Solution Manual Research Papers - Academia.edu (original) (raw)

Foreword This collection of problems and solutions is intended to aid students taking our course in Solid State Physics. Exercises are an integral part of a course and the reader is urged to attempt most of them. The problems are selected from areas usually covered in a first course and are of a type most often assigned for class work and given on examinations. No arrangement in order of complexity has been attempted and for some problems only answers are given.
Figure A.1: Scattering from (a) a single electron, (b) two electrons. (c) The scattering
vector s. Note that the vectors k0, k, and s form an isosceles triangle.
where δ is the phase lag of the wave from electron 1 behind that of electron 2 1. (The time
factor has been omitted for the sake of brevity, but its presence is implied.) Referring to
the figure, we may write
δ = (P1M − P1N) 2π/λ = (r · S − r · S0) k
where r is the vector radius of electron 2 relative to electron 1, and S0 and S are the unit
vectors in the incident and scattered directions, respectively. The expression for δ can be
set forth in the form
δ = s · r, (A.3)
where the scattering vector s is defined as
s = k (S − S0) = k − k0. (A.4)
As seen from Fig A.1c, the magnitude of the scattering vector is given by
s = 2k sin(θ) (A.5)
where θ is half of the scattering angle. Substituting the expression Eq (A.3) for δ into
Eq. (A.2), one finds
u′ = fe
A
D
eikD
h
1 + eis·r
i
(A.6)
In deriving this we have chosen the origin of our coordinates at electron 1. But it is now
more convenient to choose the origin at an arbitrary point, and in this manner treat the
two electrons on equal footing. The ensuing expression for the scattered field is then
u′ = fe
A
D
eikD
h
eis·r1 + eis·r2
i
(A.7)
where r1 and r2 are the position vectors of the two electrons relative to a new origin.
Equation (A.6) is a special case of (A.7), where r1 = 0, that is, where the origin is chosen
1The distance D to the field point is assumed to be large, otherwise the denominator D in Eq. (A.2)
would not be the same for the two electrons. This conditions simplifies the calculations, and is the reason
why the detector is usually placed far from the crystal
1. SCATTERING FROM AN ATOM 3
at electron 1, as pointed out above. The generalisation of (A.7) to an arbitrary number
of scatterers is now immediate, and the result is
u′ = fe
A
D
eikD
X
l
eis·rl
where rl is the position of the l th electron, and the sum is carried out over all the electrons.
By analogy with the case of the single electron, equation (A.1), the scattering length for
the system as a whole is now given by the sum
f = fe
X
l
eis·rl . (A.8)
That is, the total scattering length is the sum of individual lengths with the phases taken
properly into account. The intensity I of the scattered beam is proportional to the square
of the magnitude of the field, and therefore
I ∼| f |2= f2
e |
X
l
eis·rl |2 . (A.9)
Results (A.8) and (A.9) are the basic equations in the treatment of scattering and diffraction
processes, and we shall use them time and again in the following pages.
We may digress briefly to point out an important aspect of the scattering process: the
coherence property involved in the scattering. This property means that the scatterers
maintain definite phase relationships with each other. Consequently we can speak of
interference between the partial rays. By contrast, if the scatterers were to oscillate
randomly, or incoherently, the partial rays would not interfere, and the intensity at the
detector would simply be the sum of the partial intensities, that is,
I ∼ Nf2
e ,
where N is the number of scatterers. Note the marked difference between this result and
that of coherent scattering in eq. (A.9).
The scattering length of the electron is well known , and can be found in books on
electromagnetism. Its value is
fe =
h
1 + cos2(2θ)

/2
i1/2
re,
where re, the so–called classical radius of the electron, has a value of about 10−15m. 2
We can now apply these results to the case of a single free atom. In attempting to
apply eq. (A.8), where the sum over the electrons appears, we note that the electrons do
not have discrete positions, but are spread as a continuous charge cloud over the volume
of the atom. It is therefore necessary to convert the discrete sum to the corresponding
integral. This readily leads to
f = fe
X
l
eis·rl → fe
Z
ρ(r)eis·rd3r,
2For the sake of visual thinking, consider the electron to be in the form of a sphere whose radius is
roughly equal to the scattering length fe. Thus the electron ”appears” to the radius as a circular obstacle
of cross section f2
e .
4 A. X-RAY DIFFRACTION
where ρ(r) is the density of the cloud (in electrons per unit volume), and the integral
is over the atomic volume. The atomic scattering factor fa is defined as the integral
appearing in the above expression, i.e,
fa =
Z
ρ(r)eis·rd3r,
(Note that fa is a dimensionless quantity.) The integral can be simplified when the
density ρ(r) is spherically symmetric about the nucleus, because then the integration
over the angular part of the element of volume can be readily performed. The resulting
expression is
fa =
Z R
0
4πr2ρ(r)
sin sr
sr
dr, (A.10)
where R is the radius of the atom (the nucleus being located at the origin). As seen
from eq. (A.10), the scattering factor fa depends on the scattering angle (recall that
s = 2k sin θ), and this comes about from the presence of the oscillating factor (sin sr)/sr
in the integrand. The wavelength of oscillation is inversely proportional to s in Fig A.2a,
and the faster the oscillation – i.e., the shorter the wavelength – the smaller is fa, due
to the interference between the partial beams scattered by different regions of the charge
cloud. Recalling that s = 2k sin θ, Eq (A.5), we see that as the scattering angle 2θ
increases, so also does s, and this results in a decreasing scattering factor fa.
0 3 6 9 12 15
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-0.5
0