# Tag Info

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The whole volume XVII of the Proceedings of the Cambridge Philosophical Society is available on the Internet Archive at the url https://archive.org/details/proceedingsofcam1718191316camb It is p.43 as marked in the text and p.60 in the document. It is available in pdf, ePub, Kindle, Daisy and DjVu formats, as well as a dedicated (very nice) online reader....

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Assuming I've understood your setup correctly this is where the $2\theta$ comes from: Typically we think of the diffracton experiment with this geometry: But that means you have to rotate the X-ray source to change $\theta$ and this is usually inconvenient. In practice you keep the X-ray source constant and you rotate the diffracting plane by $\theta$ and ...

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Yes, in fact Compton scattering is often an annoyance in x-ray diffraction when you want to study the diffuse background from crystal defects and disorder. The amount of Compton scattering is highly dependent on the xray energy, and is highest when it reaches the energy scale of the electron mass of 512 keV. Below is a plot showing the various ...

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The Bragg equation is $2d \sin \theta=n \lambda$, so you write $d={n \lambda \over 2 \sin \theta}$ if you measure the angle $\theta$ you get a measurement of the lattice spacing $d$. You presumably know $\lambda$ very accurately so the uncertainty on the measurement of $d$ comes from the uncertainty on the measurement of $\theta$. (This assumes your ...

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Here is a link to the paper - within another doc. http://www.xtal.iqfr.csic.es/Cristalografia/archivos_10/Bragg-firstpaper-mini.pdf

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1. X-ray diffraction takes pictures of Fourier space As briefly described on pg. 34 of "Introduction to Solid State Physics" 7th edition by Kittel, the scattering amplitude for an arbitrarily-shaped object is $$F(\mathbf{k}_i,\mathbf{k}_o)=\widehat{N}(\mathbf{k}_i-\mathbf{k}_o)$$ where $\mathbf{k}_i$ is the incident wavevector, $\mathbf{k}_o$ is ...

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Generally, the local phase angle $\alpha$ of a wave described by $$\psi=e^{i\vec k\cdot \vec r}=e^{i\alpha(\vec r)}$$ is of course $$\alpha(\vec r)=\vec k\cdot \vec r$$ That is just the very definition of phase $\alpha$ of a plane wave (in stationary description, i.e. temporal phase $e^{-i\omega t}$ has been divided out of the wave). Due to that, all the ...

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When did the "Laue way" came up? Was it only named after Max von Laue or did he himself publish a paper introducing this formalism? Max von Laue submitted his equations for publication 08 June 1912. The paper is W. Friedrich, P. Knipping and M. Laue, "Interferenzerscheinungen bei Röntgenstrahlen ", Annalen der Physik, volume 346, issue 10, pages 971-988, ...

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The d is not separation between points in reciprocal lattice. Actually, they do not even have the same units. d is the separation between lattice planes, as you said. What is related to reciprocal lattice vectors is the change (before and after scattering) in the wave vector of light: change in k = reciprocal lattice vector, which is the Laue condition that ...

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In Fraunhofer diffraction, every points at the aperture (the opening) are regarded as secondary sources. Mathematically, the wave fronts are undergone Fourier transform of a step function over the aperture. We extend the idea to a crystal lattice. Waves "diffracted" from each particles (atoms, ions, molecules and so on) are regarded as point sources (at ...

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If you have a single crystal, you can probably tell the orientation of the crystal plane(s) and then move both the X-ray source and the detector symmetrically so $\theta_\text{left}$ from the source and $\theta_\text{right}$ to the detector are equal. For educational purposes there are devices that rotate the crystal and the detector simultaneously, ...

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On applying L'Hôpital's rule, we get $y = \frac{2Msin(Mx)cos(Mx)}{2sin(x)cos(x)}$. Again applying L'Hôpital's rule $\frac{sin(Mx)}{sin(x)} = M$, giving $y=M^{2}$. Just here, it is proved that $\frac{sin^{2}(Mx)}{sin^{2}x}$ has maxima at $x = n\pi$. Here n is any integer and not just even integers. In $\frac{sin^{2}(\frac{1}{2}M\vec{a_{1}}.\vec{\Delta k)... 2 X-ray diffraction in a crystal is due to superposition of the reflections from all of the planes formed by the molecules or atoms in the crystal. Those are the cleavage planes, and are also called the Bragg planes. For any given x-ray wavelength and any given spacing of the planes, there are only certain angles where the reflections are in phase with each ... 2 This is a consequence of the distributional identity $$\int_{-\infty}^{\infty} e^{i k x} \, dx = 2 \pi \delta(k).$$ In particular, this implies that $$\iiint e^{i \vec{k} \cdot \vec{x}} \, dV = \left[ \int_{-\infty}^{\infty} e^{i k_x x} \, dx \right] \left[ \int_{-\infty}^{\infty} e^{i k_y y} \, dy \right] \left[ \int_{-\infty}^{\infty} e^{i k_z z} \, dz \... 2 It has to do with optical laws, regardless of the wave you use. Bragg's law gives you the angles in which there is a maximum, but it doesn't say anything about the intensity you will detect on them. If you perform the calculations, you'll find out that the intensity of the maxima rises with N^2, but the width decreases as 1/N. You can check this in a ... 1 I think it is because the Bragg planes are not continuous planes, on the length scale of a typical X ray wavelength. Rather each Bragg plane is an array of regularly spaced scattering centres. So the 'reflection' from a Bragg plane is the result of a discrete sum of terms, in which each term is the scattering from one atom or molecule. The individual atoms ... 1 We need to calculate the phase difference between these two rays. Thus, as stated here we need to calculate$$ \frac{\Delta \phi}{2\pi} = \frac{\textrm{"distance beam 0"} - \textrm{"distance beam 1"}}{\lambda} = \frac{(p + q) - 0}{\lambda} $$Instead of using the index "zero" and "no index" I prefer to use "1" and "2". I also included some angles and draw ... 1 This is more a wording problem than a physics problem. In order to solve this, consider a simple harmonic oscillator, i.e. a particle of mass m on a weightless spring with spring constant D. This system has an eigenfrequency f_0 = \frac{1}{2\pi}\sqrt{\frac{D}{m}}, which is also called resonance frequency. If we excite a harmonic oscillator, it always ... 1 The key idea to all these equations is the superposition of several \cos-terms. So let's quickly review this: Suppose we have to waves A_1 = \cos{(\alpha)} and A_2 = \cos{(\beta)}. If we add them together, we obtain$$ A_{tot}(t, x) = A_1(t, x) + A_2(t, x) = 2 \cos{(\frac{\alpha - \beta}{2})} \cos{(\frac{\alpha + \beta}{2})} % = \cos{(k x - \omega t)... 1 Laue's theory of diffraction takes the approach you suggest. It does not consider any planes actually. It just integrate the contributions from all differential volumes of the crystal, for a given position of the detector. The condition for maxima obtained by this approach is equivalent to Bragg condition. So Bragg law diagram, with planes and Ray paths ... 1 The diffraction pattern is essentially a Fourier transform of the product of the illumination beam and the diffractive structure. The envelope of the power spectrum of the far-field diffraction pattern is determined by the width of the illumination beam. See Fraunhofer diffraction. 1 Your question actually is at the heart of the wave-particle duality, it just states that any particle (in your case a photon) can also be described as a wave, the latter feels interference in the same way you learned in high school. About your second question, you can always argue that these waves are going to interfere at infinity, where you can place your ... 1 Rays, of course, don't exist. Each one of those arrows is a representation of the same plane wave. If you like, you could think of each arrow as a thin pencil chosen (but not separated) from the plane wave. The plane wave, of course, fills all of space. A way to think about this is to imagine that the plane wave is partially reflected at the first ... 1 You can get additional peaks, or even the absence of peaks depending on the unit cell choice. There are two things at part here: the arbitrary choice of non-primitive unit cells, and the structure factors of Bragg peaks. Take a simple cubic lattice with one atom per unit cell. Nothing is stopping you from choosing a body-centered cubic unit cell, even ... 1 Here is my approach - which doesn't quite give the same answer you were looking for... From Bragg's Law, we know $$n\lambda = 2d\sin\theta$$ This tells us that the smaller the angle$\theta$, the shorter the wavelength - and the shorter the wavelength, the greater the energy, since wavelength and energy are related through the relationship:$\$E = \frac{...

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While your answer has elements of "check my work" in it, as discussed in the comments above, I think the basic questions you posed are legitimate. Yes, the question refers to the kinetic energy of the neutrons, and the de Broglie wavelength associated with them; and yes, "other energies" will appear because higher order scattering can occur between the same ...

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With the x-ray diffraction and the Bragg's equation '2dsint=n^ you can get the inter planer spacing 'd' which can also lead you to get the lattice constant which is a vital property of a crystal structure.

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