Derive planar density expressions for BCC (110) planes in terms of the atomic radius R.

Derive planar density expressions for BCC (110) planes in terms of the atomic radius R.

Expert Solution

A BCC unit cell (110) can be drawn as ( you can edit the picture as I have drawn it in power point) :

As shown there are one atom at each of the four cube corners,each shared with four adjacent unit cells, while the center atom lies entirely within the unit cell. Hence there are two atoms in that BCC (110) plane

From the above figure, the area of the rectangle is the product of a and b. The length a is the unit cell edge length, which is for BCC is . The length of diagonal c is equal to 4R. For the right angle triangle bounded by the lengths a, b and c.

b = (c² - a²)^1/2

putting values of c and a in the above equation

b = ((4R)² - (4R/ (3^1/2 ))²)^1/2 = 4R (2/3)^1/2

Thus, in terms of R, the area of this (110) plane is just

Thus are (110) = ab = (4R/ (3^1/2 ) X (4R (2/3)^1/2) = 16 R² (2)^1/2 / 3

Planar density will be equal to number of atoms centered on plane divided by area of plane.

= 2 atoms / 16 R² (2)^1/2 / 3

Planar density =

queen_honey_blossom answered 8 months ago

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