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Transcript
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: 1 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 3 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 4 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 5 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 6 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 7 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. 8 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 11 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) 12 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? 13 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. 14 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 15 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. 16 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. 17 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