Lattice length for Zincblende structures

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SUMMARY

The discussion centers on calculating the lattice length for Zincblende structures using the covalent radii of Gallium (Ga) at 1.26 Å and Arsenic (As) at 1.19 Å. The formula derived is [4/sqrt(3)] x [radius of Ga + radius of As], resulting in a lattice length of 5.66 Å. The participant questions the reasoning behind this formula and its correlation to Body Centered Cubic (BCC) packing. The calculation relies on geometric principles, specifically the arrangement of atoms within the unit cell.

PREREQUISITES
  • Covalent radii of elements, specifically Gallium and Arsenic
  • Understanding of Zincblende crystal structure
  • Basic geometry related to atomic packing
  • Knowledge of Body Centered Cubic (BCC) packing structures
NEXT STEPS
  • Research the geometric principles of atomic packing in crystal structures
  • Study the Zincblende structure and its properties in detail
  • Explore the mathematical derivation of lattice parameters in crystal systems
  • Investigate the similarities and differences between Zincblende and Body Centered Cubic (BCC) structures
USEFUL FOR

Chemists, materials scientists, and crystallographers interested in understanding crystal structures and lattice calculations, particularly those working with semiconductor materials like Gallium Arsenide.

nefizseal
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So we have the covalent Radii of Ga = 1.26 A and As = 1.19 A. And the structure is Zinc blende with arsenic occupying half of the tetrahedral sites. So its probably not close packed...

The formula the works for me is, [4/sqrt(3)] x [ radius of Ga + radius of As ]. I can get the answer ( 5.66 A ) by this but I don't know why or how it works? Can someone please explain the reasoning behind it.

Also I noticed this is like the formula for the Body centre cubic packing. Is there any corelation?
thx...
 
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Without looking at the structure - such things can be always calculated just by simple geometry. Draw the cell, assume atoms "touch" each other, draw distances between atom centers, use whatever is needed to solve the triangles present.
 

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