How do I calculate the volume density of atoms in a diamond lattice?

  • Thread starter Thread starter salman213
  • Start date Start date
  • Tags Tags
    Diamond Lattice
Click For Summary
SUMMARY

The discussion focuses on calculating the volume density of atoms in a diamond lattice structure, specifically using Silicon with a lattice constant of 5.4 x 10^-10 m. The confusion arises from the interpretation of the face-centered cubic (FCC) structure, where the total number of atoms is determined to be 8 due to the presence of two FCC unit cells. The volume density is calculated using the formula atom density = 8 / (5.4 x 10^-10)^3, which accounts for the total number of atoms in the combined volume of the two unit cells.

PREREQUISITES
  • Understanding of diamond lattice structures
  • Knowledge of face-centered cubic (FCC) unit cells
  • Familiarity with lattice constants and their significance
  • Basic principles of volume density calculations
NEXT STEPS
  • Research the properties of diamond lattice structures in materials science
  • Learn about the calculation of atomic density in different crystal structures
  • Explore the implications of lattice constants on material properties
  • Study the differences between FCC and other crystal lattice types
USEFUL FOR

Students and professionals in materials science, physicists, and chemists interested in crystallography and atomic structure calculations.

salman213
Messages
301
Reaction score
1
1.Hey!

I don't get the solution to the following question so I am hoping someone can explain!

Ok so, the question is we have Silicon, which I have read has a DIAMOND LATTICE structure.
Which basically seems to mean two face centered cubes come together. Now I need to calculate the VOLUME DENSITY of ATOMS. I know the lattice constant ex. 5.4 x 10^-10 m

Now From my knowledge: fcc in 3d has 1/8th of an atom on each of the 8 corners of the cube. And has also has 1/2 of a atom on each of the 6 faces. That makes a total of

1/8*8 = 1
1/2*6 = 3
--------------
4 atoms in total

Now the solution says there are somehow 8 atoms. I am assuming there are 8 because it is a diamond lattice structure and so there are 2 cubes meaning 4 x 2 = 8 atoms..first is this assumption correct?



if that is correct then when they calculate the volume they use

(5.4 x 10^-10)^3 = I THOUGHT THIS WAS THE VOLUME OF ONLY ONE CUBE...:O..

answer somehow is

atom density = 8/ (5.4 x 10^-10)^3

HELP!

why did they use 8 atoms and why did they use that volume for only one cube if the 8 atoms come from fcc cubes!

please help!
 
Physics news on Phys.org
ok i figured it out after some research ..
 

Similar threads

Undergrad Silicon FCC -- Why are so many atoms shown in the lattice?
  • · Replies 5 ·
Replies
5
Views
3K
How do I calculate the density of calcium from atomic radius?
  • · Replies 11 ·
Replies
11
Views
17K
Chemistry Calculating the volume of individual solution(s) [Mole/Atoms Concept]
  • · Replies 10 ·
Replies
10
Views
2K
Undergrad Why Do Different Primitive Unit Cells for Diamond Look So Different?
  • · Replies 2 ·
Replies
2
Views
7K
How to Calculate the Density of Aluminum in a Face-Centered Cubic Unit Cell
  • · Replies 2 ·
Replies
2
Views
12K
Graduate Optical phonons in fcc-lattice
  • · Replies 8 ·
Replies
8
Views
4K
Loss of atoms from the surface of the material: How does it work?
  • · Replies 15 ·
Replies
15
Views
2K
Chemistry How do we know ##BF_3## has Lewis structure but ##Al_2S_3## does not?
  • · Replies 1 ·
Replies
1
Views
2K
Bonding of Diamond: Unreactivity & Impurity
  • · Replies 2 ·
Replies
2
Views
2K
What Are the Ionic Radii of Mg2+ and O2- in MgO's Crystal Structure?
  • · Replies 4 ·
Replies
4
Views
14K