Planar Concentration of Nb Atoms in BCC Crystal

[elcotufa]

Homework Statement

Niobium Nb has the BCC crystal with a lattice parameter of a= .3294 nm

Find the planar concentrations of atoms per m^2 of the (100, (110) and (111) planes.

The 100 plane has 1 atom on the plane (1/4 times 4)

the 110 plane has 2 atoms on the plane and using pythagorean theorem the long side is a*sqrt(2)

http://www.gly.uga.edu/schroeder/geol6550/111.jpeg
best pic I could find for the 111 plane
Now for the last plane (111) this is what I did

the area would be 1/2 base * height
the base is $$a*\sqrt2$$ and

pythagorean theorem to find the height is a^2 and half of the long side from plane (110) squared $$\sqrt{a^2+\frac{2a^2}4}=.4034nm$$

$$\frac12(.4034)10^{-9}(.3294)10^{-9}$$
Now how many atoms are on the plane? and are there mistakes in what I have done so far?

And how many atoms would that (111) plane have in a FCC model?
Answer would have number of atoms over area
Thanks for the help

 

[Astronuc]

One should find diagrams and discussion in W.D. Callister, Materials Science and Engineering, An Introduction, 7th Ed., John Wiley and Sons, 2007, or later editions.

Problems with planar densities of fcc and bcc are found in this assignment.
http://maecourses.ucsd.edu/~jmckittr/mae20-wi11//Assignment 4 solutions.pdf

One calculates the number of atoms per unit area, where area may be given in terms of nm².

For the (111) plane in bcc, note that the plane does not pass through the atom in the center of the cubic structure. The base is the diagonal in one of the 6 faces, so the length is as one shows, $\sqrt{2}$a.

The height of the triangle (hypotenuse) formed by the base of the (111) plane to the upper corner of the cell is found from the Pythagorean theorem, realizing one measures from the center of the diagonal to the upper corner of the cubic cell. One's answer is correct.

For an fcc cell, note that the base of the (111) plane (the diagonal of the unit cell) does intersect an atom.