How to calculate 2D packing fractions?

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    2d Fractions
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SUMMARY

The discussion focuses on calculating the packing fractions for various planes in a Face-Centered Cubic (FCC) structure. The user initially miscalculated the packing fraction for the (100) plane as pi/2, which is incorrect. By applying the correct geometric principles, including Pythagorean theorem, the user determined the correct packing fraction for the (100) plane to be pi/4. The user also successfully calculated the packing fractions for the (110) and (111) planes using similar methods.

PREREQUISITES
  • Understanding of Face-Centered Cubic (FCC) crystal structures
  • Knowledge of packing fraction calculations
  • Familiarity with geometric principles, specifically the Pythagorean theorem
  • Basic knowledge of atomic radius and area calculations
NEXT STEPS
  • Research the packing fractions for other crystal structures, such as Body-Centered Cubic (BCC) and Hexagonal Close-Packed (HCP).
  • Learn about the implications of packing fractions on material properties.
  • Explore advanced geometric methods for calculating packing in complex crystal lattices.
  • Study the relationship between atomic arrangement and crystallography in solid-state physics.
USEFUL FOR

This discussion is beneficial for materials scientists, crystallographers, and students studying solid-state physics who are interested in understanding atomic packing and its implications on material properties.

Talvon
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I have a question regarding various planes in an FCC, and determine their packing fractions.

I searched but couldn't find anything :)

For example, one of the planes is the (100) plane, and I have said there are 2 full atoms (1 in the middle, and 4 quarters from each side), distance 'a' apart. Using the usual packing fraction equation ((Volume of atom x number of atoms)/Volume of cell), but replacing the volume with area, I calculate this to be pi/2, which is wrong because it can't be higher than one :confused:

My exact numbers were:
Area of a circle = pi x r², where r=a/2
-> (pi x a²/4) x 2 (-# of atoms) / a²
The a² cancels, as does the 4 with the 2 to leave pi over 2.

Any help would be appreciated, cheers :)
 
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Why do you assume the area of the cell to be a²?

If you have a look at the diagonal line of your cell area, you get the maximum packing fraction if the atoms exactly touch each other. So the diagonal line must have a length of:
a (full atom) + 2* a/2 (quarter of 2 other atoms) =2 a.

So using Pythagoras you will get that the side lines of your area will have a length of sqrt(2) a, which leads to a cell area of 2 a².
 
Ace, cheers :D Didn't make sense initially but then it clicked :P

Managed to work it through to get pi/4, and managed to solve the (110) and (111) plane in a similar way :smile:
 

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