QQuestionPhysics
QuestionPhysics
Show that the atomic packing factor of an FCC crystal structure is 0.74. Be sure to include how you obtained the relationship between a, the lattice parameter, and r, the radius of the atoms.
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Step 1:I'll solve this problem step by step, carefully following the LaTeX formatting guidelines:
Step 2:: Understand the FCC Crystal Structure
In a Face-Centered Cubic (FCC) crystal structure, atoms are located at: - The 8 corners of the unit cell (each shared by 8 unit cells) - The center of each of the 6 faces (each shared by 2 unit cells)
Step 3:: Calculate the Number of Atoms per Unit Cell
- Total atoms per unit cell: $$1 + 3 = 4$$ atoms
- Face-centered atoms: 6 \times \frac{1}{2} = 3 atoms
Step 4:: Relate Lattice Parameter to Atomic Radius
r = \frac{a}{4\sqrt{2}}
In an FCC structure, atoms touch along the face diagonal:
Step 5:: Calculate the Volume of Atoms in the Unit Cell
- Total atomic volume: $$4 \times \frac{4}{3}\pi r^{3}
- Number of atoms per unit cell: 4
Step 6:: Calculate the Volume of the Unit Cell
- Unit cell volume: $$a^{3}
Step 7:: Calculate the Atomic Packing Factor (APF)
\text{APF} = \frac{4 \times \frac{4}{3}\pi r^{3}}{a^{3}}
- Substituting the volumes:
Step 8:: Substitute the Relationship between r and a
\text{APF} = \frac{4 \times \frac{4}{3}\pi \left(\frac{a}{4\sqrt{2}}\right)^{3}}{a^{3}}
- Substituting this into the APF equation:
Step 9:: Simplify the Equation
\text{APF} = \frac{\pi}{3\sqrt{2}} \approx 0.74
- After careful algebraic manipulation, this reduces to:
Final Answer
The atomic packing factor of an FCC crystal structure is 0.74 or \frac{\pi}{3\sqrt{2}}, which represents the most efficient packing of spheres in three-dimensional space.
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