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        <description>Bragg Scattering

Although the proportional counter used to detect the X-rays in this experiment can distinguish different energies, its resolution is limited. Much better energy resolution can be obtained by diffracting x-rays with a crystal of known lattice spacing. Bragg reflection from a single crystal is analogous to the diffraction of visible light from an optical diffraction grating. As formulated by Bragg and von Laue, and as explained, for example, in Kittel's $n\lambda = 2d \sin\theta$…</description>
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        <description>Moseley's Law

While investigating the emission spectra of the elements, Moseley discovered the following empirical formula, relating the energy $E$ of the $K_\alpha$ line of an element to its atomic number $Z$:
  $E=0.75hcR\left(Z-1\right)^2$   (2)  
where $h$ is Planck's constant, $c$ is the speed of light and $R$$n$$E_n = \dfrac{hcRZ^2}{n^2}$$n = 2$$n = 1$$Z-1$$Z-1$$Z$</description>
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        <description>X-ray Absorption
Absorption of radiation may be considered as any mechanism which removes some radiation from a directed beam. For x-rays – those photons with energies from about 100 eV to 100 keV (higher energy than ultraviolet light, but lower than what are typically termed gamma rays), – the two most common interaction modes in the absorber for removing x-rays from a beam are the photoelectric effect and Rayleigh scattering, of which the photoelectric dominates. Since these mechanisms are ene…</description>
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        <description>X-ray fluorescence

When a material is exposed to x-rays with energy larger than an atomic electron’s binding energy, the photon may be absorbed and the electron will be ejected from the atom. The vacancy created leaves the atom in an unstable state. An electron from a higher energy state will fall into the lower state, and a photon will be emitted. The emission of this secondary photon is called fluorescence. Since the energy levels of each atomic species is unique, so is the energy of the fluo…</description>
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        <description>X-ray Production in a Metal Target

When a metal target in an x-ray tube is struck by a beam of electrons accelerated through a voltage V, two concurrent processes give rise to an x-ray emission spectrum.

First, the electrons lose kinetic energy in Coulomb interactions with nuclei in the target. This lost energy produces a continuous spectrum of photons called $E_{max} = eV - \Phi$$\Phi$$\Phi$$E = eV = hc/\lambda.$$hc$$hc \approx 1240 \textrm{eV}\cdot\textrm{nm}$$K_n = 1$$K_\alpha$$n = 2$$n = 1…</description>
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