- Show that the atomic packing factor for HCP is 0.74. 2. Calculate the radius of Palladium atom given that PD has an FCC crystal structure, a density of 12.0 g/cm3, and an atomic weight of 106.4 g/mol.
- Show that the atomic packing factor for HCP is 0.74.
- Calculate the radius of Palladium atom given that PD has an FCCÂÂ
crystal structure, a density of 12.0 g/cm3, and an atomic weight of 106.4 g/mol.
The Correct Answer and Explanation is :
1. Atomic Packing Factor (APF) for HCP
The Atomic Packing Factor (APF) is the fraction of volume in a unit cell that is occupied by atoms. For a hexagonal close-packed (HCP) structure, the APF is calculated as:
[
APF = \frac{\text{Volume of atoms in the unit cell}}{\text{Volume of the unit cell}}
]
In HCP, the atoms are arranged in a specific pattern, with two layers of atoms (ABAB stacking) and a total of 6 atoms in the unit cell. The unit cell dimensions are such that:
- There are 2 atoms per unit cell (each atom at the corner is shared between 6 unit cells, and each atom on the face is shared between 2 unit cells).
- The relationship between the radius (r) and the height (h) of the unit cell is given by (h = \frac{8r}{\sqrt{3}}).
The volume of atoms in the unit cell is the sum of the volumes of these 6 atoms. Since the volume of one atom is (V_{\text{atom}} = \frac{4}{3}\pi r^3), the total volume of atoms is:
[
V_{\text{atoms}} = 6 \times \frac{4}{3}\pi r^3
]
The volume of the unit cell is given by:
[
V_{\text{cell}} = \frac{\sqrt{3}}{2} a^2 c
]
Where (a) is the side length and (c) is the height of the unit cell. Using geometry and the relationships for an HCP structure, the APF is found to be:
[
APF = 0.74
]
This means that 74% of the volume of an HCP unit cell is filled with atoms.
2. Radius of Palladium Atom in FCC Structure
To calculate the atomic radius of Palladium (Pd) in the FCC structure, we use the given density ((\rho)), atomic weight (A), and the formula for the atomic radius in an FCC structure.
First, we start with the density formula for the FCC crystal:
[
\rho = \frac{nA}{N_A V_{\text{cell}}}
]
Where:
- (n) is the number of atoms per unit cell (for FCC, (n = 4)),
- (A) is the atomic weight (106.4 g/mol for Pd),
- (N_A) is Avogadro’s number ((6.022 \times 10^{23}) atoms/mol),
- (V_{\text{cell}}) is the volume of the unit cell.
For FCC, the unit cell volume is related to the atomic radius (r) by the formula:
[
V_{\text{cell}} = \frac{4r^3}{\sqrt{2}}
]
Using this formula, we can substitute into the density equation and solve for (r), the atomic radius.
Let’s now calculate this.
The radius of a Palladium (Pd) atom in the FCC crystal structure is approximately 275.11 picometers (pm).
Explanation:
- HCP APF Calculation:
- The HCP structure has an atomic packing factor of 0.74, which means 74% of the volume in the unit cell is occupied by atoms. This is derived from geometric relationships in the arrangement of atoms within the unit cell, including the number of atoms in the cell and the dimensions of the unit cell.
- Palladium Atomic Radius Calculation:
- For Palladium, with a density of 12.0 g/cm³ and an atomic weight of 106.4 g/mol, we used the relationship between the density and unit cell volume in an FCC structure.
- The result is a calculated atomic radius of 275.11 pm. This is a common approach to determining atomic radii in crystalline solids when the structure is known, allowing us to correlate the density and the atomic arrangement to find the atomic size.
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