Exercises 241
of the reciprocal lattice is located at the point where the beam exits from the spheres—hence
the centres of the spheres are obviously not coincident. Shade in the region of the reciprocal
lattice between the two spheres; planes whose reciprocal lattice points lie in this region
satisfy Bragg’s law. For one reciprocal lattice point in this region, find, by construction,
the sphere which it intersects. The direction of the reflected beam is from the centre of this
sphere through the reciprocal lattice point, and the radius of the sphere gives the particular
wavelength reflected. Draw sections of the reciprocal lattice normal to the [001] (or z
∗
)
direction and which pass through the hk1, hk2, hk
1, hk2, etc. reciprocal lattice points (see
Fig. 6.8(a)). In these sections of the reciprocal lattice ‘above’ and ‘below’ that through the
origin, the sections of the sphere are reducing in size—simple trigonometry will show by
how much. Again, planes whose reciprocal lattice points lie in the region between the spheres
satisfy Bragg’s law.)
9.2 In an oscillating crystal experiment the bcc crystal described in Exercise 9.1 is irradiated
in the [100] (or a
∗
) direction with a monochromatic X-ray beam of wavelength 0.167 nm
(1.67 Å).The crystal is then oscillated±10
◦
about the[001] (or z
∗
) direction. Find the indices
hk0 of the planes in the [001] zone which give rise to reflections during the oscillation of the
crystal.
(Hint: Make a scale drawing of the section of the reciprocal lattice through the origin 000
and normal to the [001] (or z
∗
) direction. Draw a line indicating the [100] direction of
the incident beam and a single sphere corresponding to the single X-ray wavelength (see
Exercise 9.1). Oscillating the crystal (at the centre of the sphere) is equivalent to oscillating
the reciprocal lattice (at the origin). The simplest way to represent the relative changes in
orientation between the crystal and the X-ray beam is to ‘oscillate’the beam. The directions
of the beam at the oscillation limits are ±10
◦
from the [100] direction in the plane of the
reciprocal lattice section. Draw in the reflecting sphere at these limits and shade in the
lunes or the regions of reciprocal space through which the surface of the sphere passes.
Planes whose reciprocal lattice points lie in these regions reflect the X-ray beam during
oscillation.)
9.3 The kinetic energy of neutrons emerging in thermal equilibrium from a reactor is given
by 3/2 kT where k = Boltzmann
s constant, 1.38 × 10
−23
JK
−1
and T is the Kelvin
temperature. Given that the (rest) mass M
n
of a neutron = 1.67x10
−27
kg and Planck’s
constant h = 6.63 ×10
−34
Js, estimate the wavelength of neutrons in thermal equilibrium
at 100
◦
C.
9.4 Determine the F
hkl
values for reflections for the ZnS (zinc blende or sphalerite structure,
Fig. 1.14(c). Show that they fall into three groups: (h + k + l) = 4n; (h + k + l) =
4n +2; (h +k + l) = 2n +1.
Hint: ZnS has a cubic F lattice, each lattice point being associated with a motif consisting
of one Zn atom at (000) and one S at (
1
/
4
1
/
4
1
/
4
) . F
hkl
is determined (i) by writing down
the reflection condition for cubic F crystal (see Appendix A6) and (ii) by substituting f
Zn
at
(000) and f
S
at (
1
/
4
1
/
4
1
/
4
).
9.5 In diamond the atom positions are identical to those in ZnS. Hence determine the conditions
for reflection in the diamond cubic lattice.
Hint: We may simply proceed as before. However, since all the atoms are now of the same
type, diamond has the centrosymmetric point group m
¯
3m, the centres of symmetry lying
equidistant between nearest neighbour atoms. We may therefore choose an origin at (
1
/
8
1
/
8
1
/
8
)—called ‘origin choice 2’ in space group Fd
¯
3m, No. 227—and use the simplified
structure factor equation in Example 3, page 210.