X-Ray Fluorescence Analytical Techniques - CNSTN : Centre ...

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(b) What is the stopping potential if the wavelength of the incident light is 400 nm? The stopping potential is calculated using Einstein's Photoelectric Equation for the light wavelength λ = 400 nm. The photon energy for the incident wavelength is: hc 1240 eV. nm E = hν = = = 3.10 eV λ 400 nm Note that a standard value of hc = 1240 eV.nm was used. Get used to using certain combinations of fundamental constants, and manipulating the value of these constants in different units. This will save a lot of time doing calculations, and will ensure that answer comes out with the correct units. The stopping potential energy is then: eVs = hν− We = 3.10 − 2.9eV eVs = 0.20 eV Therefore, the stopping potential is VS = 0.20 Volts (and is negative by definition). V. A television tube operates at 20,000 V. What is λmin for the continuous x-ray spectrum produced when the electrons hit the phosphor? The minimum emission wavelength for X-ray generation is given by the Duane-Hunt Rule. By accelerating the electron through a region of total potential V = 20,000 volts, we have given it a kinetic energy of eV = 20,000 eV = 20 keV. Therefore: VI. hc 1240 eV. nm λ min = = eV 20.000eV λ min = 0.062 nm A photon having 40 keV scatters from a free electron at rest.What is the maximum energy that the electron can obtain? The photon scattering off the electron undergoes a Compton shift. It transfers some of its energy and momentum to the electron and emerges with a reduced energy and momentum (and therefore increased wavelength λ'). The electron kinetic energy is then equal to the difference between the photon energies (by conservation of energy). The electron will have a maximum kinetic energy when the Compton shift is largest. The maximum Compton shift occurs when cosθ is a minimum (cosθ = -1). The Compton shift is then given by:
h ∆λ = ( 1 − cosθ ) = 2λ c = 0.0486 Angstroms moc Where λC is the Compton wavelength (λC = 0.0243 Angstroms). The incident photon wavelength is: hc 1240 eV. nm λ= = = 0.031nm = 0.31 A E 40000eV The scattered photon wavelength is then λ'= λ + ∆λ = 0.31+ 0.0286 A = 0.3386 A. The equivalent photon energy then: hc 1240 eV. nm E′ = = = 36.621eV ≈36.6 keV λ′ 0.03386nm Therefore, the maximum electron kinetic energy is: VII. Ke = E − E′ = 40.0 − 36.6keV = 3.4keV How much photon energy would be required to produce a proton-antiproton pair? Where could such a high-energy photon come from? According to conservation of energy, the minimum photon energy required to produce a particle-antiparticle pair is just equal to twice the rest mass energy of the particle: 2 E = 2 mpc = 2× 938.3 MeV = 1.877 GeV This energy could be obtained in a particle accelerator or a synchrotron.
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X-Ray Fluorescence Analytical Techn
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II.4 Energy Resolution III. Spectru
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Module: Title: X-Ray Fluorescence A
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very high resolution and X-ray phot
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where: c = speed of light (2.99782
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Figure I.2: Bohr’s atomic model,
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decelerated as they penetrate the m
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The mass attenuation coefficient ha
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e neglected provided that momentum
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In higher atomic number materials,
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Figure I.14: A Bremsstrahlung (Cont
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filters but more often with pulse h
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SECTION II ENERGY DISPERSIVE X-RAY
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just above the absorption edges of
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energy. Electronic noise is minimiz
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Page 65 and 66: SECTION VI QUANTITATIVE ANALYSIS Th
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Page 77 and 78: I. (The Photon Concept) EXERCICES A
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