# The structure of crystalline solids problem solutions

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CHAPTER 3

THE STRUCTURE OF CRYSTALLINE SOLIDS PROBLEM SOLUTIONS

Fundamental Concepts
3.1 Atomic structure relates to the number of protons and neutrons in the nucleus of an atom, as well as the number and probability distributions of the constituent electrons. On the other hand, crystal structure pertains to the arrangement of atoms in the crystalline solid material.

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Unit Cells Metallic Crystal Structures
3.2 For this problem, we are asked to calculate the volume of a unit cell of lead. Lead has an FCC crystal structure (Table 3.1). The FCC unit cell volume may be computed from Equation 3.4 as
VC = 16R 3 2 = (16) (0.175 × 10-9 m) 3( 2 ) = 1.213 × 10-28 m3

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3.3 This problem calls for a demonstration of the relationship a =cell shown below

4R for BCC. Consider the BCC unit 3

Using the triangle NOP
(NP) 2 = a 2 + a 2 = 2a 2

And then for triangle NPQ,
(NQ) 2 = (QP) 2 + ( NP) 2

But NQ = 4R, R being the atomic radius. Also, QP = a. Therefore,
(4R) 2 = a 2 + 2a 2

or
a =

4R 3

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3.4 We are asked to show that the ideal c/a ratio for HCP is 1.633. A sketch of one-third of an HCPunit cell is shown below.

Consider the tetrahedron labeled as JKLM, which is reconstructed as

The atom at point M is midway between the top and bottom faces of the unit cell--that is MH = c/2. And, since atoms at points J, K, and M, all touch one another,
JM = JK = 2R = a

where R is the atomic radius. Furthermore, from triangle JHM,
(JM ) 2 = ( JH ) 2 + ( MH ) 2

or

⎛ c ⎞2 a 2 = (JH )2 + ⎜ ⎟ ⎝2 ⎠
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Now, we can determine the JH length by consideration of triangle JKL, which is an equilateral triangle,

cos 30° =

a /2 = JH a 3

3 2

and
JH =

Substituting this value for JH in the above expression yields ⎛ a ⎞2 ⎛ c ⎞ 2 a2 c2 a2 = ⎜ ⎟ + ⎜ ⎟ = + ⎝ 3⎠ ⎝2 ⎠ 3 4

and, solving for c/a
c = a 8 = 1.633 3

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