Atomic Chain in BCC - # Atoms/Unit Length & Area

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i'd like to know what is the number of atoms per unit length for an atomic chain in the [1 1 1] direction of a BCC structure.

i can figure out the unit length, however I'm not sure how many atoms to use. Is it 1/8+1/8+1? That means 1/8s for the two corner atoms and 1 atom for the middle? or is it just 3 atoms?

part of the problem is i cannot find anywhere the meaning of atomic chain in the BCC structure.

likewise, what if i had to find the number of atoms per unit area in the (1 1 1) plane? the area part is easy, but again is the number of atoms 4 or 1/8 * 3 + 1?
 
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What does one mean by 'atomic chain'? We usually refer to an 'atomic lattice', in which the atoms are located at specific points in a repeated crystal structure.

In a BCC cube, there is one whole atom at the center of the unit cube, and 1/8 atom at each corner of the cube.
 
i'm not sure but that's what i was asked.

so if it's atomic lattice, then i would be right to say the corners are 1/8 and the middle is 1 atom? but isn't the middle atom shared by other diagnoal lattice?
 
madeinmsia said:
i'm not sure but that's what i was asked.

so if it's atomic lattice, then i would be right to say the corners are 1/8 and the middle is 1 atom? but isn't the middle atom shared by other diagnoal lattice?
In a sense yes. One could shift the unit cell diagonally in any of the 8 directions of the BCC unit cell, and the corner atom becomes a center atom, and center atom is now a corner atom.

The unit cell is a 3D concept, as opposed to a 2D-plane.

The diagonal length is BCC unit cell is R + 2R + R, where R is the atomic radius, because the corner atoms are tangent to the center atoms. So the cubic unit cell must have a face side length of less than 2R.

http://www.science.uwaterloo.ca/~cchieh/cact/c123/bcc.html

Nice pictures here - http://www.ndt-ed.org/EducationResources/CommunityCollege/Materials/Structure/metallic_structures.htm

http://genchem.chem.wisc.edu/lab/winss/metal_cells/body_centered_cubic_metals/dim_bbc_unit_cell.htm
 
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To solve this, I first used the units to work out that a= m* a/m, i.e. t=z/λ. This would allow you to determine the time duration within an interval section by section and then add this to the previous ones to obtain the age of the respective layer. However, this would require a constant thickness per year for each interval. However, since this is most likely not the case, my next consideration was that the age must be the integral of a 1/λ(z) function, which I cannot model.
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