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

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c = 3 x 10 8 m/s (i) 1 2 mV = eV 2 ⇒ V = = 2eV −19 3 2× 1.6× 10 × 18× 10 m −31 9× 10 − 19+ 3+ 31−1 2 16 10 2 ⇒ V = × × × ⇒ V = 64× 10 ⇒ V = 8× 10 m ⋅ (ii) min 7 14 −1 s hc eV = λ min hc ⇒ λ min = eV −34 8 6.63× 10 × 3× 10 ⇒ λ min = −19 3 1.16× 10 × 18× 10 ⇒ λ ≈ III. -11 6.9 x 10 m An X-ray tube is operated with an anode potential of 10 kV and an anode current of 15 mA. Calculate i.the number of electrons hitting the anode per second, ii.the rate of production of heat at the anode stating any assumptions made, iii.the frequency of the emitted X-ray photon of maximum energy. e = 1.6 x 10 -19 C h = 6.6 x 10 -34 Js Solution I = 15 x 10 -3 A V = 10 x 10 3 V e = 1.6 x 10 -19 C h = 6.6 x 10 -34 Js (i)
charge = 15× 10 Cs second −3 -1 charge −19 = 1.6× 10 C electron −3 electron 15× 10 ⇒ = second −19 1.6× 10 electron 16 ⇒ = 9.375× 10 / second second (ii) Power = V × I -3 -3 Power = 10×10 × 15×10 Power = 150W Entire range of wavelengths produced gives heat. (iii) hfmax = eV eV ⇒ fmax = h −19 3 1.6× 10 × 10× 10 ⇒ fmax = −34 6.63× 10 18 ⇒ fmax = 2.4× 10 Hz IV. (a) What is the threshold frequency for the photoelectric effect on lithium (We = 2.9 eV)? The lithium emitter has a work function of We = 2.9 eV. The important thing to remember is to not get confused with the units of energy (1 eV = 1.6 x 10 -19 Joules). The threshold frequency gives the photon energy (hν) that equals the work function of the emitter (We): We 2.9 eV ν o = = h −15 4.136× 10 eV. s 14 ν o = 7.0× 10 Hz Note the value of Planck's constant in eV.s was used instead of J.s to ensure that the units are correct.
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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
Page 29 and 30: energy. Electronic noise is minimiz
Page 31 and 32: E is the photon energy, ε is the a
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Page 37 and 38: Figure II.11: Schematic diagram of
Page 39 and 40: factor of 2, because also the refle
Page 41 and 42: IV. Instrumentation The major compo
Page 43 and 44: One of the inherent advantages in T
Page 45 and 46: Figure III.6: Sample preparation me
Page 47 and 48: III.1 gives an overview of various
Page 49 and 50: crystal mounted on a 2θ goniometer
Page 51 and 52: Figure IV.4: Collimators with diffe
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Page 57 and 58: II.4.2 Scintillation Counters The s
Page 59 and 60: When using other counting gases (Ne
Page 61 and 62: SECTION V SAMPLE PREPARATION XRF an
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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
Page 79: I. Solution SOLUTIONS An FM radio s
Page 83: h ∆λ = ( 1 − cosθ ) = 2λ c =
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