Math Problem Statement

Give the breakdown in solving fcc bcc scc relations between a and r for different ionic crystals

Solution

In crystallography, the relationships between lattice parameter (aa) and atomic radius (rr) differ for face-centered cubic (FCC), body-centered cubic (BCC), and simple cubic (SCC) crystal structures. Here is a detailed breakdown for each:

1. Simple Cubic Crystal Structure (SCC)

In a Simple Cubic (SC) crystal structure:

  • Each atom is located at the corners of the cube.
  • The edge length aa is directly related to the atomic radius rr, as atoms only touch each other along the edges.

Relation between aa and rr:

a=2ra = 2r Explanation:

  • Each edge of the cube has two atoms touching each other.
  • Therefore, the edge length aa is twice the atomic radius rr.

2. Body-Centered Cubic Crystal Structure (BCC)

In a Body-Centered Cubic (BCC) structure:

  • Atoms are located at the corners and one atom at the center of the cube.
  • The atoms touch each other along the body diagonal, not along the cube edge.

Relation between aa and rr:

a=4r3a = \frac{4r}{\sqrt{3}} Explanation:

  • The body diagonal of the cube, which spans across two corner atoms and one central atom, is 3a\sqrt{3}a.
  • Since this diagonal covers four radii (two from each corner and two from the center atom), we have 3a=4r\sqrt{3}a = 4r.
  • Solving for aa gives a=4r3a = \frac{4r}{\sqrt{3}}.

3. Face-Centered Cubic Crystal Structure (FCC)

In a Face-Centered Cubic (FCC) structure:

  • Atoms are located at each corner and the centers of each face of the cube.
  • The atoms touch each other along the face diagonal, not the edges or body diagonal.

Relation between aa and rr:

a=4r2a = \frac{4r}{\sqrt{2}} Explanation:

  • The face diagonal of the cube, which spans across two atoms at adjacent corners and one atom in the center of the face, is 2a\sqrt{2}a.
  • This face diagonal covers four radii (two from each corner and two from the face-centered atom), so 2a=4r\sqrt{2}a = 4r.
  • Solving for aa gives a=4r2=22ra = \frac{4r}{\sqrt{2}} = 2\sqrt{2}r.

Summary Table of aa and rr Relations:

Crystal StructureRelation between aa and rr
Simple Cubic (SCC)a=2ra = 2r
Body-Centered Cubic (BCC)a=4r3a = \frac{4r}{\sqrt{3}}
Face-Centered Cubic (FCC)a=4r2a = \frac{4r}{\sqrt{2}}

These relations help in calculating one parameter if the other is known, which is crucial in material science and solid-state physics when analyzing crystal structures of ionic crystals.

Would you like to go deeper into the applications or assumptions involved in these calculations?


Additional Questions for Further Exploration:

  1. How does the atomic packing factor (APF) differ among SCC, BCC, and FCC structures?
  2. Why is the FCC structure often more dense than the BCC and SCC structures?
  3. How do the number of atoms per unit cell vary for SCC, BCC, and FCC structures?
  4. What are examples of materials that crystallize in each of these structures?
  5. How does the coordination number differ between SCC, BCC, and FCC structures?

Tip:

Understanding these relations helps predict material properties like density and stability, essential for engineering and materials science applications.

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Math Problem Analysis

Mathematical Concepts

Crystallography
Lattice Geometry
Atomic Radius
Lattice Parameter

Formulas

Simple Cubic (SCC): a = 2r
Body-Centered Cubic (BCC): a = 4r / √3
Face-Centered Cubic (FCC): a = 4r / √2

Theorems

Geometric relations in cubic structures

Suitable Grade Level

Undergraduate (Material Science, Physics)