Practice Problems Set II
MIME260
P1. For the HCP crystal structure,
(a) show that the ideal c/a ratio is 1.633;
(b) show that the atomic packing factor for HCP is 0.74.
Solution:
(a) A sketch of one-third of an HCP unit cell is shown below.
Consider the tetrahedron labeled as JKLM, which is reconstructed as
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Practice Problems Set II
MIME260
(b) The APF is just the total sphere volume-unit cell volume ratio. For HCP, there are the equivalent of six spheres
per unit cell, and thus
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Practice Problems Set II
MIME260
P2. Assuming hard-sphere model. Calculate the radius of a palladium atom, given that Pd has an FCC crystal
structure, a density of 12.0 g/cm3, and an atomic weight of 106.4 g/mol.
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Practice Problems Set II
MIME260
Solution:
P3. (Assuming hard-sphere model.) Titanium has an HCP crystal structure and a density of 4.51 g/cm3.
(a) What is the volume of its unit cell in cubic meters?
(b) If the ratio is 1.58, compute the values of c and a.
(c) Denote the radius of Titanium as r, do we have 2r > a or 2r < a or 2r = a?
Solution:
(a)
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Practice Problems Set II
MIME260
(b)
(c) 2r < a
P4. (Assuming hard-sphere model.) Below are listed the atomic weight, density, and atomic radius for three
hypothetical metals. For each determine whether its crystal structure is FCC, BCC, or simple cubic and then justify
your determination. A simple cubic unit cell is shown in figure below.
Solution:
ABCC
BSimple cubic
CBCC
P5. (Assuming hard-sphere model.) Methane (CH4) has a tetrahedral structure similar to that of SiO2 (Figure
below), with a carbon atom of radius 0.77 × 10−8 cm at the center and hydrogen atoms of radius 0.46 × 10−8 cm at
four of the eight corners. Calculate the size of the tetrahedral cube for methane.
Solution:
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Practice Problems Set II
MIME260
3a
 rC  rH
2
3a  2  (0.77 108  0.46 108 )
P6 (Assuming hard-sphere model.) Titanium has an HCP unit cell for which the ratio of the lattice parameters c/a is
1.58. If the radius of the Ti atom is 0.1445 nm and atomic weight is ATi = 47.87 g/mol, (a) determine the unit cell
volume, and (b) calculate the density of Ti and compare it with the literature value, 4.51 g/cm3.
REMARK: please note that the Callister solution for this problem IS WRONG. So please DO NOT FOLLOW
Callister solution manual (in case you have one).
Solution:
One important thing to note is that c/a value is smaller than the ideal one, 1.633. As a result, the six atoms on the
bottom layer WILL NOT touch each other (also the three atoms in the middle layer WILL NOT touch each other).
BUT for any atom in the middle layer, it will still touch three atoms in the bottom or top layer. You can imagine this
scenario when you compress the HCP along the c-axis. As a result (denoting the radius of Ti as R),
a  2R!
Instead, we have a  2R
Also follows we have
2
 3   c 2
2
a       2R 

3
2

  
From the above we have
a
2.044 R
And the unit cell volume is
VC 
3 3a 2 c
 1. 0577? 022 cm 3 /unit cell
2
From which we can have

4.51 g / cm3
The same as the value in literature.
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Practice Problems Set II
MIME260
P7. Zinc has an HCP crystal structure, a c/a ratio of 1.856, and a density of 7.13 g/cm3. Compute the atomic radius
for Zn.
REMARK: please note that like P6, Callister’s solution is wrong in its reasoning, though the final answer remains
unchanged.
Solution:
One important thing to note is that c/a value is larger than the ideal one, 1.633. As a result, the six atoms on the
bottom layer WILL touch each other (also the three atoms in the middle layer WILL touch each other). BUT for
any atom in the middle layer, it WILL NOT touch any atoms in the bottom or top layer. You can imagine this
scenario when you elongate the HCP along the c-axis. As a result (denoting the radius of Ti as R),
The relation
a  2R remains valid. And we have
VC =
3 3a 2c
 6 R 2c 3
2
Also follows we have
R  0.133 nm
P8. Within a cubic unit cell, sketch the following directions:


(a)
[1 10 ] ,
(e) [1 1 1] ,
(b)
[1 2 1] ,
(f) [1 22] ,
(c)
[01 2] ,
(d)
[13 3] ,


(g) [12 3 ] ,
(h) [1 03] .


Solution: The directions asked for are indicated in the cubic unit cells shown below.


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Practice Problems Set II
MIME260
P9. Determine the indices for the directions shown in the following cubic unit cell:
Solution :
Direction A is a [430] direction;
Direction B is a [ 23 2] direction;

Direction C is a [13 3] direction;

Direction D is a [136 ] direction.


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Practice Problems Set II
MIME260
P10. What are the indices for the two planes drawn in the sketch below?
Solution :
Plane 1 is a (020) plane. (Please note this is not reduced to (010). Please see the footnote in Callister, P64
for more information.)
Plane 2 is a (221) plane.
P11. Determine the Miller indices for the planes shown in the following unit cell:

Solution
Plane A: it is a (322) plane;
For plane B it is a (1 01) plane.

P12. Convert the (010) and (101) planes into the four-index Miller–Bravais scheme for hexagonal unit cells.

Solution
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Practice Problems Set II
MIME260
For (010), h = 0, k = 1, and l = 0, the value of i is equal to
i   (h  k)   (0  1)  1
Therefore, the (010) plane becomes (011 0) .

Now for the (101) plane, h = 1, k = 0, and l = 1, this leads to

such that (101) becomes (10 1 1) .
i   (h  k)  [1  0]  1


P13. Determine the indices for the planes shown in the hexagonal unit cells below:
Solution
(a) This plane passes through the origin of the coordinate axis system; therefore, we translate this plane
one unit distance along the x axis, per the sketch shown below:
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Practice Problems Set II
MIME260
This is a (10 1 0) plane.

(b) For this plane, intersections with the a1, a2, and z axes are –a, a, and c. In terms of a and c these
intersections are –1, 1, and 1. Using i   (h  k ) , this is a (1 101) plane.

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