Session #18: Homework Solutions

Session #18: Homework Solutions Problem #1 (a) In a diffractometer experiment a specimen of thorium (Th) is irradiated with tungsten (W) L α radiation. Calculate the angle, θ , of the 4th reflection. (b) Suppose that the experiment described in part (a) is repeated but this time the incident beam consists of neutrons instead of x-rays. What must the neutron velocity be in order to produce reflections at the same angles as those produced by x-rays in part (a)? Solution (a) ν = 1 5 = (74 − 7.4)2 R → λ = 1.476 × 10-10 m 36 λ Th is FCC with a value of Vmolar = 19.9 cm3 ∴ 4 a3 = NA Vmolar 1/3 ⎛ 4 × 19.9 ⎞ → a= ⎜ ⎟ ⎝ 6.02 × 1023 ⎠ λ = 2d sinθ ; d = = 5.095 × 10-8 cm a 2 h + k2 + l2 4th reflection in FCC: 111; 200; 220; 311 → h2 + k2 + l2 = 11 λθ = 2a sinθ h2 + k2 + l2 ⎛ λ h2 + k2 + l2 → = sin-1 ⎜ ⎜ 2a ⎝ ⎞ ⎛ ⎞ ⎟ = sin-1 ⎜ 1.476 11 ⎟ = 28.71o ⎜ ⎟ ⎟ ⎝ 2 × 5.095 ⎠ ⎠ (b) λneutrons = λ x-rays λneutrons = h h h 6.6 × 10−34 = , ∴ v= = = 2.68 × 103 m / s p mv mλ 1.675 × 10−27 × 1.476 × 10−10 Problem #2 A Debye-Scherrer powder diffraction experiment using incident copper (Cu) Kα radiation gave the following set of reflections expressed as 2θ : 38.40°; 44.50°; 64.85°; 77.90°; 81.85°; 98.40°; 111.20°. (a) Determine the crystal structure. (b) Calculate the lattice constant, a. (c) Assume that the crystal is a pure metal and on the basis of the hard-sphere approximation calculate the atomic radius. (d) Calculate the density of this element which has an atomic weight of 66.6 g/mol. Solution Follow the procedure suggested in lecture: (1) Start with 2θ values and generate a set of sin2 θ values. (2) Normalize the sin2 θ values by generating sin2 θn /sin2 θ1 . (3) Clear fractions from the “normalized” column. (4) Speculate on the hkl values that would seem as h2+k2+l2 to generate the sequence of the “clear fractions” column. (5) Compute for each θ the value of sin2 θ /(h2+k2+l2) on the basis of the assumed hkl values. If each entry in this column is identical, then the entire process is validated. (a) For the data set in question, it is evident from the hkl column that the crystal structure is FCC (see table below). (b) λ2 4a2 = λCu Kα sin2 θ h2 + k2 + l2 = 0.0358, = 1.5418 Å, ∴ a = (c) In FCC, 2a = 4r, ∴ r = 1.5418 (4 × 0.0358)1/2 = 4.07 Å 2 × 4.07 Å = 1.44 Å 4 (d) Here we’ll use atomic mass and atomic volume. ρ= 6.02 × 1023 m 4 atoms NA atoms × (4.07 × 10−8 cm)3 = 10.15 cm3 ; ∴ Vmolar = = 3 4 V V a molar ∴ ρ= 66.6 g/mol 10.15 cm3 / mol = 6.56 g/cm3 Data Reduction of Debye-Scherrer Experiment: 2 sin2 normalized 38.40 44.50 64.85 77.90 81.85 98.40 111.20 0.108 0.143 0.288 0.395 0.429 0.573 0.681 1.00 1.32 2.67 3.66 3.97 5.31 6.31 clear fractions 3 4 8 11 12 16 19 (hkl)? sin 2
h + k2 + l2 111 200 220 311 222 400 331 0.0360 0.0358 0.0359 0.0358 0.0358 0.0358 0.0358 2 Problem #3 The following diffractometer data (expressed as 2θ ) were generated from a specimen irradiated with silver (Ag) Kα radiation: 14.10°; 19.98°; 24.57°; 28.41°; 31.85°; 34.98°; 37.89°; 40.61°. (a) Determine the crystal structure. (b) Calculate the lattice constant, a. (c) Assume that the crystal is a pure metal and on the basis of the hard-sphere approximation calculate the atomic radius. (d) At what angle θ would we find the first reflection if, instead of Kα radiation, we used silver L α radiation to illuminate the specimen? Solution We follow the same approach as described in the answer to Problem 2. (a) See table below. It is evident that the crystal structure is BCC. Look at the hkl column. (b) λ2 2 4a = sin2 θ 2 2 2 h +k +l (c) In BCC, = 7.53 × 10−3 , λ Ag Kα 3a = 4r ∴ r = (d) λ = 2 dhkl sin θ , dhkl = λL given by: ν = λ-1 = α = 0.574 Å ∴ a = 0.574 4 × 7.53 × 10-3 = 3.31 Å 3 × 3.31 Å =1.43 Å 4 a a ⎛ λ ⎞ = ∴ θ = sin-1 ⎜ ⎟ 2 ⎝ 2a ⎠ h2 + k2 + l2 5 R(Z 36 - 7.4)2 = 5 36 × 1.1 × 107 (47 - 7.4)2 = 2.40 × 109 m-1 ⎛ 4.17 ⎞ o → λ = 4.17 Å ∴ θ = sin-1 ⎜ ⎟ = 63.0 ⎝ 2 × 3.31 ⎠ Data Reduction of Diffractometer Experiment: incident x-ray AgK 2 sin2 14.10 19.98 24.54 28.41 31.85 34.98 37.89 40.61 0.0151 0.0301 0.0452 0.0602 0.0753 0.0903 0.1054 0.1204 normalize d 1.00 1.99 2.99 3.99 4.99 5.98 6.98 7.97 clear fractions try again 1 2 3 4 5 6 7 8 2 4 6 8 10 12 14 16 α → λ =0.574 Å hkl 110 200 211 220 310 222 321 400 103 sin
h 2 + k 2 + l2 2 7.550 7.525 7.533 7.525 7.530 7.525 7.529 7.525 Problem #4 What is the maximum wavelength ( λ ) of radiation capable of second order diffraction in platinum (Pt)? Solution The longest wavelength capable of 1st order diffraction in Pt can be identified on the basis of the Bragg equation: λ = 2d sin θ . λmax will diffract on planes with maximum interplanar spacing (in compliance with the selection rules): {111} at the maximum value θ (90°). We determine the lattice constant a for Pt, and from it obtain d{111}. Pt is FCC with a value of atomic volume or Vmolar = 9.1 cm3/mole. Vmolar = NA 3 a ; a= 4 3 9.1 × 10−6 × 4 = 3.92 × 10−10 m NA If we now look at 2nd order diffraction, we find 2 λ = 2d{111} sin90° a ∴ λmax = d{111} = 3 = 3.92 × 10−10 3 = 2.26 × 10−10 m Problem #5 What acceleration potential V must be applied to electrons to cause electron diffraction on {220} planes of gold (Au) at θ = 5°? Solution We first determine the wavelength of particle waves ( λp ) required for diffraction and then the voltage to be applied to the electrons: λ = 2d{220} sin θ = 2 aAu = λ= 3 4 × 10.2 × 10−6 6.02 × 1023 2 × 4.08 × 10-10 8 eV = λP = h 2meV 8 sin5° = 4.08 × 10−10 m sin5° = mv2 , ∴ v= 2 h = mv a 4.08 × 10−10 2 × 0.087 = 0.25 × 10−10 m = λp 2eV m , ∴ V= h2 2λ2me = 2415 V Problem #6 How can diffraction on {110} planes of palladium (Pd) be used to isolate Kα radiation from the “white” spectrum of x-rays emitted by an x-ray tube with a copper (Cu) target? Rationalize your answer and provide an appropriate schematic drawing. Solution {110} planes of Pd cannot be used to isolate K α radiation from the x-rays emitted by a tube with a Cu target. Pd has FCC structure and any reflection on {110} planes are destructively interfered with by corresponding {220} planes, composed of “center” atoms. X O X d(220) O O X O X X O X X O X X X X X X O = corner atoms X = center atoms X O X d(110) O X O d{220} = 1 1 d → Δx for {220} reflections = Δx for {110} reflections 2 {110} 2