass units (amu) to kilograms is our next step. The atomic weight of platinum is given as 195.1 amu. To convert this to kg, you need to know the conversion factor: 1 amu equals approximately 1.66054 x 10^{-27} kg.

You perform the conversion using:

• Mass of one platinum atom = 195.1 amu x 1.66054 x 10^{-27} kg/amu
• This gives: 3.24 x 10^{-25} kg per atom.

This conversion is essential for getting the mass in the correct unit to be used in the following steps of density and volume calculations. By converting to kg, we align our units properly for subsequent calculations involving density which is given in kg/m^3.

Density is a key concept in this exercise. The density formula is expressed as:

equation: \(\rho = \frac{M}{V}\)

Where

• \(\rho\) represents the density.
• M represents the mass of the unit cell.
• V represents the volume of the unit cell.

To find the volume, you rearrange the formula:

equation: \(V = \frac{M}{\rho}\)

Given the density of platinum as \(21.4 \times 10^{3} \text{ kg/m}^3\), and the mass of the unit cell previously calculated as \(1.296 \times 10^{-24} \text{ kg}\), you substitute these values into the formula:

equation: \(V = \frac{1.296 \times 10^{-24}}{21.4 \times 10^{3}} \text{ m}^3\)

This calculation yields the volume of the unit cell as \(6.06 \times 10^{-29} \text{ m}^3\). Knowing both the mass and the density allows for the identification of the unit cell volume, an important step towards determining the side length of the unit cell.

After finding the volume of the unit cell, you need to calculate the side length (a). Since a unit cell in an fcc structure forms a cube, its volume can be represented as \(\text{side length}^3\) or a^3.

To find the side length:

• We have the volume \(V = 6.06 \times 10^{-29} \text{ m}^3\).
• We solve for a: \(a = \root{3} (6.06 \times 10^{-29}) \text{ m}\).
• This results in \(a = 3.93 \times 10^{-10} \text{ m}\).

Finally, convert the side length from meters to Angstroms (Å), knowing that 1 meter equals 10^10 Å:

Thus, \a = 3.93 Å.

Understanding this conversion is crucial, as crystallographic dimensions are often conveniently represented in Angstroms in the realm of chemistry and materials science.