Chapter 11: Problem 5
In the body centered cubic unit cell and simple cubic unit cell, the radius of atom in terms of edge length (a) of the unit cell is respectively: (a) \(\frac{a}{2}, \frac{a}{2 \sqrt{2}}\) (b) \(\frac{a}{\sqrt{2}}, \frac{a}{2}\) (c) \(\frac{a}{2 \sqrt{2}}, \frac{a}{2}\) (d) \(\frac{\sqrt{3} a}{4}, \frac{a}{2}\)
Short Answer
Step by step solution
Understanding the Body-Centered Cubic (BCC) Unit Cell
Deriving the Body-Centered Cubic (BCC) Radius Formula
Understanding the Simple Cubic (SC) Unit Cell
Deriving the Simple Cubic (SC) Radius Formula
Matching the Options to the Calculated Results
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Body-Centered Cubic
- The body diagonal contains the diameter of two atoms.
- Relationship: The body diagonal \[ \sqrt{3}a = 2 imes (2r) \] leads to the formula for atomic radius, \[ r = \frac{\sqrt{3}a}{4} \].
- Useful in determining packing efficiency and density of materials.
Simple Cubic Unit Cell
- Atoms touch along the cube's edge.
- Relationship: The edge length is equal to twice the atomic radius, \[ a = 2r \] resulting in the atomic radius, \[ r = \frac{a}{2} \].
- Examples include polonium which crystallizes in this structure.
Atomic Radius
- For BCC structures: \[ r = \frac{\sqrt{3}a}{4} \].
- For SC structures: \[ r = \frac{a}{2} \].
- Atomic radius affects material properties such as strength and ductility.
Edge Length
- For BCC: Relates to the atomic radius as \[ \sqrt{3}a = 4r \].
- For SC: Simply twice the atomic radius, or \[ a = 2r \].
- Critical for calculating the weight and volume of crystalline materials.
Crystallography
- Essential for material science, helping in the development of new alloys and compounds.
- Informs on the physical and chemical behavior of crystalline substances.
- Leads to the discovery of materials with targeted properties for specific applications.