Question

For a given primitive cell volume V, which monoatomic crystal will have its nearest neighbours furthest away, between simple cubic, BCC, and FCC?

Answer #1

In the bcc structure each atom has 8 nearest neighbours (coordination number) at a distance of

**d = 2r = sqrt(3a/2)a** approximately
**0.866a** and 6 next-nearest neighbours at a distance
of

**d= a 2.3r** approximately **1.15 d**
. It is remarkable that there is a smaller number of nearest

neighbours compared to the close-packed structures but for the
**bcc structure the next-nearest**

**neighbours are only slightly further away** which
makes it possible for those to participate in bonds.

--Given Values--
Atomic Radius (nm) = 0.116
FCC Metal = Gold
BCC Metal: = Sodium
Temperature ( C ) = 1017
Metal A = Tin
Equilibrium Number of Vacancies (m-3) = 6.02E+23
Temperature for Metal A = 369
Metal B = Gallium
1) If the atomic radius of a metal is the value shown above and it
has the face-centered cubic crystal structure, calculate the volume
of its unit cell in nm3? Write your answers in
Engineering Notation.
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