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Solutions to Homework _2

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					                     Solutions to Homework #2
材導                                                                           陳智 10/90

3.20 This problem asks that we calculate the unit cell volume for Re which has an HCP crystal
    structure. In order to do this, it is necessary to use a result of Problem 3.7, that is


                                                  2
                                            V = 6R c 3
                                             C

       The problem states that c = 1.615a, and a = 2R. Therefore
                                                               3
                                    V = (1.615)(12 3 )R
                                     C

                                       -8      3            -23   3           -2  3
          = (1.615)(12 3 )(1.37 x 10        cm) = 8.63 x 10     cm = 8.63 x 10 nm


                                                   _
3.26 (a) This portion of the problem asks that a [01 1] direction be drawn within a monoclinic

    unit cell (a ≠ b ≠ c, and  =  = 90≠ ). One such unit cell with its origin at point O is
    sketched below. For this direction, there is no projection along the x-axis since the first
    index is zero; thus, the direction lies in the y-z plane. We next move from the origin
    along the minus y-axis b units (from point O to point R). Since the final index is a one,
                                                                                _
    move from point R parallel to the z-axis, c units (to point P). Thus, the [01 1] direction

    corresponds to the vector passing from the origin to point P, as indicated in the figure.


                                                   z




                        -                                                              c
             P        [011]

                                                                                              y
                                                          

                                                   O
                                                                             a

     -y          R
                                                           b
                              x
    (b) A (002) plane is drawn within the monoclinic cell shown below. We first remove the
    parentheses and take the reciprocals of the indices; this gives   and 1/2. Thus, the
                                                                     , ,
    (002) plane parallels both x- and y-axes, and intercepts the z-axis at c/2, as indicated in
    the drawing.


                                      z



                                                                            c


                                                                                y
                                             

                                      O
                                                                  a


                                              b
                   x


                       _
3.30 Direction A is a [4 30] direction, which determination is summarized as follows. We first

    of all position the origin of the coordinate system at the tail of the direction vector; then in
    terms of this new coordinate system


                                                   x                 y                z
                                                    2a              b
    Projections                                   - 3               2                 0c

    Projections in terms of a, b,
                                                    2               1
             and c                                - 3               2                 0


    Reduction to integers                          -4               3                 0
                                                                   _
    Enclosure                                                     [4 30]

                                _
             Direction B is a [23 2] direction, which determination is summarized as follows.

    We first of all position the origin of the coordinate system at the tail of the direction vector;
    then in terms of this new coordinate system
                                                   x                 y                z
                                                  2a                                 2c
    Projections                                    3                -b               3

    Projections in terms of a, b,
                                              2                                   2
        and c                                 3                 -1                3

Reduction to integers                          2                -3                2
                                                                _
Enclosure                                                     [23 2]

                           __
        Direction C is a [133 ] direction, which determination is summarized as follows.

We first of all position the origin of the coordinate system at the tail of the direction vector;
then in terms of this new coordinate system
                                               x                 y                z
                                              a
Projections                                   3                 -b                -c

Projections in terms of a, b,
                                              1
        and c                                 3                 -1                -1

Reduction to integers                          1                -3                -3
                                                                __
Enclosure                                                     [133 ]

                            _
        Direction D is a [136 ] direction, which determination is summarized as follows.

We first of all position the origin of the coordinate system at the tail of the direction vector;
then in terms of this new coordinate system


                                               x                 y                z


                                              a                 b
Projections                                   6                 2                 -c

Projections in terms of a, b,
                                              1                 1
        and c                                 6                 2                 -1

Reduction to integers                          1                 3                -6
                                                                 _
Enclosure                                                     [136 ]
           _            _
3.36 The (11 01) and (112 0) planes in a hexagonal unit cell are shown below.



                     z                                           z




                          a                                          a
                           2                                          2



a                                          a
    3                                          3                          a
                                   a
                                       1                                      1
                    _                                            _
                  (1101)                                      (1120)



3.45 Planar density, PD, is defined as


                                                        A
                                                          c
                                                   PD = A
                                                         p


        where A is the total plane area within the unit cell and A is the circle plane area within
               p                                                  c
        this same plane. For the (100) plane in BCC, in terms of the atomic radius, R, and the unit
        cell edge length a


                                                        2
                                               4R 2 16R
                                       A =a =   = 3
                                           2
                                        p      3


        Also, upon examination of that portion of the (100) plane within a single unit cell, that there
        resides a single equivalent atom--one-fourth from each of the four corner atoms.
        Therefore,


                                                             2
                                                   A = R
                                                    c

          Hence
                                                         2
                                                    R
                                           PD =          2 = 0.59
                                                   16R
                                                     3
            That portion of a (110) plane that passes through a BCC unit cell forms a rectangle
    as shown below.




                                                 R                   4R
                                                                      3




                                              4R 2
                                                 3


    In terms of the atomic radius R, the length of the rectangle base is 4R 2    /   3 , whereas
                        4R
    the height is a =      . Therefore, the area of this rectangle, which is just A is
                         3                                                         p


                                                       2
                                       4R 24R = 16R 2
                                 A =       
                                  p     3  3       3



    Now for the number equivalent atoms within this plane. One-fourth of each corner atom
    and the entirety of the center atom belong to the unit cell.          Therefore, there is an
    equivalent of 2 atoms within the unit cell. Hence


                                                      2
                                              A = 2(R )
                                               c

      and
                                                    2
                                               2R
                                       PD =       2   = 0.83
                                              16R 2
                                                 3

3.49 Although each individual grain in a polycrystalline material may be anisotropic, if the
    grains have random orientations, then the solid aggregate of the many anisotropic grains
    will behave isotropically.


3.50 From the table, molybdenum has a BCC crystal structure and an atomic radius of 0.1363
    nm. Using Equation (3.3), the lattice parameter a may be computed as
                        4R   (4)(0.1363 nm)
                   a=      =                = 0.3148 nm
                         3           3


Now, the interplanar spacing d         maybe determined using Equation (3.10) as
                                 111


                              a                    0.3148
             d    =                            =          = 0.1818 nm
              111         2       2        2          3
                        (1) + (1) + (1)

				
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