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Solution Exercise Sheet 1
JCNS neutron labcourse 2015
Excercise 1.2
Exercise Sheet 1
Exercise 1.1 Multiple Choice
• visible light
• 0.1nm
• 10fm
• 0.1nm
• of course neutrons
• X-ray in case of nanoparticles and electrons in case of thin layers
• 107
Exercise 1.2 Comprehension
a scattering: understanding of microscopic and tomic structure of matter
imaging: investigation of surface
b X-ray scattering is determined by the density of electrons within a crystal. A crystal structure is composed of a pattern, a set of atoms arranged in a particular way, and a lattice exhibiting
long-range order and symmetry. Glass is a non-crystalline solid material.
c Neutrons have a magnetic moment
d: Scattering Probes
As in science the particle–wave–dualism was accepted the real breakthrough for structure studies of
condensed matter systems begun. In the following picture different scattering probes are illustrated
by their qualitative location of interaction with condensed matter:
As in science the particle–wave–dualism was accepted the real breakthrough for structure studies of
condensed matter systems begun. In the following picture different scattering probes are illustrated
by their qualitative location of interaction with condensed matter:
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Solution Exercise Sheet 1
JCNS neutron labcourse 2015
Excercise 1.2
In general, for scattering experiments using any kind of scattering probe it really depends on which
part of the spectrum the (particle) wave interacts with matter. For instance, the electromagnetic
spectrum is arranged into different types of radiation characterized by their wavelength.
Figure 1: The electromagnetic spectrum.
An EM wave is described by the physical properties: frequency ν, wavelength λ and energy E all
physical quantities are combined in the given formula:
E =~·ω =h·ν =
h·c
λ
Light
Light has wavelengths between 350 nm ≤ λ ≤ 750 nm. The length scales become much larger than
the diameter of an atom. Thus, the atomic structure is negligible. According to Maxwell’s equations
light is characterized in matter by the dielectric and polarisation tenser. In general, light scatters at
density and concentration fluctuations
δα = 2 · 0 · nsolvent ·
∂nsolution M
·
∂cm
NA
where δα is the polarisation and n the refractive index, ∂nsolution
refractive increment, M the molecular
∂cm
mass, NA Avogadro number and 0 the vacuum permittivity. For small isotropic particles in solution
the intensity of unpolarized light with an wavelength λ scattered by a single particle is given by
Is
α2 · 8π 4
= 2 4 · (1 + cos2 θ)
I0
r ·λ
where θ is the scattering angle and α the molecular polarizability describing the charge displacement
of a molecule in an electric
field.
r2
The ratio RΘ = Is (θ)
is
called
Rayleigh ration relates the scattered intensity by taking a Virial
I0
expansion to the properties of the scattering object due to its shape, form factor P (Q), molecular
weight Mw and interactions between the surrounding particles A2 depending on the concentration
c.
2
∂n
Mw
2π 2 · n2solvent
∆RΘ = RΘ,solution − RΘ,solvent =
·
·
· c · (1 + cos2 θ)
4
λ
∂c
NA
= K · c · Mw · (1 + cos2 θ)
K · c · (1 + cos2 θ)
1
=
+ 2A2 · c + . . .
∆RΘ
Mw · P (Q)
2
Solution Exercise Sheet 1
JCNS neutron labcourse 2015
Excercise 1.2
The Rayleigh theory only holds for dimensions of the particle smaller than the wavelength of incoming
beam (a ≤ λ/20). Hence, the particles can be assumed as point-like and the scattered intensity is
isotropic. For wavelengths λ/20 ≤ a ≤ λ/2 the shape of the scattering object becomes important
and the radiation is scattered at different positions in the molecule (Mie-scattering).
X-rays
X-ray radiation is typically in the range of 1 pm < λ < 10 nm corresponding to energies in the range
E = 10 keV for λ ≈ 1Å (slit experiments, on the length scale of atomic distances!) and E = 2 eV for
λ ≈ 5000Å. The related quantum mechanical particle of the EM wave is a photon with zero mass,
spin one and zero charge. The interaction of a photon is limited to the Coulomb interactions within
the electron cloud of an atom. The penetration depth of the x-ray radiation clearly depends on the
its energy. For instance, hard X-rays (10 keV to 120 keV) can easily penetrate solid matter while
X-rays of lower energies hardly penetrate matter, e.g. 600 eV x-rays only have an attenuation length
of about several nanometer.
In contrast to neutron scattering the scattering amplitude for X-rays dependent on the atomic number
Z meaning the more electron an element has the bigger is the surrounding electron cloud. The X-ray
scattering amplitude is than given as a function of the atomic number f (x − ray) ∝ Z · rel where
rel is the classical electron radius. X-ray scattering is nowadays rather easy to handle especially the
production of X-rays is relatively simple in contrast to neutron production, using a typical X-ray tube
high fluxes are achievable yet. But there is a disadvantage of X-ray scattering: radiation damage
Electrons
In contrast to photons the electron is an elementary particle carrying a negative electric charge
of q = 1.602 × 10−19 C and has a spin of 1/2. According to its low mass approximately me =
9.109 × 10−31 kg at high energies the electron’s speed almost approaches the speed of light. Considering Einstein, the kinetic energy for a relativistic electron is given by
r
h2 · c2
− me− · c2
relativistic kinetic energy
E = m2e− · c4 +
λ2
h2
E=
classical limit
2 me− · λ2
Some important features of electron radiation used for condensed matter studies:
• interaction with matter trough electrostatic mechanism
• strong scattering on highly dense matter with large atomic numbers (mostly interaction with
electron cloud)
• at very high energies also interaction contributions with nucleus (scattering depends on the
incident electron velocity or energy)
• with increasing penetration depth the beam becomes more divergent, however at large atomic
numbers there is scattering saturation (penetrating radiation depth R depends on the density
R ∼ %−1 and the energy of the electrons R ∼ Ee )
• advantage: can be deflected using electric or magnetic fields, easy in instrumentation
• disadvantage: only thin sample volumes, multiple scattering effects at large scattering angles,
Bremsstrahlung (safety guarantee)
α-particles 24He2+
The α-particle is a doubly-ionized helium nuclei with two protons and two neutrons. The particle
carrying two positive charges (q = 2e), its mass is mα = 6.645 × 10−27 kg (m = 4 u) and its net spin
2
is zero. The classical kinetic energy E = 2 mh4u ·λ2 is in the range of some MeV. According to a sharp
defined low penetration depth, a simple sheet of paper of some cm thickness is in general enough
for total shielding. α’s are not dangerous for the human body unless a source of α radiation is not
incorporated. Because of the short range of absorption they become extremely unhealthy in case of
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Solution Exercise Sheet 1
JCNS neutron labcourse 2015
Excercise 1.3
inhalation or food intake. Finally, the biological effect can be quantized as 20 times higher than an
equivalent of beta or gamma radiation.
e: CO2 Probe - Molecular Beam Scattering
Replacing the neutron by a CO2 beam has many consequences for the scattering due to the properties
of a molecular beam. The CO2 is a rather big in size compared to a neutron (44 u) - rotational,
translational, vibration states are possible. This leads to an interaction of the molecule with the
electron cloud only on the surface of the sample. The penetrating depth will be nearly zero due to
strong Coulomb forces since neutrons can easily go through. We also have the problem of creating
a defined beam in terms of collimation and coherence. Then the CO2 molecule is non isotropic and
molecule can no longer be assumes as point-like. All together makes calculations very complicated
and many surprising and unexpected result will come out of such an experiment.
Exercise 1.3 Huygens Principle and Coherence
From Babinet’s principle the interference pattern is connected to the double slit experiment with
infinite small slit sizes. The theorem states that the diffraction pattern of a particle is rather identical
to diffraction at a hole of the same size and shape, e.g. compare diffraction of a isotropic particle
with a spherical diaphragm. The intensity distribution of the double slit interference pattern is given
by
2
2
sin β
sin β
2 πL
2 ∆φ
= 4I0 cos
·
sin θ ·
I(θ) = 4I0 cos
2
β
λ
β
π × slit size
β=
sin θ
λ
In conclusion the intensity maximum diminishes at higher distances from the maximum of zeroth
order.
a.) Where are the interference maxima?
The picture below explain the situation:
If the condition ∆s = L sin θ = m λ, m ∈ Z is fulfilled, than constructive interference occur. From
a.) and b.) (ψ = 0) we can calculate the position of the maximum using
tan α =
xmax
D
In the far-field limit all angles become small leading to the simplification:
tan α ≈ α
sin θ ≈ θ
θ≈α
⇒ sin θ ≈ tan α
and the interference maximum is given by
xmax =
D
D·m·λ
∆s =
≈ Dθ
L
L
So the angles θ, where interference maxima occur in the far field limit are:
θ≈
xmax
D
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Solution Exercise Sheet 1
JCNS neutron labcourse 2015
Excercise 1.3
Furthermore, the phase difference ∆φ of the scattered waves are related to the path length difference
∆s:
2π
2π
∆s =
L sin θ = 2π m, m ∈ Z
∆φ =
λ
λ
b.) and c.) What happens with the interference pattern?
1. broad wavelength distribution by well defined constant propagation direction:
• The total intensity as a function of the scattering angle θ is defined as (infinite slit size)
2π L
2 πL
2 ∆φ
= 4I0 cos
sin θ = 2I0 1 + cos
sin θ
I(θ) ∼ 4I0 cos
2
λ
λ
Thus, the maximum width will become broader and the resolution of the pattern will
become worse (R = λ/∆λ). For each wavelength of the incoming beam the frequency of
the cosine is changed. If the initial beam consists of a wavelength distribution, than the
superposition of each diffraction pattern resulting from each wavelength lead to wiping
out the integral diffraction pattern.
2. same wavelength (monochromatic waves) but different incident beam angles:
• Considering a more general formula of Bragg’s equation see figure b.) (ψ 6= 0):
m·λ
L (sin θ − sin ψ) = m λ = ∆sψ + ∆sθ
xmax = D ·
+ sin ψ
L
then we know that constructive interference occurs if the phases of the waves are an integer
multiple of the wavelength and if the incident waves are coming under the angle ψ, then
the total path length difference is additive. The result on the scattering pattern is
2 ∆φ
2 πL
I(θ) ∼ 4I0 cos
= 4I0 cos
(sin θ − sin ψ)
2
λ
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Solution Exercise Sheet 1
JCNS neutron labcourse 2015
Excercise 1.3
The important difference to the previous question is that now the frequency is constant
but the phase change results in a shift of the diffraction pattern. This effect is typically
observed for an extended sources. The interference pattern are almost eliminated due to
the loss of spatial coherence.
d.) Using a normal light bulb for scattering experiment.
The light source (bulb) is an extended object. Thus, the emission happens at different positions on
the glow wire. Therefore, the emitted waves are highly incoherent. Their phases are hardly correlated
because of a glow wire consists of many atoms stochastically emitting the light (no fixed relationship
over the coherence time - spatial incoherent). But, interference structures are only observed within
the coherence volume. Apart from coherence condition, the emitted light is polychromatic. The
wavelength spectra of a light bulb covers a range from 400 nm to 700 nm being rather shifted to the
infra-red. The characteristic color temperature of a light bulb is about 2300 K to 2900 K compared to
daily light 5000 K to 7000 K. For instrumentation we need to select one wavelength in the order of the
length scale of our object. To select a certain wavelength from the light spectrum a monochromator
is needed for our instrument. The monochromator uses either the phenomenon of optical dispersion
in a prism, or that of diffraction using a diffraction grating to split the white beam into its colors.
By special collimation of this light it is possible to chose one wavelength. In addition, diaphragms
(a small slit) define the propagation direction of the incident beam. Using lenses a rather parallel
beam can be created. The collimation at all ensures a homogeneous and controllable illumination of
the object. For observing a diffraction pattern it is important that the particle sizes are smaller than
the wavelength of the our beam.
e.) longitudinal, transverse coherence and resolution
Definition 1 (Coherence) Waves are coherent if the time dependence of their electric fields is
equal unlike a small phase shift τ .
The coherence length lc = c τ is defined with τ ≈ 10−8 s (characteristic time for emitting a photon). If
the path length difference is greater than the lc , than the interference pattern follows from scattered
waves out of phase.
Definition 2 (longitudinal coherence) The longitudinal coherence is a measure of the distance
over which two wave emitted with slightly different wavelength from the same source completely dephase. The name is referred to the propagation direction of the wave.
Definition 3 (transverse coherence) The transverse coherence measures the lateral distance along
two wavefronts with the same wavelength coming from the same source but at different points completely dephase.
The resolution A of our instrument will be good as far as the longitudinal coherence length is high
lc =
λ
λ2
=
·λ=A·λ
∆λ
∆λ
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Solution Exercise Sheet 2
JCNS neutron labcourse 2015
Excercise 2.2
Exercise Sheet 2
Exercise 2.1 Multiple Choice
• 1 nm
• 1 fm
• 3 He
• Cd
• 100%58 Ni
• non of above
• The phase problem does not allow one to determine the atomic position directly by a simple
mathematical procedure.
Exercise 2.2 Bragg Scattering
a The conditions required for the achievement of coherent superposition are:
• The incident angle has to be equal to the scattering angle.
• The difference in the path length between two waves reflected by two adjacent planes has to
be equal to an integer number of wavelength.
b If the condition ∆s = nλ(1) with n ∈ Z is fulfilled, then constructive interference occurs.
∆s = DE + EF = 2 · DE (DE = EF, see sketch)
DE = d · sin Θ
⇒ ∆s = 2 · d · sin Θ
(1) ⇒ n · λ = 2d · sin Θ
c One can easily show the equivalence of Bragg and Laue with the construction of the well-known
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Solution Exercise Sheet 2
JCNS neutron labcourse 2015
Excercise 2.4
Ewald sphere (see sketch):
From the lecture one knows
2π
d = ~
G
~ =n·G
~ = ~k − k~0 (Laue condition). Furthermore one can see
and Q
~
n · G
/2
.
sin Θ =
|k|
With k =
2π
λ
(1)
(2)
and d we get the Bragg Equation and, thus, the equivalence is shown.
Exercise 2.3 Neutron Scattering from Ti-Zr alloys
a The coherent Bragg reflection is zero if the coherent scattering length is zero
−x < b >Ti = (1 − x) < b >Zr
< b >Ti = −3.4 fm < b >Zr = 7.1 fm
x = (< b >Zr )/(< b >Zr − < b >Ti ) = 0.67
Alloys stoichiometry: Ti0.67 Zr0.33
b The disadvantage of this chamber can be related to the high absorption cross section of Titanium
Exercise 2.4 Neutron Absorption
a v = 2200m/s
h
λ = v·m
N
λ = 1.798 Å
0.231 barn : 1.798 Å
(3)
x : 1Å
(4)
⇒ σa = 0.129barn
b n − γ resonance does not occur, due to the very small absorption cross section of Aluminium.
One has to consider Diffraction as well.
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Solution Exercise Sheet 2
c
JCNS neutron labcourse 2015
2
σa = 0.129 barn = 0.129 · 10−8 Å
Aluminium has a face centered cubic crystal structure
4 atoms/unit cell
2
σa · 4 = 0.516 barn = 0.51610−8 Å
3
Vcell = (4.049)3 = 66.41Å
2
3
−1
µ = (0.516 · 10−8 Å )/66.41Å = 7.7699 · 10−11 Å
d x = 10 cm = 1 · 109 Å
I
I0
= exp(−µ · x) ∼ 0.9252
9
Excercise 2.4
Solution Exercise Sheet 3
JCNS neutron labcourse 2015
Excercise 3.3
Exercise Sheet 3
Exercise 3.1 How are neutrons characterized
Kinetic energy of a free neutron as a function of its momentum and its velocity:
Ekin =
p2
v2 m
=
2
2m
λ=
h
h
=√
p
2mEkin
=⇒ Ekin =
h2
v2 m
=
2mλ2
2
=⇒ v =
h
mλ
Solutions to the given wavelengths:
λ [Å] Ekin [meV ]
1
82
1.8
25
5
3.3
v [ ms ]
3956
2198
791
• It should be advised, that the usage of decimal places is disputable due to a common wavelength
resolution of about 2 - 3% (up to 20% @ KWS-2) and energy resolution of about 4-6%.
• It should be pointed to the temperature equivalent of the unit [eV]: 1 eV =
b 11604.5 K, so that
b
25 meV ≈ 300 K .
Exercise 3.2 How many neutrons are produced?
For instance the FRM-II has a power of 20 MW and assuming that the flux maximum is displaced
10 cm from a point-like core the neutron flux density φ can be calculated:
235
U + n −→ fragments + 2.52n + 180 MeV;
yield ≈ 1
n
fission
20 MW = x · 180 · 1.602 · 10−19 MWs
20
MW
n
x=
= 6.936 · 1017
−19
180 · 1.602 · 10
MW · s
s
surface area of a sphere with a radius r = 10 cm:A = 4π (10cm)2 = 1256.64cm2
φ=
x·1 n
6.936 · 1017 · 1
n
n
14
=
5.519
·
10
=
A
1256.64
s · cm2
s · cm2
Now, assuming a hypothetical spallation source with the same thermal power of 20 MW. What is its
neutron flux density?
238
Uspallation −→ 50
x=
φ=
MeV
;
n
yield ≈ 40
n
spallation
20
MW
1
= 6.242 · 1016
−19
40 · 50 · 1.602 · 10
MW · s
s
x · 40n
6.242 · 1018 · 40
n
n
=
= 1.987 · 1015
2
S
1256.64
s · cm
s · cm2
10
Solution Exercise Sheet 3
JCNS neutron labcourse 2015
Excercise 3.3
Exercise 3.3 How do the neutrons come to your experiment?
1. How is the neutron flux reduced, when you build the diffractometer/spectrometer at larger
distance without a neutron transport system?
• beam divergence: reduction with 1/r2
2. When is it advantageous to have the instrument close to the neutron source?
• when there is no neutron guide material for the neutron wavelength you need (hot neutrons)
• when you need highest flux and your experiment is relatively unsensitive for background
• at a pulsed source with short pulses, the resolution can become too good if your instrument
is too far away from the source, so that the wavelength band becomes too small.
3. What reasons can you imagine to separate the instrument from the neutron source?
• lower γ background (curved neutron guide)
• ’instrument density’ is lowered, more space available for larger instruments
• if you want to focuss with an elliptic neutron guide
• if you need a large collimation path (e.g. SANS)
• if you need a good wavelength/energy resolution (e.g. chopper systems)
• for ’colder’ neutrons
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Solution Exercise Sheet 4
JCNS neutron labcourse 2015
Excercise 4.1
Exercise Sheet 4
Exercise 4.1
a) The lattice points uvw = 030, -120, 1-20, and 450
b) The lattice directions [uvw] = [100], [210], [-2-10] and [-250]
c) The traces of the lattice planes (hkl) = (100), (210), (-210), and (140)
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Solution Exercise Sheet 4
JCNS neutron labcourse 2015
Excercise 4.2
Exercise 4.2
a) Draw the positions of all atoms (Y, Ba, Cu, O) into the above given
projection.
b) Given the space group P2/m2/m2/m: What is the crystal system and
the Bravais lattice type?
Orthorhombic; Bravais lattice type P (primitive)
c) How many formula units are in one unit cell?
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Solution Exercise Sheet 4
JCNS neutron labcourse 2015
Excercise 4.4
1 formula unit per unit cell
d) Give the coordination numbers and describe the polyhedra of oxygen
around atoms Cu1 and Cu2
Cu1: coordination number: 4; planer square.
Cu2: coordination number: 5; square pyramid (pentahedron).
e) The atom sits on an inversion centre:
Cu1; Y; O4.
f ) Calculate the interplanar spacings d(hkl) for the lattice planes (100),
(200), (020), (002), (00-2)
2
For orthorhombic crystal system: dhkl = ( ha2 +
2
1
2
1
k2
b2
+
l2 − 12
)
c2
1
d100 = ( (3.858
+ 0 + 0)− 2 = 3.858 Å;
Å)2
2
+ 0 + 0)− 2 = 1.929 Å;
d200 = ( (3.858
Å)2
d020 = (0 +
22
(3.846Å)2
1
+ 0)− 2 = 1.923 Å;
d002 = d00−2 = (0 + 0 +
22
− 12
)
2
(11.680Å)
= 5.86 Å
g and h) List all symmetry equivalent lattice planes with identical dspacing for the following types of lattice planes:(h00), (00l), (0kl), (hkl)
and their multiplicity factor M of reflections
d(h00) = d(-h00); multiplicity factor M(h00) = 2
d(00l) = d(00-l); multiplicity factor M(00l) = 2
d(0kl) = d(0-kl) = d(0k-l) = d (0-k-l); multiplicity factor M(0kl) = 4
d(hkl) = d(-hkl) = d(h-kl) = d(hk-l) = d(-h-kl ) = d(-hk-l) = d(h-k-l) = d(-h-k-l); multiplicity factor
M(hkl) = 8
Exercise 4.3 Types of scattering Experiments
a) Discuss/define the following terms:
A. Elastic scattering: there is no energy transfer between incoming and scattered beams (Energy
conservation of the particle or quantum during the scattering process).
B. Inelastic scattering: there is a transfer of both momentum and energy between incoming and
scattered beams (there is a loss or gain of particle or quantum energy during the scattering event).
C. Coherent scattering: which involves the interference of waves. There is a well-defined relationship
between the phase of the incoming wave and the phase of the outgoing wave.
D. Incoherent scattering: scattering without interference. There is no well-defined such relationship
is called incoherent scattering.
What does the term “diffraction” correspond to in this context?
elastic, coherent scattering
Exercise 4.4 Ewald Constrution
(a) Sketch the Ewald-construction for a single crystal experiment.
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Solution Exercise Sheet 4
JCNS neutron labcourse 2015
Excercise 4.5
A most useful means to understand the occurrence of
diffraction spots is the Ewald construction. We draw a
sphere of radius 1/λ, in the center of which we imagine the
real crystal. The origin of the reciprocal lattice lies in the
transmitted beam, at the edge of the Ewald sphere.
We know already that diffraction maxima (reflections,
diffraction spots) occur only when the 3 Laue equations,
or equivalent, the Bragg equation in vector form, are satisfied. This condition occurs whenever a reciprocal lattice
point lies exactly on the Ewald sphere.
As you may have assumed already, the chance for this to
occur is modest, and we need to rotate the crystal in order
to move more reciprocal lattice points through the Ewald
sphere. In the following, a reciprocal lattice is drawn in the
origin, and we rotate it along the vertical axis of the drawing. We actually accomplish this by rotating the crystal
along the same axis.
b) Ewald-construction for a beam with zero divergence but non-vanishing
wavelength-spread ∆λ/λ
Exercise 4.5 Filtering
a) the purpose of a beryllium (graphite) filter for neutron diffraction
Filters have been used to minimize the λ/2 contamination which suppress the higher orders stronger
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Solution Exercise Sheet 4
JCNS neutron labcourse 2015
Excercise 4.7
than the desired wavelength. Polycrystalline beryllium and graphite are frequently used as filter for
cold neutron experiments. Due to their unit cell dimensions, they block higher orders of wavelengths
smaller than about 3.5 Åand 6 Å, respectively.
b) discuss how it works
These filters use the fact that there is no Bragg diffraction
if λ > 2dmax , where dmax is the largest interplanar spacing
of the unit cell. As can be seen in the sketch, for long
wavelengths the Ewald sphere is too small to be touched by
any reciprocal lattice point, so that the desired wavelength
can pass through the filter without being attenuated by
Bragg diffraction. The shorter wavelengths are diffracted
sideways out of the main beam into a neutron absorbing
material such as B4C. The higher order reflections can be
suppress very effectively.
Exercise 4.6 Structure determination & refinement
a) Describe, in simple terms, the “phase problem of crystallography”
When a crystal is irradiated with a beam of X-rays the resulting interference effect gives rise to the socalled diffraction pattern which is uniquely determined by the crystal structure. Only the intensities
of the scattered rays can be measured; the phases, which are also needed in order to work backwards,
from diffraction pattern to the atomic positions, are lost in the diffraction experiment. However,
owing to the known atomicity of real structures and the large number of observable intensities, the
lost phase information is in fact contained in the measured intensities. The problem of recovering
the missing phases, when only the intensities are available, is known as the phase problem.
A scattering experiment is equivalent to performing a Fourier transform of the scattering object
followed by taking the square of the resulting complex amplitude (the diffracted intensity I ∼ A2 ).
Because our detector can only measure the magnitude (the absolut value) of a diffracted wave but
are completely insensitive to its phase.
b) ”structure refinement” & ”structure determination”
Structure refinement: Once we have estimated the intensities and phases of the reflections, we can
calculate the electron density within the crystal and build an approximate model by placing atoms
into the density as appropriate. Use the factor formula to calculate diffraction intensities from the
model. To see how well your model explains the data, you ”back” calculate the diffraction pattern
you expect to see, given your model, and compare it to the diffraction you actually see. The model
is iteratively improved by cycling back and forth between structure factors and electron densities
(by Fourier and inverse Fourier transformations) to give an optimum match between observed and
calculated intensity.
Structure determination: use some methods to reconstruct the missing phase information from
the measured magnitudes and from a-priori information (e.g. type of atoms, number of atoms,
symmetry...) about the scattering object.
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Solution Exercise Sheet 4
JCNS neutron labcourse 2015
Excercise 4.8
Exercise 4.7 Structure factor equation
a) Write down the structure factor equation
F (τ ) =
X
bj · exp[2πi(τ · r j )] · Tj (τ ) = |F (τ )| · exp[iφ(τ )]
(5)
j
b) Identify and discuss all parameters in the formula
The structure factor F (τ ) is the Fourier transform of the scattering density within the unit cell
containing the complete structural information.
τ : reciprocal lattice vector
bj (τ ) = bj = const. : the scattering length of nucleus j.
r j = xj a 1 + yj a 2 + zj a 3 : atomic coordinates
Ti (τ ) : Debye-Waller factor including information about dynamical and static displacements (site
occupations and the thermal vibrations) of the nucleus j from its average position r j in the unit
cell
c) Under which conditions does this formula hold (kinematical diffraction
conditions)?
This formula holds only for elastic scattering. The magnitude of the incident wave is the same at
all points in the specimen (this implies a small sample size, weak interaction between radiation and
matter, no multiple diffraction and negligible absorption) and that the diffracted beams are much
weaker than the primary beam.
Exercise 4.8 Neutron diffractometers
a) What is the purpose of a monochromator?
to select a particular wavelength band (λ ± ∆λ) out of the ”white” beam
b) How does it work?
The wavelength fulfill the Bragg condition 2dhkl sinθhkl = nλ with 2θhkl = 2θmonochromator of a crystal
monochromator can be diffracted and the other wavelengths will be transmitted and absorbed.
c) What does the term ”collimation” mean?
Collimation defines the beam direction and divergence. In order to increase the intensity of the
monochromatic beam at the sample position the monochromator crystal is often bent in vertical
direction perpendicular to the diffraction plane of the experiment. In this way the vertical beam
divergence is increased leading to a loss of resolution in reciprocal space. The diffracted intensity
from the sample is measured as a function of the scattering angle 2θ and the sample orientation
(especially in case of a single crystal). 2θ is again defined by collimators
d) What is the resolution function of a diffractometer?
The resolution function describes the instrumental resolution. It shows the reflection half width as a
function of scattering angle. It defines the smallest observable features.
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Solution Exercise Sheet 4
JCNS neutron labcourse 2015
Excercise 4.9
e) Why is it important?
The resolution function describes the instrumental resolution. During the structure refinement the
instrument resolution has to be considered to obtain the correct peak width of the sample.
f ) What is the purpose of a hot neutron source?
measure with hot neutrons results a lower absorption of the sample; provide high Q; provide a snap
shot of the atomic order, which needs short interaction time and high velocity of the neutrons
g) How does it work?
Hot neutrons result from collision of thermal neutrons with hot graphite. In doing so they gain energy,
so they become wamer and faster. Depending on the temperature of the moderator (graphite, water,
fluid deuterium), one gets hot, thermal and cold neutrons.
Exercise 4.9 Rietveld Refinement
A.) Discuss the basic problem of refining crystal structures from powder diffraction data
In a powder sample all crystallite orientations are at once. Even with optimized resolution, the severe
overlap of reflections on the 2θ-axis often prohibits the extraction of reliable integrated intensities
from the experiment.
B.) Sketch the fundamental idea to solve this problem.
Compare the experimental data with some known composition;
use contraints of space group symmetry and composition;
proceed to reciprocal space structure solution: then, real space structure solution
C.) What kind of data can be obtained from a Rietveld refinement? (collect a list and sort into
categories: Structural parameters, instrumental parameters, others)
Instrumental parameters: - scale factor; - background; - line broadening and shape; zero shift;
Structural parameters: - scale factor; - lattice parameter; - atomic coordinates; - temperature factors;occupancies
18
Solution Exercise Sheet 5
JCNS neutron labcourse 2015
Excercise 5.2
Exercise Sheet 5
Exercise 5.1 Fraunhofer far field for grating
The first task is to rewrite the transmission function
1 + cos
T (x) =
2
2π
x
a
The cosine term can also be written as
cos(θ) =
exp(iθ) − exp(−iθ)
2
which leads to
x − 21 exp −i 2π
x
1 + 21 exp i 2π
a
a
T (x) =
2
which leads to answers a, a, a for the first three questions.
The above equation for the transmission can be seen as a varying scattering length in x direction.
The first term is a constant, which would correspond to a microscopically isotropic sample in terms
of scattering length. Since scattering is dependent on differences in scattering lengths, the first term
is consistent with a transmission (or an infinitely large object). So answer b is correct here.
In the x direction, the setup is similar with a grating with slit separation a. A grating results in
. Since the transmission function
a diffraction pattern of sharp peaks at multiple of q values of 2π
a
is not built of step functions, the result is different in this case. To get the scattering function, we
have to do a Fourier Transform. Since a Fourier Transform results in spectra of frequencies present
in a curve, we can immediately conclude that there will be two sharp peaks, since the transmission
function consists of a single oscillation frequency. The Fourier transform of a cosine function cos(kx)
which leads to answers c and c.
leads to δ(q−k)+δ(q+k)
2
The modulation of the Transmission T(x) does not contain any terms including y. As such, T (x, y) =
T (x, 0) = T (x), which means there is no modulation in y direction. The result is similar to the
constant term in the first task. Answer a.
Since there are no features in y direction, the scattered intensity will once again fall back into the
primary beam, so answer b is right in this case.
Exercise 5.2 the pin-hole camera
SANS-Instrument
19
Solution Exercise Sheet 5
JCNS neutron labcourse 2015
Excercise 5.3
The picture above shows the geometry of the SANS instrument. The distance between the aperture
and the pin-hole is the same as the distance between the pin-hole and the detector. Simple usage of
intercepting line theorems provides the solutions.
• What is the ratio of the entrance aperture and the pin-hole dimensions?
The entrance aperture has double the diameter/diagonal/side length compared to the pin hole.
Answer a.
• What is the ratio of the areas?
Since area scales quadratically with length, the entrance aperture has four times the area of
the pin-hole. Answer c.
Eye
• What would be the minimal object size that could be resolved at this distance?
Since the distance between the object and the aperture (D1 = 1km) is much larger that
the distance between the aperture and the detector/retina (D2 = 2cm), the parallel wave
approximation is valid. The result is that a single point on the retina is roughly as big as the
aperture. The question we need to solve is: How big is the distance in the object plane that
shifts the resulting point at the retina by 1 mm? The intersecting line theorems once again
provide an easy solution.
D1
dobject
=
dretina
D2
with dretina = 1mm, D1 = 1km and D2 = 2cm.
The result is dobject = 50 m. According to this result, we would be able to distinguish objects,
that are at least 50m apart at a distance of 1 km(Answer b).
• Why do we see better?
The eye does not merely act as a pinhole camera. The lense further focuses the incoming light
on a much smaller area, which provides a much better resolution than a simple pin-hole. The
drawback is a finite depth of field, which means that one has to focus on an object. Answer b.
Exercise 5.3 Understanding of the Manuscript
The first task is to find the value for k that transforms
Γ(~r) = (ρ − hρi)2 · exp(−|~r|/ξ)
to
20
Solution Exercise Sheet 5
JCNS neutron labcourse 2015
Excercise 5.4
sin(kr)
Γ(~r) = (ρ − hρi)2 · exp(−|~r|/ξ) ·
kr
Since sin(kr) ≈ kr for kr << 1, this equality is given for k = 0. The right answer is c.
In the structure factor, one has to build the ensemble average by integration. Since the particles
cannot penetrate, the minimum distance allowed for two particles of the same size is always 2R. The
right answer is thus b.
For the last question, answer c is the right one. A powerlaw of α = 2 is usually observed for polymers
(gaussian segment distribution) as well as α = 4 for a hard sphere. Answer (b) is wrong since a rod
(Dimensionality=1) shows a powerlaw of α = 1. Furthermore, polymers are a fractal structure, and
so have no intrinsic integer dimensionality. Answer (a) comes into play when there are deviations
from the powerlaws given above. Spheres with surface roughness usually show α > 4, which indicates
a fractal structure. The same is true for polymers, which are in principle a long tube (Dimensionality
= 1, α = 1) and show α = 2.
Exercise 5.4 spherical form factor
The form factor amplitude is given by the equation
Z
3
~
d r exp −iQ~r ρ(~r)
V
with ρ(~r) being the scattering length density profile.
ρ for |~r| ≤ R
ρ(~r) = ρ(r) =
0 for |~r| > R
Z
~ r ρ(~r) =
d3 r exp −iQ~
A(Q, R) =
V
Switch to spherical coordinates.
Z R
Z
2
=
drr ρ
0
Make the substitution
d cos(θ)
dθ
Z
=
π
2π
Z
dθ sin(θ) exp(−iQr cos(θ))
dφ =
0
0
= − sin(θ).
R
Z
2
1
drr ρ
Z
−1
0
2π
d cos(θ) exp(−iQr cos(θ))
dφ =
0
Expand the exponential and do the φ integral (= 2π).
Z
= 2π
R
2
Z
1
d cos(θ)(cos(Qr cos(θ)) − i sin(Qr cos(θ))) =
drr ρ
0
−1
The integration of the sine term is zero since sine as an odd function.
Z R
sin(Qr)
= 4π
drr2 ρ
Qr
0
21
Solution Exercise Sheet 5
JCNS neutron labcourse 2015
Excercise 5.4
is the
This expression is equivalent to a radial Fourier Transformation, which means that sin(Qr)
Qr
Fourier term in this expression. Integration by parts and normalizing to the volume of the sphere as
well as setting ρ = 1 provides the normalized form factor amplitude:
RR
4π 0 drr2 sin(Qr)
Qr
K(Q, R) =
4
3
πR
3
=3
sin(QR) − QR cos(QR)
(QR)3
22
Solution Exercise Sheet 6
JCNS neutron labcourse 2015
Excercise 6.1
Exercise Sheet 6 Macromolecules (structure)
Remark For the chemically interested: in reality, PEP does not dissolve in DMF :-)
Exercise 6.1
a) The coherent scattering length density ρ is the sum over all coherent scattering lengths bj (the
summation
is over all atoms constituting the molecule) divided by the volume of the single molecule
Mm
vm = d·NA , with NA Avogadros number:
ρ=
bi
X
i
Mm
d·NA
h-PEP:
P
bi = 5 · 6.65 · 10−13 + 10 · (−3.741 · 10−13 )[cm] = −4.176 · 10−13 cm
i
g
1
/( cmg 3 mol
)] = 1.383 · 10−22 cm3
vh−P EP = (5 · 12 + 10 · 1)/(0.84 · 6.022 · 1023 )[ mol
−13
−22
3
ρh−P EP = −4.176 · 10 /1.383 · 10 [cm/cm ] = −3.02 · 109 cm−2
h-DMF:
P
bi = 3 · 6.65 · 10−13 + 7 · (−3.741 · 10−13 ) + 1 · 9.36 · 10−13 + 1 · 5.80 · 10−13 [cm] = 8.923 · 10−13 cm
i
g
1
vh−DM F = (3 · 12 + 7 · 1 + 1 · 14 + 1 · 16)/(0.95 · 6.022 · 1023 )[ mol
/( cmg 3 mol
)] = 1.276 · 10−22 cm3
ρh−DM F = 8.923 · 10−13 /1.276 · 10−22 [cm/cm3 ] = 6.99 · 109 cm−2
Finally, the contrast factor is:
∆ρ2
(−3.02 · 109 − 6.99 · 109 )2 (cm−2 )2
mol
=
= 1.66 · 10−4
23
−1
NA
6.022 · 10
mol
cm4
b) Assuming the background is only arising from the solvent h-DMF one first has to calculate its
dΣ
dΩ tot
total macroscopic scattering cross section
dΣ
dΩ
=
tot
dΣ
dΩ
(scattering intensity in units of cm−1 ):
+2·
coh
dΣ
dΩ
inc
For an incoherent scatterer all scattering intensity is distributed uniformly into the full solid angle
dΣ
4π therefore ( dΩ
) is given by:
P
dΣ
σi
=
/vm
dΩ
4π
Mm
2
With the scattering cross section σ = 4πb and the molecular volume vm = d·NA = 1.28 · 10−22
X
σi =
X
4πb2i = 4π
23
X
b2i
Solution Exercise Sheet 6
JCNS neutron labcourse 2015
Excercise 6.1
For h-DMF this gives:
Coherent scattering cross section:
) = {3 · (6.65 · 10−13 )2 + 7 · (−3.741 · 10−13 )2 + 1 · (9.36 · 10−13 )2 + 1 · (5.80 · 10−13 )2 } = 1.205 ·
( dΣ
dΩ coh
−23
10 cm2
Incoherent scattering cross section:
dΣ
( dΩ
)inc = {3 · 02 + 7 · (2.53 · 10−12 )2 + 1 · 02 + 1 · 02 } = 3.257 · 10−22 cm2
Total scattering cross section:
) = 1.205 · 10−23 + 2 · 3.257 · 10−22 = 3.257 · 10−22 cm2 (rule of thumb!!!)
( dΣ
dΩ tot
And finally the total macroscopic scattering cross section, i.e. the background arising from the solvent
gives:
dΣ
= 3.257 · 10−22 /(1.276 · 10−22 ) = 2.553cm−1
dΩ
The coherent scattering from the polymer at Q = 0 (i.e. the forward scattering) should be five times
larger than the scattering from the solvent in the polymer solution, which contains only (1 − φpolymer )
h-DMF:
I(Q = 0) = 5 · 2.553cm−1 = 12.765cm−1
At Q = 0 the form factor P (Q) reaches his asymptotic limit 1, i.e. the forward scattering is only
given by:
2
I(Q) = ∆ρ
φ
V
NA polymer w
Q→0
2
3
φ
= 12.765/(1.66 · 10−4 · 0.01) = 7.69 · 106 cm
⇒ Vw = I(Q)/ ∆ρ
polymer
NA
mol
Q→0
Finally, the molecular weight is:
Mw = Vw · d = 7.69 · 106 · 0.84 = 6.46 · 106
g
mol
6.1 c) Good solvent conditions means excluded volume interactions are active which gives a scaling
relation between Rg and Mw :
0.6
Rg = 0.01 nm·mol
· Mw0.6 [nm]
g 0.6
For the molecular weight calculated in b) this results in the following value for Rg :
0.6
g 0.6
· (6.46 · 106 mol
) = 121.9nm = 1219Å
Rg = 0.01 nm·mol
g 0.6
Requested was the Q-vector where the signal from the polymer vanishes in the background resulting
from the solvent in the polymer solution:
2
dΣ
I(Q) = ∆ρ
φ
V
P
(Q)
=
polymer
w
NA
dΩ h−DM F
The forward scattering I(Q = 0) =
I(Q = 0) = 5 · dΣ
dΩ h−DM F
∆ρ2
φ
V
NA polymer w
was given to be 5 times the background level:
2 2
Q R
I(Q) = I(Q = 0) · P (Q) = I(Q = 0) · exp − 3 g
(ignoring the structure factor or assuming S(Q) = 1 here!!!)
2 2
( dΣ )
Q R
I(Q)
P (Q) = exp − 3 g = I(Q=0)
= 5· dΩdΣ h−DM F = 15
( dΩ )h−DM F
exp
Q2 Rg2
3
Q2 Rg2
3
2
= ln5 = 1.61
Q =
=5
1.61·3
Rg2
s
Q=
1.61 · 3
=
Rg2
r
1.61 · 3
= 0.0018Å−1
12192
24
Solution Exercise Sheet 6
JCNS neutron labcourse 2015
Excercise 6.2
6.1 d) Starting from the definition of the overlap volume fraction φ∗ (volume of single molecule
divided by volume of sphere with radius Rg ):
Mw
φ =
/
d · NA
∗
4π 3
R
3 g
and the scaling relation between Rg and Mw for the good solvent limit, see 1 c), where the explicit
value 0.01 is for convenience replaced by a constant k2 :
Rg = k2 · Mw0.6
One yields the following relation between φ∗ and the molecular volume Vw :
Vw
Vw
4π
4π 3 1.8
∗
0.6
0.6 3
1.8
φ =
/
(k2 · d · Vw ) =
/
k · d · Vw
NA
3
NA
3 2
φ∗ =
3
3
Vw
=
V −0.8
3 1.8
1.8
Vw 4πNA k2 d
4πNA k23 d1.8 w
Expressing the volume fraction φp in terms of the overlap volume fraction φ∗ , φp = k1 φ∗ , and inserting
the above derived relation in the well-known equation for the forward scattering, see b) gives:
I(Q = 0) =
∆ρ2
3
∆ρ2
3
∆ρ2
−0.8
k1 φ∗ Vw =
k1
V
V
=
k
V 0.2
w
3 1.8 w
2 1
NA
NA 4πNA k2 d
NA 4πk23 d1.8 w
4π NA2 k23 1.8
d
3 ∆ρ2 k1
5
4π NA2 k23 1.8
Vw = I(Q = 0)
d
3 ∆ρ2 k1
Vw0.2 = I(Q = 0)
Inserting the numerical values
2
mol
23
−1
= 1.66 · 10−4 cm
,
I(Q = 0) = 12.765cm−1 , ∆ρ
4 , NA = 6.022 · 10 mol
NA
g
nm·mol0.6
−9 cm·mol0.6
k2 = 0.01 g0.6 = 10
, k1 = 0.1, d = 0.84 cm3
g 0.6
gives: h
i
10−9·3
0.841.8
0.1
0.6·5
−0.2·5 0·5
23
6.022·10
Vw = 12.765 4π
3 1.66·10−4
5
4
0.6·3
cm−1 cm
mol−1 cm3 mol
mol
g 0.6·3
g 1.8
cm3·1.8
5
3
Vw = (1417356)5 [cm mol
g ] = 5.72 · 1030 cm
mol
Finally, the molecular weight is:
Mw = d · Vw = 4.8 · 1030
g
mol
Remark Unfortunately the numbers give huge values, so practically it is nonsense, but we can exchange to deuterated solvent, only the numbers will change.
Exercise 6.2
It is A
25
Solution Exercise Sheet 6
JCNS neutron labcourse 2015
Excercise 6.3
Exercise 6.3
a) The scattering length density ρ is the sum over all coherent scattering lengths bi (the summation
goes over
all atoms constituting the monomer) divided by the molecular monomer volume vm =
Mm
, with NA Avogadros number:
d·NA
X bi
ρ=
Mm
d·NA
i
h-PEP:
P
bi = 5 · 6.65 · 10−13 + 10 · (−3.741 · 10−13 ) = −4.176 · 10−13 cm
i
g
1
/( cmg 3 mol
)] = 1.383 · 10−22 cm3
vh−P EP = (5 · 12 + 10 · 1)/(0.84 · 6.022 · 1023 )[ mol
ρh−P EP = −4.176 · 10−13 /1.383 · 10−22 [cm/cm3 ] =−3.02 · 109 cm−2
h-PEO:
P
bi = 2 · 6.65 · 10−13 + 4 · (−3.741 · 10−13 + 1 · (5.803 · 10−13 ) = 4.133 · 10−13 cm
i
g
1
vh−P EO = (2 · 12 + 4 · 1 + 1 · 16)/(1.12 · 6.022 · 1023 )[ mol
/( cmg 3 mol
)] = 6.494 · 10−23 cm3
ρh−P EO = 4.133 · 10−13 /6.494 · 10−23 [cm/cm3 ] = 6.36 · 109 cm−2
b) The average scattering length density ρ̄ of the isotopic solvent mixture is given by sum
P of the
scattering length densities of the isotopes weighted by their volume fractions φi = Vi /
Vi and
i
φH2 O = (1 − φD2 O ).
The match point in terms of φH2 O is given by:
φH2 O =
ρP olymer − ρD2 O
ρH2 O − ρD2 O
Scattering length densities:
H
P2 O:
bi = 2 · (−3.741 · 10−13 ) + 1 · (5.803 · 10−13 )[cm] = −1.679 · 10−13 cm
i
g
1
vH2 O = (2 · 1 + 1 · 16)/(1.0 · 6.022 · 1023 )[ mol
/( cmg 3 mol
)] = 2.989 · 10−23 cm3
ρH2 O = −1.679 · 10−13 /2.989 · 10−23 [cm/cm3 ] =−5.617 · 109 cm−2
D
P2 O:
bi = 2 · (6.67 · 10−13 ) + 1 · (5.803 · 10−13 )[cm] = 1.914 · 10−12 cm
i
g
1
vD2 O = (2 · 2 + 1 · 16)/(1.1 · 6.022 · 1023 )[ mol
/( cmg 3 mol
)] = 3.019 · 10−23 cm3
ρD2 O = 1.914 · 10−12 /3.019 · 10−23 [cm/cm3 ] =6.340 · 1010 cm−2
Finally, using the calculated values one obtains the following match points for:
h-PEP: φD2 O = 0.037
h-PEO: φD2 O = 0.173
26
Solution Exercise Sheet 6
JCNS neutron labcourse 2015
Excercise 6.5
Exercise 6.4
a) The first minimum in the form factor of a solid (or compact) sphere occurs at
Qmin R = 4.49
therefore the radius of the micellar core is
4.49
R = 4.49/Qmin = 0.12
= 37.4Å
Å−1
giving a core volume:
Rc3 = 219130Å3
Vc = 4π
3
From the given degree of polymerisation Dp,P EP = 15 the molecular weight of the PEP-block
g
g
= 1050 mol
Mw,P EP = Dp,P EP · MM,P EP = 15 · 70 mol
and the molecular volume
3
g
Vw,P EP = Mw,P EP /DP EP = 1050 mol
/0.84 cmg 3 = 1250 cm
mol
can be calculated. This transforms into a volume per molecule
3
/6.022 · 1023 mol−1 = 2.076 · 10−21 cm3 = 2076Å3
vP EP = Vw,P EP /NA = 1250 cm
mol
Assuming a compact PEP core (full segregation) the micellar aggregation number Nagg can be simply
calculated by
Vc
219130Å3
Nagg =
=
= 106
vP EP
2076Å3
b) The experimentally observed forward scattering I(Q = 0) should correspond to the one calculated from Nagg . Or Nagg can be derived from I(Q = 0), which is a measure for the molecular mass.
Remark The volume fraction φ is needed for this, but not given in this exercise!
Exercise 6.5
For calculating the forward scattering I(Q = 0) the virial expansion reduces to:
∆ρ2
φ
NA
I(Q = 0)
=
1
+ 2A2 φ
Vw
⇒ I(Q = 0) =
∆ρ2
φ
NA
1
Vw
+ 2A2 φ
Contrast factor (core contrast = solvent matches PEO):
2
∆ρ2
(ρh−P EP − ρh−P EO )2
(−3.02 · 109 cm−2 − (−6.36 · 109 cm−2 ))
mol
=
=
= 1.85 · 10−5 4
23
−1
NA
NA
6.022 · 10 mol
cm
3
mol
Inserting the numerical values: Vw = Vw,P EP = 1250 cm
, A2 = 2 · 10−4 cm
3
mol
−3
φ1 = 1 · 10 :
1.85 · 10−5 · 1 · 10−3
mol mol
I(Q = 0) =
/
= 2.31 · 10−5 cm−1
1250−1 + 2 · 2 · 10−4 · 1 · 10−3 cm4 cm3
27
Solution Exercise Sheet 6
JCNS neutron labcourse 2015
Excercise 6.6
φ2 = 5 · 10−3 :
1.85 · 10−5 · 5 · 10−3
mol mol
I(Q = 0) =
= 1.15 · 10−4 cm−1
/
1250−1 + 2 · 2 · 10−4 · 5 · 10−3 cm4 cm3
φ3 = 7.5 · 10−3 :
1.85 · 10−5 · 7.5 · 10−3
mol mol
I(Q = 0) =
/ 3 = 1.72 · 10−4 cm−1
−1
−4
−3
4
1250 + 2 · 2 · 10 · 7.5 · 10
cm cm
Exercise 6.6
A volume fraction of 0.25 corresponds to a number density Nz , i.e. number of particles per unit
volume [cm−3 ]:
4π
Nz = 0.25/( (250 · 10−8 cm)3 ) = 3.82 · 1015 cm−3
3
For each particle therefore a volume of
1
1
cm3 = 2.62 · 10−16 cm3
=
Nz
3.82 · 1015
is available, which transforms into a mean distance D between particles of
r
1
1
−1
= Nz 3 = (3.82 · 1015 cm−3 )− 3 = 6.40 · 10−6 cm = 640Å
D= 3
Nz
The first structure factor peak is expected finally to occur at
Q=
2π
2π
= 0.0098Å−1
=
D
640Å
28
Solution Exercise Sheet 7
JCNS neutron labcourse 2015
Excercise 7.1
Exercise Sheet 7
Exercise 7.1 Coherent and incoherent scattering cross section*
The coherent and incoherent scattering lengths are given by:
b c = p + b+ + p − b− ,
b2i = p+ p− (b+ − b− )2
with the propabilities for realizing the states with neutron and nuclear spin (I) paralell p+ or antiparallel p− .
I +1
2I + 1
I
p− =
2I + 1
p+ =
The coherent and incoherent cross sections are therefore:
σcoh = 4πb2c
σinc = 4πb2i
The isotope incoherent scattering cross section is:
X
iso
σinc
= 4π ·
pa (bac − hbc i)2 = 4π (b − hbi)2
a in all isotopes
Remark: (b − hbi)2 = hb2 i − hbi2 = p+ b2+ − p− b2− − (p+ b+ + p− b− )2 = p+ p− (b+ − b− )2
The nuclear spin incoherent cross section for a single isotope is:
nuc spin
σinc
=4π b2 all spin states per isotope − hbi2all spin states per isotope =
X
4π ·
pa (ba − hbi)2 = p+ p− (b+ − b− )2
a in all spin states
The total nuclear spin incoherent is the weighted sum:
X
nuc spin
spin
σinc
=
pa (σa )nuc
inc
total
a in all isotopes
And the total incoherent scattering is the sum of the nuclear spin incoherent and the isotope incoherent cross sections.
nuc spin
iso
σinc = σinc
total + σinc
Using this formulas its possible to calculate the following cross sections.
1
H
σcoh
= 4πb2c = 175.99 f m2 = 1.76 barn
2
H
σcoh
= 4πb2c = 559.7 f m2 = 5.60 barn
1
H
σinc
= 4πp+ p− (b+ − b− )2 = 7991 f m2 = 79.91 barn
1
1
2
2
H
H
H
iso
+ σinc
= 79.83 barn + 0 = 79.83 barn
σinc
= pH σinc
+ pH σinc
29
Solution Exercise Sheet 7
JCNS neutron labcourse 2015
Z Symb
A
p
I
bc
b+
b−
σcoh
1H
1
1H
2
1H
55
25 M n
59
27 Co
1.76
5.60
1.43
1.64
-3.61
10.28
28 N i
61
28 N i
10.61
Excercise 7.2
σinc
79.83
79.91
0.37
5.05
5.13
σabs
4.46
2.58
55
25 M n
σcoh
= 4πb2c = 143.14 f m2 = 1.43 barn
55
25 M n
σinc
= 4πp+ p− (b+ − b− )2 = 36.77 f m2 = 0.37 barn
59
Co
= p+ b+ + p− b− = −3.61 f m
b27
c
59
27 Co
σcoh
= 4πb2c = 163.77 f m2 = 1.64 barn
59
27 Co
= 4πp+ p− (b+ − b− )2 = 505.88 f m2 = 5.05 barn
σinc
The coherent scattering length and the absorption cross section of an element are the average of the
scattering lengths of its isotopes.
X
Ni
=
pi · bic = 10.28 f m
b28
c
i
28 N i
σinc
X
= 4π[
Ni
p(b28
− ba )2 ] +
c
nuc spin
σinc
total
= 511 f m2 + 0.02 barn = 5.13 barn
a in all isotopes
28 N i
=
σabs
X
i
pi · σabs
= 4.46 barn
i
The spin dependent scattering length for
b c = p + b+ + p − b− → b+ =
σinc
61
28 Ni
can be derived with the following equations:
bc − p − b−
p+
2
bc − p − b−
σinc
2
= p+ p− (b+ − b− ) = p+ p−
− b−
=
→ =
4π
p+
p−
p−
p−
σinc
=
· (bc − p− b− − p+ b− )2 =
· (bc − b− (p− + p+ ))2 =
· (bc − b− )2 =
p+
p+
p+
4π
s
r
p+ σinc
p+ σinc
5/8 190 f m2
2
→ (bc − b− ) =
→ b − = bc −
= 7.6 f m −
= 2.58 f m
p− 4π
p− 4π
3/8 4π
4πb2i
b2i
7.6 − 83 · 2.58 f m
bc − p − b −
→ b+ =
=
= 10.61 f m
p+
5/8
30
Solution Exercise Sheet 7
JCNS neutron labcourse 2015
Excercise 7.2
Exercise 7.2 Neutron contrast*
To constuct a sample chamber of Zr1−x T ix the coherent scattering length should be zero to avoid
coherent scattering.
xZr 7.16 + xT i (−3.37) = 0, with xZr + xT i = 1
7.16
= 0.68 ∧ xZr = 0.32
(1 − xT i )7.16 − 3.37xT i = 0 → xT i =
7.16 + 3.37
The disadvantage is the high incoherent and absorption cross-section of Ti.
31
Solution Exercise Sheet 7
JCNS neutron labcourse 2015
Excercise 7.4
Exercise 7.3 Precession*
The velocity of neutrons with λ = 5.4 Å is given by:
v=
p
h
6.62
=
=
· 103 m/s. = 734 m/s
m
m·λ
1.67 · 5.4
With this velocity the time of the neutron inside of the magnetic field of the coil is given by:
t=
l
2.21m
=
= 3 · 10−3 s
v
734 m/s
Inside of the coil, the spin of the neutron rotates with the larmor frequency:
ω = −γB = −2π · −2916.4
Hz
· 1000 Oe = 2π2.916 · 106 Hz
Oe
Meaning, the neutron rotates by
n=ω·t=ω·
mλ
l
= ωl
= 2π2.916 · 106 Hz · 3 · 10−3 s = 2π · 8748
v
h
during its flight through the coil. This number depends highly of the used values, which means that
it is not clear where the spin points afterwards.
With a bandwidth of 10% or 20 % in ∆λ/λ, the width of the wavelength is ∆λ = 0.1·5.4 Å = 0.54 Å.
The difference in rotations is therefore n = ωl m∆λ
= 2π · 560. With a bandwidth of 10 %, the beam
h
has no polarization, when leaving the coils of a spin echo spectrometer.
Exercise 7.4 Flipping**
The velocity of the neutrons with λ = 3.4 Å is given by:
v=
h
6.62
=
· 103 m/s. = 1166 m/s
m·λ
1.67 · 3.4
With a length of the coil of 1 cm, the time of the neutron inside the coil is
t=
l
0.01 m
=
= 8.576 · 10−6 s
v
1166 m/s
Taking the larmor frequency of the procession of the magnetic moment ω = −γH, the spin of the
neutron rotates by ω · t inside the field of the coil. To get a π-flip it has to be ω · t = π.
π = ω · t = −γH · t → H = −
π
π
=−
= 20 Oe
γ·t
−2π · 2916.4 Hz/Oe · 8.576 · 10−6 s
This solution is not unique, because of 3π-flips, 5π-flips in general (2n + 1)π-flips also corresponds
to a π-flip of the polarization. But taking a bandwidth of 5% into account, the smallest field is the
best choice. The neutrons with a different velocity have a different velocity then necessary for a
π-flip. This means the polarization of the neutron beam is lost and with bigger field this effect is
more dominant.
32
Solution Exercise Sheet 7
JCNS neutron labcourse 2015
Excercise 7.6
Exercise 7.5 Flipping ratio and corrections*
The flipping ratio is defined by fN SF = IN SF /ISF , which IN SF = 19000 and ISF = 1000 we get a
flipping ratio of fN SF = 19. The polarization of the beam is therefore:
P =
18
fN SF − 1
=
= 0.9
fN SF + 1
20
Vanadium as a spin incoherent scatterer has a contribution of 2/3 and 1/3 in the spin-flip and
non-spin-flip channels respectively, leading to the following Intensities:
1
IN SF = P · I0 +
3
2
ISF = P · I0 +
3
2
(1 − P ) · I0
3
1
(1 − P ) · I0
3
The flipping ratio is therefore:
fN SF =
1
P + 23 (1 − P )
IN SF
2−P
=
= 32
= 0.58
1
ISF
1
+
P
P
+
(1
−
P
)
3
3
Is is better to determine the flipping ratio with a coherent scatterer than with a spin incoherent
scatterer, because multiple scattering can occur. With multiple scattering two spin-flips can occur,
which looks like a non spin-flip.
Exercise 7.6 Magnetic scattering***
The background can be derived from Figure 7.13a (lecture script) as bg = 60. The SF and NSF
intensities for different polarizations can be taken from the maxima in 7.13a (normally you would
have to integrate, but that is not practicable for the tutorial). The separation into spin-coherent,
spin-incoherent and magnetic scattering can be achieved with the use of the given equations in Table
1.
Figure 2: Polarization analysis of the scattering by M nF2
33
Solution Exercise Sheet 7
JCNS neutron labcourse 2015
Polarization/Field
PkxkQ
Pk z ⊥ Q
Excercise 7.6
Spin-flip
Non spin-flip
⊥
⊥
dσ Mz
2 dσ
dσ My
+
+
bg
+
3 dΩ inc
dΩ mag
dΩ mag
⊥
2 dσ
dσ My
+ bg + dΩ mag
3 dΩ inc
dσ
dσ
+ 31 dΩ
+ bg
dΩ coh
inc
⊥
dσ
1 dσ
dσ Mz
+
+
bg
+
dΩ coh
3 dΩ inc
dΩ mag
Table 1: Scattering intensities, with background subtracted
⊥
Because it is powder
dσ Mx
dΩ mag
⊥
=
dσ My
dΩ mag
Polarization/Field
PkxkQ
Pk z ⊥ Q
⊥
=
dσ Mz
dΩ mag
⊥
=
dσ M
dΩ mag
leads to the following equations:
Spin-flip
Non spin-flip
⊥
2 dσ
dσ M
= 305 − 60 = 245
+ 2 dΩ
3 dΩ inc
mag
⊥
M
2 dσ
dσ
+ dΩ
= 200 − 60 = 140
3 dΩ inc
mag
dσ
dσ
+ 13 dΩ
= 105 − 60 = 35
dΩ coh
inc
⊥
M
dσ
dσ
dσ
+ 31 dΩ
+ dΩ
= 200 − 60 =
dΩ coh
inc
mag
140
Table 2: Intensities from Figure 7.13a with background subtracted
dσ
From the non spin-flip equations a magnetic scattering cross section of dΩ
= 140 − 35 = 105 can
mag
be derived. Inserting the magnetic cross section in the spin-flip equations leads to the incoherent
dσ
dσ
dσ
= 3/2 · (140 − 105) = 52.5. With dΩ
+ 13 dΩ
= 35 the coherent scattering
cross section: dΩ
inc
coh
inc
dσ
cross section is dΩ coh = 35 − 52.5/3 = 17.5.
M⊥
It is important to notice that the total magnetic scattering cross section is given by σmag = 2 · σmag
,
dσ
leading to a magnetic scattering cross section of dΩ mag = 2 · 105 = 210.
The scattering cross sections are therefor:
•
dσ
dΩ coh
= 17.5 relatively: 1
•
dσ
dΩ inc
= 52.5 relatively: 3
•
dσ
dΩ mag
= 210 relatively: 12
Regarding Figure 7.13d (lecture script)
a) In the figure the spin-flip-channel is shown and mainly the magnetic scattering is visible.
b) The magnetic interaction has a bigger extent than the interaction with the core in real space and
therefore is smaller in reciprocal space. Because of this the form factor of the magnetic scattering is
visible, leading to a decrease of the scattered intensity at higher angles.
c) From the data it is possible to derive the form factor of the magnetic scattering.
34
Solution Exercise Sheet 8
JCNS neutron labcourse 2015
Excercise 8.1
Exercise Sheet 8
Exercise 8.1 Displacement Parameters
The Debye-Waller-factor Tj (τ ) enters the structure factor formula as the exponential factor exp[B ·
(sin2 θ/λ2)].
a) Discuss the physical origin of this factor.
The Debye Waller factor is usually expressed as
e−W = e−2Bs
with B = 8π 2 hu2 i and s =
2
(6)
sinθ
.
λ
The physical origin of the Debye Waller factor stems from the roots of the solid state physics. In
first approximation, let us assume that we have two atoms forming a quantum oscillator separated
at equilibrium by a distance r. The total energy U (r) of the system would have two parts:
U (r) = U attractive (r) + U repulsive (r)
(7)
repulsive
With U attractive (r) = − rA
= rBm > 0. In order to have stable equilibrium
n < 0 and U
2
dU (r)
U (r)
= 0 and d dr
> 0 m > n with A and B creating the equilibrium distance r0 . The min2
dr
imum of the total energy is not symmetric around which means that the average position of the atom
is changing versus temperature (thermal expansion) and on the other hand the atoms are bounced
to vibrate between the limits of the potential surface in 3-d.
b) Describe the overall effect of this displacement factor on the diffracted intensities.
It is clear from the extraction of the Debye Waller factor that the effect of the lattice vibrations is not
to broaden out the Bragg peaks.
The Bragg peaks remain perfectly sharp but their overall intensity
~
is diminished as function of Q
or 2θ by a factor e−W .
c) It is generally said, that neutron diffraction yields much more precise displacement parameters
than X-ray diffraction. Correct? If so: Why?
The Debye – Waller formalism holds in the case of X-ray as well as in neutron scattering experiments.
Based on the different physical origin of scattering between neutrons and x-rays one can more precisely
measure atomic displacement parameters using neutrons because of the absence of neutron form factor
fall-off.
d) What are anisotropic displacement parameters and how can they be visualized?
As it was proven previously the Debye-Waller factor is given by the formula: e−1/2 . This equation
takes different forms according to the basis vectors it refers. For instance:
~ = ha~∗ + k b~∗ + lc~∗
Q
(8)
~ = ∆x~a + ∆y~b + ∆z~c
u(t)
(9)
35
Solution Exercise Sheet 8
JCNS neutron labcourse 2015
Excercise 8.2
Where a~∗ , b~∗ , c~∗ reciprocal unit vectors and ~a, ~b ,~cP
unit
in real space. The Debye-Waller
P vectors
factor can be rewritten in the form of a tensor exp(−
hi β ij hj ) with β ij = h∆xi ∆xj i taking into
~ = P ∆xj aj andaj ai = δij . Thus the atomic displacement parameters
~ = P hi ai ,u(t)
account that Q
are reduced to a tensor B with its elements given bellow:


B11 B12 B13
B21 B22 B23
(10)
B31 B32 B33
where for symmetry reasons the Bij elements are coupled with the Bji . Thus, only 6 independent
elements exist. The atomic displacement parameters can be visualized by ellipsoids. For example an
isotropic ellipsoid would have all the off-diagonal elements zero Bji = 0 and all the diagonal elements
equal Bii = Bjj = Bkk .
e) Is it correct, that all atoms in cubic crystals have to vibrate isotropically? (Yes/No, Why?)
The anisotropy of vibration depends on the site symmetry at the position of the atom. In cubic space
groups there are special positions with cubic site symmetry with the vibration being isotropic. There
are also other positions with lower symmetry (e.g. the lowest symmetry of any position is triclinic)
where the vibration is anisotropic.
f ) Discuss the non-zero values of the displacements factors for T = 0 K in fig. 8.8. (Is it real? An
artifact? Why?)
In the case of a quantum oscillator even at T = 0 K the total energy of the system (ground state)
is not zero but it has a finite value. Thus the displacement parameters should not have zero values
at T = 0 K. We should also point out that for some compounds (like diamond) zero point motion
contributes significantly to the observed displacements even at room temperature.
Exercise 8.2 Diffraction contrast and site occupancies
a) Assume you have grown a compound containing both Pb and Bi. Which kind of diffraction
experiment is better suited to distinguish Pb and Bi: X-ray or neutron? Why?
Bismuth and lead are next to each other in the PSE, so they differ by just one electron, which makes
their interaction with x-ray very similar. Their coherent neutron scattering lengths are significantly
different with: bBi = 8.532f m and bP b = 9.405f m. This make a neutron experiment the better
choice to distinguish Pb and Bi.
b) Assumed Bi and Pb sit on the same site in your structure and this site is also supposed to contain
vacancies. Is one diffraction experiment sufficient to uniquely determine the occupation probabilities?
(Yes/No, Why?)
No. One experiment would only be sufficient to determine the occupation probabilities of two elements on one site and no vacancies on the same. This makes two equations with just two unknown
values (Occelement1 and Occelement2 ):
Occelement1 · belement1 + Occelement2 · belement2 = btotal
(11)
Occelement1 + Occelement2 = 1
(12)
and
36
Solution Exercise Sheet 8
JCNS neutron labcourse 2015
Excercise 8.4
for extra vacancies on the same site the second equation is no longer true and the other equation
with two unknown values can not be solved anymore.
Exercise 8.3 Choice of neutron wavelengths
a) Magnetic neutron diffraction experiments are usually done with rather long wavelengths (see
chapter 8.7: λ = 1.87 Å): Why?
In the case of fig. 13 the magnetic Bragg peaks are only visible at low diffraction angles. This
means that the use of a shorter wavelength would not help for two reasons. The first reason is that
using a shorter wavelength the total number of reflections appearing in the diffractogram would be
higher according to Bragg law. Consequently, the resolution would be lower. The second reason is
related to magnetic form factor. The magnetic form factor decreases at high angle in analogy with
the electronic form factor of x-rays.
b) Diffraction experiments aiming at obtaining precise atomic coordinates and displacements are
done with much shorter wavelengths (see chapter 8.8: λ = 0.552 Å): Why?
A shorter wavelengths is good to separate slight differences in peak positions and gives an overall
better resolution for the best precision.
c) Powder diffraction experiments usually use longer wavelengths than single crystal experiments:
Why?
A short wavelength in powder diffraction where reflexes are measured only dependent on one rotation
angle (2θ-angle) would cause the spectrum to stretch out and drastically reduce the number of
observed reflexes.
Exercise 8.4 Hydrogen bonded crystals
Assume you have grown a new hydrogen-bonded compound in the form of a single crystal and you
want to know how the hydrogen bonds are arranged within the structure.
a) Collect arguments Pro & Con the usage of a single crystal x-ray- vs. single crystal neutron
diffraction experiment to study your new crystal.
Consider, for instance, factors like: Availability / costs of the experiment; time and effort required
to get beam time; required size of the crystal; scattering power of hydrogen; expected precision of the
H- position; absorption & incoherent scattering; additional effort needed for deuteration etc.
For many standard scattering experiments x-ray are much more suitable than neutrons because of
the high flux, the better collimation and the less divergence and of course because they are really
much cheaper. However, for some topics of research neutron scattering is superior. One example is
the investigation of compounds which contain hydrogen. In this case the electronic density is very
low and also not centered on the proton but somewhere in between the hydrogen and the bonded
partner. In the hydrogen compounds the x-ray contrast becomes very small. In contrast, for neutrons
the scattering length density strongly depends on the isotopes of the elements. Therefore by using
deuterated compounds the scattering contrast can easily be tuned without changing the chemical
properties of the samples. Another difference between neutron and x-ray scattering experiments is
related with the size of the sample. In principle, neutrons do not interact strongly with the matter.
Thus, a larger amount of mass should be used in order to reduce the duration of the experiment given
the lower neutron flux compared to x-rays. In summary, x-rays interact with the atomic electrons
37
Solution Exercise Sheet 8
JCNS neutron labcourse 2015
Excercise 8.4
and neutrons with the nuclei. This difference makes them complementary techniques for studying
most of the physical properties.
38
Solution Exercise Sheet 8
JCNS neutron labcourse 2015
Excercise 8.5
Exercise 8.5 Density maps from diffraction experiments
a) How can one obtain (from diffraction) the bonding electron density map? (discuss the experiment(s), the necessary calculations and the information obtained)
The bonding electron density map gives details of the chemical bonding. In order to obtain such a
map a combination of neutron and x-ray experiments are required. Specifically, the Fourier transform
(Fourier synthesis) of the x-ray data is calculated in order to have the total electron density in real
space. The resulting map can be significantly improved by taking the atomic positions and the atomic
displacement parameters from more accurate neutron diffraction
b) Discuss the difference between the bonding electron density map and a magnetization density
map. (which kind of data is used, what is the specific information?)
On one hand, the electron density provides information on the total electron density. On the other
hand, the magnetization density provides information on the density of the unpaired electrons, which
is usually a small fraction of the total and even of the bonding electron density. In other words, the
magnetization density map illustrates the density of magnetic moments within the unit cell. The
experiment is performed by polarized neutron diffraction on a single crystal using the flipping ratio
method (allows to separate nuclear and magnetic contribution). Given the flipping ratios and the
nuclear structure factors, the magnetic structure factors can be calculated which are then Fourier
transformed to give the spatially resolved magnetization density.
39
Solution Exercise Sheet 9
JCNS neutron labcourse 2015
Excercise 9.1
Exercise Sheet 9
Exercise 9.1
The relation between the angle of total reflection is given in the lecture as:
r
λ2
ρ·b
sin Θc =
π
Together with the definition of Q
Q=
4π
sin Θ
λ
we obtain the relation for the critical q:
⇒
Q2c = ρ · b · 16π
Q2c
ρ·b =
16π
We can now read the position of Qc from the plot shown in the exercise as 0.15 nm−1 =0.015 Å−1 .
Now we can calculate the scattering length density ρ · b of the measured material:
−1 2
0.015 Å
−2
ρ·b=
= 4.476 · 10−6 Å
16π
Comparing this with the list of scattering length densities given in the exercise it is clear that the
measurement was perfomed on Gold.
When approaching the critical angle of total reflection, the transmittance rises until it finally reaches
a value of 4 meaning that the wave amplitude reaches a value of 2. This result is very surprising on
the first view. However, if one calculates the energy flow using the Poynting vector, one can show
that the energy is preserved. The value of 4 for the transmittance is found because of constructive
interference: Incident and reflected wave add up coherently to a standing wave field near the surface.
This field’s amplitude is double the amplitude of the incident wave and drops inside the material
exponentially.
40
Solution Exercise Sheet 9
JCNS neutron labcourse 2015
Excercise 9.2
Exercise 9.2
a) From Bragg’s law we know that
∆Q =
n · 2π
with n = 0, 1, . . .
d
from which we can easily deduce that
d=
2π
∆Q
where ∆Q is the distance between two neighboring maxima.
Picture 9.2:
Red curve: ∆Q = 0, 29 nm−1 ⇒ d = 21, 4 nm
Blue curve: ∆Q = 0, 073 nm−1 ⇒ d = 85, 8 nm
b) For a multilayer the sharp, intense peaks have their origin in the constructive interference between
all bilayers and thus corresponds to the bilayer periodicity. The Kiesing fringes between those peaks
depend on the full multilayer size. Thus both dimensions can be derived as described in 9.2:
Picture 9.3: For the bilayer: ∆Q = 0, 61 nm−1 ⇒ d = 10, 2 nm
For the multilayer: ∆Q = 0, 065 nm−1 ⇒ d = 96, 1 nm
So, the multilayer is composed of approx. 9,5 bilayers.
41
Solution Exercise Sheet 10
JCNS neutron labcourse 2015
Excercise 10.1
Exercise Sheet 10
Exercise 10.1
a) The nuclear scattering length density ρ can be calculated from the scattering length D and the
atomic density N via:
ρ=N ·D
This can be obtained from the mass density NM asse [g/cm3 ], the atomic weight a[g/mol] and the
Avogadro constant NA [Atoms/mol]:
N
= NA ·
⇒
NM asse
a
−2
8.90
mol/cm3 = 91.29 nm−3 → ρNN i = 9.403 · 10−6 Å
59.71
−2
7.86
= NA ·
mol/cm3 = 84.75 nm−3 → ρNF e = 8.085 · 10−6 Å
55.85
−2
21.4
= NA ·
mol/cm3 = 66.06 nm−3 → ρNP t = 6.276 · 10−6 Å
195.09
NN i = NA ·
NF e
NP t
For the N i2 F e layer this gives a nuclear and magnetic scattering length density of:
−2
ρN (N i2 F e) = 8.963 · 10−6 Å
−2
ρM (N i2 F e) = 2.72 · 10−6 Å
b) Important for the reflectivity of the two polarization channels is the contrast of the two compounds for the nuclear scattering length density with added or subtracted magnetic scattering length
density. For Ni2 Fe we assume that the density of the alloy won’t vary a lot from that of the pure met−2
als, so we can calculate the weighted mean as ρges = hρnuc ± ρmag i. For spin-up this is 11.69 · 10−6 Å
−2
and 6.25 · 10−6 Å for spin-down neutrons. This means that the contrast for spin-down neutrons
in the multilayer is almost zero. This leaves the first two plots as candidates. A deeper look into
these plots reveals at most a difference in the angle of total reflection for spin-up and spin-down. As
the spin-up scattering length density of Ni2 Fe is much larger than the spin-down scattering length
density the critical angle is larger, too. Thus the second image is measured on this compound.
c)
I. image 3 - as contrast of the ++ channel is zero
II. image 1 - the only difference to image 2 is the plateau of total reflection, which is equal for both
channels and thus results from a nonmagnetic surface layer
III. image 5 - both channels are equal, this means there is no magnetic contribution to the scattering
~ ⊥ k P~ = 0)
(M
IV. image 4 - the channels are different but both show multilayer peaks
42
Solution Exercise Sheet 10
JCNS neutron labcourse 2015
Excercise 10.2
Exercise 10.2
For polarization analysis we have to consider the polarized neutron selection rules for measuring
magnetic moments:
~ ⊥ Q
~ is measureable
M
~ ⊥ P~ → spin-flip scattering
M
~ k P~ → non spin-flip scattering
M
I. image 3 - the ++ and −− channels are equal as there is no netto magnetization parallel to the
guide field and there is a lot of spin-flip scattering due to the perpendicular magnetization
II. image 1 - There is no spin-flip scattering as no magnetization perpendicular to the neutron
polarization is present. ++ is larger than −− due to the higher scattering length density of
the layer for a neutron polarization parallel to the magnetization.
III. image 4 - There is spin-flip scattering due to the magnetization component perpendicular to the
polarization and the ++ channel shows a larger magnetization parallel than antiparallel to the
guide field.
IV. image 2 - Both spin-flip channels are different (+− is damped more than −+) which is impossible.
43
Solution Exercise Sheet 11
JCNS neutron labcourse 2015
Excercise 11.1
Exercise Sheet 11
Exercise 11.1 Scattering triangle
1.
Since cosine of
π
2
is 0, and with the relation ∆E = ~ · ω, the formula (11.2) is simplified to
r
Q=
8π 2 2m∆E
+
λ2
~2
which gives:
s
Q=
J
2 · 1.6749 · 1027 kg · 5 · 10−3 eV · 1.6022 · 10−19 eV
−1
8 · π2
+
= 2.3Å
−10
2
−34
2
(5.1 · 10 m)
(1.0546 · 10 Js)
In the elastic case, we get from (11.3):
Q=
4π
4π
π
sin Θ =
sin = 1.7Å−1
−10
λ
5.1 · 10 m
4
2.
Q
k'min
k
k'max
The maximum energy gain and loss is found at 2Θ = 0.
k=
−1
2π
2π
=
= 1.232Å
λ
5.1A
−1
0
kmax
= k + Q = 1.232Å
−1
0
kmin
= k − Q = 1.232Å
Therfore we can calculate the energy with E =
−1
+ 1Å
−1
− 1Å
−1
= 2.232Å
−1
= 0.232Å
~2 k2
:
2m
−1
Emax
(1.0546 · 10−34 Js)2 (2.232Å )2
J
=
= 1.6594 · 10−21 J · 1.6022 · 10−19
= 10.32meV
27
2 · 1.6749 · 10 kg
eV
Emin
(1.0546 · 10−34 Js)2 (0.232Å )2
J
=
= 1.7878 · 10−23 J · 1.6022 · 10−19
= 0.1115meV
27
2 · 1.6749 · 10 kg
eV
−1
−1
(1.0546 · 10−34 Js)2 (1.232Å )2
J
E0 =
= 5.0394 · 10−22 J · 1.6022 · 10−19
= 3.145meV
27
2 · 1.6749 · 10 kg
eV
And therefore
~ωmin = −3.034meV
44
Solution Exercise Sheet 11
JCNS neutron labcourse 2015
Excercise 11.2
~ωmax = 7.18meV
For larger values of ~ω one needs to use shorter wavelengths.
3.
The dispersion relation of phonons (sound waves) at low Qs is ω ≈ v ·|k|. The momentum and energy
of the phonon will be transfered to the neutron. We get:
~ω =
2500 ms · Q · ~
J
1.6022 · 10−19 eV
~ω = 24.7meV
Taking formula (11.2) again and isolating cos(2Θ) we get
−Q2 +
q
cos(2Θ) =
4π
λ
(−1.5 · 1010 m−1 )2 +
q
4π
1·10−10 m
8π 2
(1·10−10 m)2
4π
(1·10−10 m)2
±
±
8π 2
λ2
4π
λ2
±
±
2mω
~
=
2mω
~
J
2·1.6749·1027 kg·24.7·10−3 eV ·1.6022·10−19 eV
−34
2
(1.0546·10
Js)
J
2·1.6749·1027 kg·24.7·10−3 eV ·1.6022·10−19 eV
−34
2
(1.0546·10
Js)
This results in cos(2Θ1 ) = 0.9821 and 2Θ1 = 10.9◦ and cos(2Θ2 ) = 0.9838 and 2Θ2 = 10.4◦
Challenges for this measurment:
Energy resulution is not a problem.
E=
~2 k 2
~2 4π 2
(1.0546 · 10−34 Js)2 4π 2
1
=
=
= 81.81meV
J
2m
2mλ2
2 · 1.6749 · 1027 kg · (1 · 10−10 m)2 1.6022 · 10−19 eV
|~ω|
= 0.3
E
• Since 1meV corresponds to 11.7K, neutrons with a temperature of nearly 1000K are needed.
This means hot neutrons or the tail of the maxwell distribution (low intensity!) is needed.
• Since the scattering angle is small, the reflected beam is close to the primary beam. This means
the detector can only cover a small solid angle and the statistics will be low.
• Neutrons cover a different and wider Q-range in the dispersion relation of phonons than photons.
• Since phonons are a collective motion of atoms, coherent scattering is needed.
• Only longitudinal phonons are observable.
Exercise 11.2 Q dependence of characteristic time
The reason for the relation is the Gaussian approximation (11.61) which is valid for diffusion and
assumed in the Rouse and Zimm model.
Q2 < r 2 >
Iinc (Q, t) = exp −
6
If on the other hand
Iinc (Q, t) = exp −
45
t
τ (Q)
β !
Solution Exercise Sheet 11
JCNS neutron labcourse 2015
Excercise 11.3
is found, the exponent β inside the potential implies that < r2 >= F · tβ . This yields:
Q2 F tβ
Iinc (Q, t) = exp −
6
comparing both exponentials yields
Q2 F tβ
=−
−
6
t
τ (Q)
β
Q2 F
1
=−
6
(τ (Q))β
β1 β1
−2
6
6
=
Qβ
τ (Q) =
2
QF
F
Exercise 11.3 Jump diffusion in confined space
Small Q: large l ≈
2π
Q
In this situation the length scale on which the diffusion is observed does not depend on Q, but on R.
Therefore, the diffusion time is ∼ D−1 R2 (not ∼ D−1 Q−2 ), meaning there is no dependence on Q if
.
Q >> 2π
r
46
Solution Exercise Sheet 11
Large Q: small l ≈
JCNS neutron labcourse 2015
Excercise 11.3
2π
Q
In this situation the particle “diffuses” out of the length l for every single jump. This result does not
depend on l as long as l << a or Q >> 2π
.
a
In conclusion, it is important to choose the right Q-range for a measurement. Looking at the above
picture, we see:
• For very low Q, the resulting spectrum is too slim to be resolved.
• At very high Q, the resulting spectrum has become too broad. It is outside of the instruments’
energy range.
• The range between both plateaus, has a Q-dependant broadening of the spectrum. This is the
Q-range, where the instrument will collect data that can be evaluated.
(A very easy and vivid description was given by a student in the 2015 course: The large Q case is like
watching a moving truck with a pinhole camera, the truck will ”diffuse” out of the watched lengsth
scale. The small Q case is like watching a lady bug on the ground from the second floor of a house
with a wide-angle lens, one won’t see awfully much.)
47
Solution Exercise Sheet 12
JCNS neutron labcourse 2015
Excercise 12.2
Exercise Sheet 12
Exercise 12.1 Electronic structure and Mott transition
a) In modeling the electronic structure of crystalline solids, what is the typical starting assumption
to separate the electronic structure from the lattice dynamics, and why does it work?
(pg. 12.3) - Adiabatic or Born-Oppenheimer approximation: Electrons follow nuclei.
Reason: very different mass scale leading to very different time scale: nuclei are slow.
b) In which of the three simplest models of electrons in a solid are the electronic correlations taken
into account at least approximately?
(pg. 12.4) Fermi liquid
c) Neglecting electronic correlations, would you predict NaCl to be an insulator or a metal? Why?
Insulator. Electronic configuration: Na: [Ne] 3s1 ; Cl: [Ne] 3s2 3p5 . Total number of electrons per
primitive unit cell is 8 , i.e. and even number. Na+ and Cl – both have a complete outer shell (noble
gas configuration).
For more information: pg. 12.6 - same calculation done for CoO.
d) The competition of which two contributions to the total energy of the electrons is crucial for
the Mott-transition? Which further contributions to the total energy are neglected in the simplest
model?
(pg. 12.6, 12.7, fig. 12.5) Kinetic energy favors delocalization and Coulomb potential favors localization.
Neglected contribution: long range Coulomb potential, entropy.
Energy sufficient only at 0 K. At finite temperature: Free energy / free enthalpy! Entropic contributions, Fermi distribution. Higher temperature mean that orbitals with higher energy are partly
occupied, for example electrons that go to the conduction band.
e) Assume that a particular material is a Mott-insulator, but just barely so (i.e. the relevant energy
contributions are almost equal). What would you predict to happen when sufficiently high pressure
is applied, and why?
The material becomes conductive: electrons become delocalized. (has been tried out experimentally
in MnO)
PS: Temperature is important: the electronic order can be broken: material become conductive.
Exercise 12.2 Electronic ordering in correlated-electron materials
a) List and briefly explain three “electronic degrees of freedom”, which can become ordered.
(pg.12.16, fig. 12.13)
Correlation-effects provide tendency towards localization of the electrons which acquire “atomic-like”
properties. They still have a finite intensity at neighboring sites, so they can still “talk” to each other,
facilitating ordering process of spin, charge and orbital.
Spin order: the spins are ordered in a fixed direction relative to each other.
48
Solution Exercise Sheet 12
JCNS neutron labcourse 2015
Excercise 12.3
Charge order: ions with different oxidation numbers are ordered in the material.
Orbital order: A distortion of the oxygen octahedral breaks the symmetry of the transition metal d
orbitals, giving a preferred direction of the orbitals (Jahn-Teller effect). These distortions occur in a
certain pattern and so do the orbitals.
b) To order of which of the electronic degrees of freedom is neutron scattering directly sensitive, and
to which not?
Directly: spin order
Indirectly: charge, orbital order
c) For those electronic degrees of freedom, to which neutron is not directly sensitive, neutron scattering can still be used to deduce an ordered arrangement: How and why? Is there a more direct
scattering method than neutron scattering?
Charge and orbital order can be detected as lattice distortion, or implicit from spin information:
solution with material constrains.
Direct scattering method: soft resonant X-ray diffraction.
e) What, if any, connection is there between orbital order and orbital magnetic momentum?
If you have orbital ordering with non-degenerate states with real wave functions you will have pureimaginary expectation values for the Angular momentum operator. Since it describes an observable
it has real eigenvalues, so they have to be zero. So you will not have an angular momentum.
Here the proof:
Let T be the time reversal operator with T −1 T = 1 and L be the angular momentum operator. Since
L contains a generalized velocity it is odd under time inversion T L = −L. Since T only acts on
the angular momentum part of the wave function T Ψ = Ψ∗ . Since the Hamiltonian as observable is
hermitian Ψ and Ψ∗ have the same real Eigenvalues. For a non-degenerate state the wave functions
must be linearly dependent and can only differ by a phase Ψ∗ = exp{(iφ)}Ψ with real φ.
The expectation value of L for states hn| and |mi is:
hn|L|mi = hn|T −1 T LT −1 T |mi = − hn|T −1 LT T −1 T |mi = − hn|T −1 LT |mi = − exp{i(φn − φm )} hn|L|mi∗
If you have a non-degenerate eigenstate hn| it follows:
hn|L|ni = − hn|L|ni∗
This means hn|L|ni is pure imaginary.
Since L is an observable it has real eigenvalues, so hn|L|ni = 0.
d) Discuss why electronic correlations favor ordering processes of these electronic degrees of freedom.
Charge order only makes sense for localized electrons; interaction via coulomb potential. Spin order
requires partial overlap of neighboring orbitals. Orbital order is related to the overlap of occupied
oxygen p orbitals with occupied or unoccupied d orbitals of the octahedrally coordinated transition
metal. Collective interaction with lattice distortions.
To have charge ordering not only atomic like states are necessary, but also localization. Although
too much localization would not allow the different states to “see” each other.
49
Solution Exercise Sheet 12
JCNS neutron labcourse 2015
Excercise 12.3
Exercise 12.3 Crystal field
1.Fe has atomic number 26 and in oxides typically has oxidation states 2+ and 3+.
a) Determine the electronic configuration of free Fe2+ and Fe3+ ions (hint: The outermost s- electrons
are lost first upon ionization).
Ans: The electronic configuration of neutral Fe is [Ar]4s2 3d6
When iron loses an electron to form an ion, it loses 4s atomic orbital first -before losing electrons
from its 3d atomic orbital. Hence the electronic configuration of Fe2+ and Fe3+ ions are:
Fe2+ =[Ar]3d6
Fe3+ =[Ar]3d5
b) From Hund’s rules determine the values of the spin S, orbital angular momentum L, and total
angular momentum J of Fe2+ and Fe3+ ions.
(Hund’s rules:
1. S max.
2. L max consistant with 1
3. J=|L-S| for a less than half filled shell,
J=|L+S| for a more than half filled shell)
Ions
n
Fe2+
Fe3+
6
5
d-shell(l=2)
lz =2, 1, 0, -1, -2
↑↓ ↑ ↑ ↑ ↑
↑↑↑↑↑
S
P
L=| lz |
J=L+S
2
5/2
2
0
4
5/2
Spectroscopic notatin
5
6
D4
S5/2
c) The effective moment µef f p
of a magnetic ion can be determined experimentally by the curie weiss
law, and is given by µef f = gJ
J(J + 1)µB , where Lande factor is gJ =
3
2
+
S(S+1)−S(S+1)
.
2∗J(J+1)
Calculate the expected effective moment in units of µB of Fe2+ and Fe3+ ions, i) assuming S, L and J as
detemined in b) and ii) setting L=0 (’quenched orbital momentum’). Compare with the experimental
values of ∼ 5.88µB for Fe3+ and ∼ 5.25 − 5.53µB for Fe2+ .
Ans: The effective number of Bohr magneton is given by
µef f = gJ
p
J(J + 1)µB
(13)
i)For the total moment in the case of
• Fe2+ (S=2, L=2, J=L+S):
gJ =
3 2(2 + 1) − 2(2 + 1)
+
= 1.5µB
2
2 ∗ 4(4 + 1)
(14)
Therefore
p
µef f = 1.5 4(4 + 1) = 6.7µB
• Fe3+ (S=5/2, L=0, J=5/2):
50
(15)
Solution Exercise Sheet 12
JCNS neutron labcourse 2015
Excercise 12.3
3 5/2(5/2 + 1) − 0
+
= 2µB
2
2 ∗ 5/2(5/2 + 1)
(16)
p
µef f = 2 5/2(5/2 + 1) = 5.92µB
(17)
gJ =
Therefore
ii) For the pure spin moment case (i.e., L=0 hence J=S),
In the case of Fe2+ :
3
2(2 + 1)
+
= 2µB
2 2 ∗ 2(2 + 1)
p
Therefore equation 1 reduces to µef f =2 2(2 + 1)=4.89µB
gJ =
(18)
In most of the transition metal ions the experimentally determined moment value is found to be
in agreement with the spin only value of the moment. This phenomenon is known as quenching of
orbital angular moment. It can be noted that in case of Fe2+ the calculated spin -only moment value
(4.89µB ) is smaller than the experimental value of moment (∼ 5.25 − 5.53µB ), this shows that the
orbital angular moment is not totally quenched.
12.3 d) The negatively charged oxygen ions surrounding the Fe ions in an oxide solid influence the
energy of the different orbital. Plot the expected energy level diagram for the case of an octahedral
environment of nearest-neighbour O2− . How dos the total spin moment of Fe2+ change between weak
and strong crystal field splitting( relative to intra-atomic ”Hund’s” exchange)?
Ans: To be able to understand the effect of crystal fields on the energy levels in transition metals it
is essential to have a clear picture of the shapes (angular dependence functions) of the d-orbitals, as
shown in figure 3.
When a free metal ion is placed in an uniform spherical field, all the d-orbitals experience equal field
so their energies are raised equally. In case of octahedral field, i.e., when there are six symmetrically
placed ligands which resides on the axes, it can be noted from the figure 3 that dz2 and dx2 −y2 orbitals
pointing directly towards the ligand, thus experience more field than dxy , dyz and dxz orbitals. This
will result in the splitting of energy level in to two, called eg and t2g separated by the Crystal
Field Splitting Energy (CFSE) ∆o , as shown in figure 4a. The low-spin and high-spin electronic
configuration of Fe2+ in the octahedral field of the ligand, O2− is as shown in figure 4b and 4c
respectively.
It should be noted that in the case of low-spin configuration as shown in figure 4b, the electrons does
not obey Hund’s rule owing to the large ∆o , compared to smaller pairing energy (or intra atomic
exchange, inherent repulsion of electrons which supports Hund’s first rule). This state is called as
low-spin state. But when the ion is surrounded by weak ligand, which means in weak crystal fields,
the CFSE (∆o ) may be smaller compared to CFSE. In this case electrons tend to obey Hund’s rule
resulting a high-spin state as shown in the figure 4c.
12.3 e) In a tetrahedral environment the energy levels of the orbitals are reversed compared to
an octahedral environment. Determine the spin moment of F e2+ in a tetrahedral environment with
strong field splitting. Is an orbital angular moment possible in this case? How about when a Jahnteller-distortion leads to a further splitting of the energy levels?
Ans: In tetrahedral case the ligands are placed on the opposite corners of a cube enclosing the metal
ion as shown in the Fig. 5 (M+ is the metal ion, say F e2+ in the present case, gray balls represent
the ligand, O2 −, for instance).
51
Solution Exercise Sheet 12
JCNS neutron labcourse 2015
Excercise 12.3
Figure 3: (a) An octahedral array of negative charges approaching a metal ion. (b-f) The orientations of
the d-orbitals relative to the negatively charged ligands. Notice that the lobes of the dz 2 and
dx2 −y2 orbitals (b and c) point toward the charges, the lobes of the dxy , dyz , and dxz orbitals
(d-f) point between the charges.
(a) The energy level diagram of d-orbitals subjected to (b) The energy level diagram for d-orbitals of Fe2+ , in lowoctahedral field
spin state, when subjected to octahedral field
(c) The energy level diagram for d-orbitals of Fe2+ , in (d) The energy level diagram for d-orbitals of Fe2+ , with
high-spin state, when subjected to octahedral field
strong field splitting, when subjected to tetrahedral
field
Figure 4: Energy level schemes
52
Solution Exercise Sheet 12
JCNS neutron labcourse 2015
Excercise 12.4
Figure 5: Four ligands arrangement in tetrahedral case
It is clear from figure 4d (recalling the shape of d-orbitals from previous answer) that none of the
d-orbitals point exactly towards the ligands. The three d-orbitals dxy , dyz and dxz are pointing
close to the direction in which ligands are approaching. As a result energy of these three orbitals
increases much more than the other two d-orbitals (dz2 and dx2 −y2 ). The d-orbitals thus split (with
the splitting energy ∆t )and the possible electronic configuration in case of F e2+ in tetrahedral field
is as shown below. Note that for F e2+ in tetrahedral crystal field there is only one possible electronic
configuration independent of the value tetrahedral crystal field splitting energy ∆t . (If somebody
asks: The orbital names come from group theory. Since there is no inversion center in the tetrahedral
field the orbitals are now named t2 and e. e= double degenerate, t= triple degenerate and 2 means
antisymmetric to C2-rotation, and g stands for ”gerade” which is German for even under inversion
)
Exercise 12.4 Orbital and Magnetic order in LaMnO3 (Optional!)
a) Why is there no charge order?
In this compound Mn is +3. To have charge order it would have to be splitted into Mn+2 and Mn+4 .
That is energetically not favorable in the crystal.1
b) What are the smallest unit cells that can describe:
By connecting equivalent lattice sites:
i) Magnetic order: a x b x 2c
i) Orbital order: 2a x 2b x c (face centered)
i) Combined: 2a x 2b x 2c (face centered)
c) Make a plot of reciprocal space in the a∗ -c∗ -plane indicating the positions, where you expect
nuclear, orbital and magnetic Bragg peaks to occur.
For a simple antiferromagnetic structure with a cell doubled in c direction the magnetic propagation
~ ± ~k = (h, k, l + 1 ) with Q
~ a vector of
vector is ~k = (00 21 ). So magnetic reflections are expected at Q
2
1
Splitting Mn(III) into Mn(II) and Mn(IV) is actually favorable in an aqueous environment! Mn(III) has a tendency
to disproportionate.
53
Solution Exercise Sheet 12
JCNS neutron labcourse 2015
Excercise 12.4
Figure 6: Left: answer to c); right: answer to d). The indices and axes correspond to the cubic cell/setting,
as displayed in Fig. 12.7 left and also to the outline drawn in Fig. 12.9 left.
the structural reciprocal lattice.
d) As c) but for the a∗ -b∗ -plane
Orbital cell is of sqrt 2 x sqrt 2-type, with reflections at (1/2,1/2,0). In analogy to the spin-case, the
nuclear positions are excluded as the orbital order reflections that are supposed to be described can
be thought as the deviation from the average orbital occupation, which would be already part of the
nuclear peaks. See Fig. 6 right.
54
Solution Exercise Sheet 13
JCNS neutron labcourse 2015
Excercise 13.2
Exercise Sheet 13 Polymer Dynamics
Exercise 13.1 Scaling arguments
a)
Re2 = N l2 = N 0 l02
with l0 = αl
N l 2 = N 0 α2 l 2
thus N 0 = N/α2
Diffusion (equation 13.2)
bT
= ζk0bNT0
D = kζN
thus ζN = ζ 0 N 0
and ζ 0 = ζN/N 0
N
2
with the result obtained above ζ 0 = ζ N/α
2 = ζα
Viscosity (equation 13.1)
0 02 0 0
2
η = ζl 36ρN = ζ l 36ρ N
thus ζl2 ρN = ζ 0 l02 ρ0 N 0
and ρ0 = ζ/ζ 0 × l2 /l02 × N/N 0 × ρ
ρ0 = 1/α2 × 1/α2 × α2 × ρ
ρ0 = ρ/α2
Mean-square displacement (equation 13.33)
with the relationships derived above, obviously the α2 in the fraction cancels
b)
In order to obtain a realistic look both the length-scale and the time-scale need to be modified in an
appropriate way according to the scaling law. Collapsing buildings or falling objects give a cue for
the real size via the falling time and the gravity constant. s = g2 t2 . Hence, the scale-factor applied
to the time is the square root of the scale-factor applied to the length. Thus, if the monster is scaled
down 1:25 the slow-motion factor needs to be 1:5.
Exercise 13.2 Length and time scales of reptation
a)
Regime 1
hr2 i = C1 t1/2
hr2 i reaches d2 at τe
1/2
Thus: C1 τe
τe =
= d2
d4
C12
Regime 2
55
Solution Exercise Sheet 13
JCNS neutron labcourse 2015
Excercise 13.2
Figure 7: Schematic view of the mean-square displacements predicted by the reptation theory. The dashed
line indicates the MSD of the Rouse model.
hr2 i = C2 t1/4
as h∆r2 i (t) is a continuous function, this relation starts in point (τe , d2 )
1/4
d2 = C2 τe
C2 =
d2
1/2
1/4
τe
= dC1
Regime 3
1/2 1/4
starts when t = τR , i.e. in point (τR , dC1 τR )
1/2 1/4
τR , dC1 τR = d
=
4 1/4
3
π
12kB T l2
πζ
1/4 4 1/4
3
π
dN 1/2 l =
ζN 2 l2
3π 2 kB T
1/4
dRe
hr2 i = C3 t1/2
1/2 1/4
1/2
τR , dC1 τR = C3 τR
1/2 −1/4
C3 = dC1 τR
h∆r2 i (t) reaches Re2 at τη
C3 τη = Re2
56
Solution Exercise Sheet 13
τη =
Re4
C32
Excercise 13.2
N 2 l4
−1/2
d2 C1 τR
=
= N 2 l4 d−2
JCNS neutron labcourse 2015
q
πζ
l−1
12kB T
q
ζ
Nl
3π 2 kB T
= N 3 l3 d−2 kBζ T 6√1 π
Regime 4
hr2 i = C4 t
starts in point (τη , Re2 )
C4 τη = Re2
C4 =
Re2
τη
√
= N l2 × N −3 l−3 d2 kBζ T 6 π
√
= N −2 l−1 d2 kBζ T 6 π
D=
C4
6
b)
1/4
τR
τe
=
=
√
πN −2 l−1 d2 kBζ T
dRe
d2
=
Re
d
=
42
4.8
= 8.75
τR = 8.754 × 7 ns = 41 µs
τη
τR
1/2
=
Re2
dRe
=
Re
d
= 8.75
τη = 8.752 × 41 µs = 3.1 ms
Basically, τR could as well be calculated using the equation given in the appendix. For this approach we require the number of monomers N , the monomer length l, and the friction coefficient ζ.
However, no equation is given for τη .
N = M/Mmonomer ≈ 6780 (6786 with Mmonomer = 28 g/mol and 6774 with the exact value of
Mmonomer = 28.05 g/mol)
√
Re2 = N l2 ⇒ l = Re / N ≈ 5.1 Å
τe =
d4 ζ
3π 2 l2 kB T
⇒ζ=
τe 3π 2 l2 kB T
d4
≈ 3.43 × 10−5 ns/Å2 × 3π 2 kB T
With these values the Rouse time τR can be calculated.
τR =
ζN 2 l2
3π 2 kB T
≈ 41 µs
In a recent paper in Macromolecules all four regimes could be accessed by combining field-cycling
and field-gradient 1 H NMR.
(http://pubs.acs.org/doi/abs/10.1021/acs.macromol.5b00855)
57
Solution Exercise Sheet 14
JCNS neutron labcourse 2015
Excercise 14.1
Exercise Sheet 14
Exercise 14.1
Figure 8: Slit Collimator (Ex. 14.1.1). The collimation with the two slits is the easiest one to build. The
resulting beam has a trapezium shape. The width of the inner plateau is equal to the slit width,
the overall width is determined geometrically.
Figure 9: Soller Collimator (Ex. 14.1.2). The soller collimator is basically a number of slits as in 14.1.1.
If the width and the distance of the slits is chosen correctly, again a trapezium shape beam will
result. This time, the width of the plateau equals the overall size of the collimator, the flanks
are determined by the width of one slit only. Thus, the resulting beam is a lot narrower than in
14.1.1. The transmission, however, is lower, because the absorbing planes have a finite thickness
d.
Figure 10: Neutron Guide (Ex. 14.1.3). The neutron guide uses a totally different principle of collimation:
Here, neutrons with too high a divergency are not absorbed, but reflected, if they hit the
wall below the critical angle of total reflection. Here, the transmission is a lot higher than in
14.1.1 and 14.1.2. Since a flat angular distribution is given, the outcoming beam has again a
trapezium shape. (θC is usually very small!). In real cases, however, where the incident beam
has a Gaussian intensity distribution, the nice trapezium shape with a wide constant plateau is
not achieved. The outcoming beam has then a Gaussian shape, which is cut at its flanks.
58
Solution Exercise Sheet 14
JCNS neutron labcourse 2015
Excercise 14.2
Exercise 14.2
a) The velocity of neutrons of 10 Å wavelength is
v(λ) =
4000
h
m
≈ h i [s] ⇒ v(10Å) = 400
mn λ
s
λ Å
10◦ at 10 cm equals 10 cm · tan(10◦ ) = 1, 76 cm So the drum has to turn for 1, 76 cm in the time it
takes the neutron to pass 10 cm.
f=
1, 76 cm · 400 ms
v
ω
=
=
= 187 Hz
2π
2πr
2π · 6 cm · 10 cm
For the wavelength spread, we will consider the fastest neutrons to pass through the drum. In this
case, the drum has not rotated 1, 76 cm, but 0, 76 cm. The time the neutrons have to pass through
the length of the durm is the shortened to 1, 08 · 10−4 s. The wavelength of neutrons at this velocity
is 4, 3 Å. This equals a wavelength spread of 57 %.
b) calculate the Bragg-angle for the PG (0 0 2) monochromator with a d-spacing of d = 3.343Å :
nλ = 2d sin(ϑ)
λ
⇒ ϑ = arcsin
2d
⇒ ϑ2.4Å = arcsin
⇒ ϑ10Å = arcsin
2.4Å
!
= 21.04◦
2 · 3.343Å
10Å
2 · 3.343Å
!
= arcsin (1.496) =?
The PG(0 0 2) monochromator with its d = 3.343Å d-spacing is not suitable to select these long
wavelengths. Crystal monochromators cannot select wavelengths longer than twice the d-spacing of
the operating bragg reflection. Calculating the wave length spread caused by the ∆ϑ = 400 ≈ 0.66◦
mosaicity of the monochromator crystal:
nλ = 2d sin(ϑ)
∆λ
⇒
= 2d cos (ϑ)
∆ϑ
⇒ ∆λ = 2d cos (ϑ) · ∆ϑ ·
2π
360◦
= 2 · 3.343Å · cos (21.04◦ ) · 0.66◦ ·
The wavelength spread for λ = 2.4Å is about 3%.
59
2π
= 0.072Å , 3%
360◦
Solution Exercise Sheet 14
JCNS neutron labcourse 2015
Excercise 14.2
14.2 c) A phase shift of 100 ◦ means that the choppers turn around 100 ◦ in the time it takes the
neutrons to travel 3 m.
100◦ 1
s = 1, 4 · 10−3 s
t=
◦ ·
360 200
3m
m
⇒v=
=
2160
1, 4 · 10−3 s
s
4000
⇒λ=
= 1, 85 Å
2160[ ms ]
For the wavelength we will consider the slowest neutrons, which have an additional time to pass
through the chopper until the 1 cm window has closed again. The speed of the chopper disks at the
given radius is
m
v = ω · r = 2πf r = 2513
s
The time for the chopper disks to turn for 1 cm is then 3, 98 · 10−5 s. So altogether, those neutrons
have 1, 4 · 10−3 s + 3, 98 · 10−5 s to pass the chopper, which equals a wavelength of 1, 92 Å. The
wavelength spread is so about 4%.
60
Solution Exercise Sheet 14
JCNS neutron labcourse 2015
Excercise 14.3
Exercise 14.3
a)
v
hmi
s
=
4000
λ[Å]
⇒ v(1 Å) = 4000
m
s
0
t=
s+s
3m
=
= 7, 576 · 10−4 s
v
3960 ms
b)
m(s + s0 )2
mv 2
=
E=
⇒t=
2
2t2
t0 =
r
m(s + s0 )2
2E
s
s0
+ 0
v1Å v
s0
0
⇒ ∆t = t − t = q
⇒ ∆E = E1Å −
2(E1Å −∆E)
m
!2
0
s
∆t +
s0
v1Å
s0
−
v1Å
m
= E1Å −
2
s0
0
∆t + s mλ
h
!2
m
2
Example: E(1Å) = 81, 8 meV, an Energy loss of 10% means 8, 18 meV
∧
∆t ≈ 5.89 · 10−5 s = 7% of t.
c)
~ω = E − E 0 =
~2
k2 − k02
2m
|~Q| = ~ |k − k0 |
2
2
⇒ |Q|2 = |k − k0 | = |k|2 + |k0 | − 2 |k| |k0 | · cos(2ϑ)
√
2m 0
0 · cos(2θ)
E
+
E
−
2
E
·
E
=
~2
61
(19)
Solution Exercise Sheet 14
JCNS neutron labcourse 2015
Excercise 14.3
14.3 d)
2
(δ∆E) =
δ∆E
δλ
2
2
(δλ) +
δ∆E
δ∆t
2
2
(δ∆t) +
δ∆E
δs0
2
(δs0 )2
with m being the neutron mass, s0 the pathlength of the neutron from sample to detector, t the time
the neutron needs from the chopper opening to the reach the detector and λ the wavelength. δλ is
the bandwidth of the selected neutron wavelength. δs0 is the flightpath uncertainty which is caused
by a finite sample size in cm range and a detection position uncertainty in the detector in mm range.
δ∆t is the timing uncertainty and is caused by a finite chopper opening and the time resolution of
the detection. Choosing small wavelengths has the biggest effect in reducing the energy resolution.
Others are more instrument specific, like: long chopper-sample-detector distances. This will increase
s + s0 but also t which dominates and therefore will reduce δ∆E. Increasing the chopper frequencies
will decrease δ∆t the time uncertainty caused by a finite chopper opening.
62