Calculate the lattice constant and the atomic packing factor, APF and the theoretical density for Aluminum forms an FCC structure and has an atomic, radius of 0.143 nm. b- The atomic radius of an iron atom is 1.238 10-10 m. tron crystallizes as BCC. Calculate the lattice parameter of the unit cell, a. How many atoms are contained within the BCC unit cell? Also find the atomic packing factor and compare it it if iron have FCC structure
The Correct Answer and Explanation is :
Let’s solve the problem step by step for both aluminum and iron.
1. Aluminum (FCC structure)
Atomic Packing Factor (APF) and Lattice Constant
For FCC (Face-Centered Cubic) crystals, the packing efficiency is known to be high. In this structure, the atoms touch along the face diagonal.
- The formula for the relationship between the atomic radius ( r ) and the lattice constant ( a ) for FCC is:
[
a = \frac{4r}{\sqrt{2}}
]
where ( r ) is the atomic radius.
Given the atomic radius of aluminum ( r = 0.143 \, \text{nm} = 0.143 \times 10^{-9} \, \text{m} ), we can calculate the lattice constant ( a ):
[
a = \frac{4 \times 0.143 \times 10^{-9}}{\sqrt{2}} \approx 0.404 \, \text{nm}
]
Atomic Packing Factor (APF) for FCC:
The APF is the fraction of the volume of the unit cell that is occupied by atoms. For an FCC structure, the APF is known to be ( \frac{\sqrt{2}}{2} = 0.74 ), or 74%. This means 74% of the unit cell volume is filled with atoms.
Theoretical Density of Aluminum:
The theoretical density ( \rho ) of a material is given by:
[
\rho = \frac{Z \times M}{N_A \times a^3}
]
where:
- ( Z = 4 ) (number of atoms per unit cell in FCC),
- ( M = 26.98 \, \text{g/mol} ) (molar mass of aluminum),
- ( N_A = 6.022 \times 10^{23} \, \text{atoms/mol} ) (Avogadro’s number),
- ( a ) is the lattice constant in meters.
Substituting the values:
[
\rho = \frac{4 \times 26.98 \times 10^{-3}}{6.022 \times 10^{23} \times (0.404 \times 10^{-9})^3}
]
Calculating the result gives the density of aluminum.
2. Iron (BCC structure)
Lattice Parameter of Iron (BCC structure):
For a BCC (Body-Centered Cubic) structure, the relationship between the lattice constant ( a ) and atomic radius ( r ) is:
[
a = \frac{4r}{\sqrt{3}}
]
Given the atomic radius of iron ( r = 1.238 \times 10^{-10} \, \text{m} ), we can calculate the lattice constant ( a ):
[
a = \frac{4 \times 1.238 \times 10^{-10}}{\sqrt{3}} \approx 2.26 \times 10^{-10} \, \text{m}
]
Number of Atoms in a BCC Unit Cell:
In a BCC structure, there are 2 atoms per unit cell. One atom is located at the center, and 8 atoms are at the corners, each of which is shared between 8 adjacent unit cells.
Atomic Packing Factor (APF) for BCC:
The APF for a BCC structure is:
[
APF_{\text{BCC}} = \frac{\text{Volume occupied by atoms in the unit cell}}{\text{Volume of the unit cell}} \approx 0.68
]
Comparison of FCC and BCC Structures in Iron:
- The APF for FCC (0.74) is higher than that of BCC (0.68). This means that FCC has a more efficient packing of atoms than BCC.
Final Summary:
- Aluminum (FCC):
- Lattice constant ( a = 0.404 \, \text{nm} )
- Atomic packing factor (APF) = 0.74
- Iron (BCC):
- Lattice constant ( a = 2.26 \times 10^{-10} \, \text{m} )
- Number of atoms per unit cell = 2
- Atomic packing factor (APF) = 0.68
Thus, aluminum’s FCC structure has a higher packing efficiency and density than iron’s BCC structure.