Calculating the Use of Data Mining Techniques - Assignment Example

Assignment 1 Name Subject Professor Date Solution 1 a. For a Body-centered cubic, the calculations of a bcc are not as hard as expected. Therefore, this involves a certain procedure which will help us achieve attain the goal much easier. We take an assumption of the origin lying parallel with the lattice at the ( 0, 0, 0 ) of the third co-ordinated xyz plane. We set the other lattice point at the ( 1 / 2, 1 / 2, 1 / 2 ) mark. From here, these values are then substituted with the ( x, y, z ) values in the equation below. Fhkl = ∑ I fi e2xi ( hxi + kyi + lzi ) Where: f = the atomic scattering factor. h, k, l = are all integer values. From here, we get the Fhkl = fi { 1 + exi ( h + k + l ) } Since h, k, l are all integer values therefore, we proceed to define the sum h + k+ l = N. the exponential in the equation is normally a representation of the two possible values that would come forth; the + 1 (where the N value is an even number) and the a – 1 (which is for odd numbers) Therefore, we can proceed to say that, F = 2f if h + k + l is even F = 0 if h + k + l is odd Having these as the possible results, therefore, there are no other possible results. The product of the resulting index is therefore shown represented as the below fig 1. This type of a lattice of allowed reflections is also known as the face centered cubic. It is seen different from the fcc lattice as this has all its indices in the reciprocal space as integers. b. The intensities of the reflection are estimated to change with regard to the crystalline orientation. Therefore, the we consider the intensity that has been scattered by the atom which is given by I ἁ ǀ F ǀ2 = [ f1 cos2∏ ( hx1 + ky1 + lz1 ) + f1 sin 2∏ ( hx1 + ky1 + lz1 ] N Where= x1, y1, z1 are co-ordinates and the hkl are the reflection indices. Since we are considering the position coordinate at the (h, k, z )= ( 0,0,0), therefore, the equation I ἁ ǀ F ǀ2 = [ f1 cos2∏ ( hx1 + ky1 + lz1 ) + f1 sin 2∏ ( hx1 + ky1 + lz1 ] Becomes I = [ f1 cos2∏ ( hx1 + ky1 + lz1 ) + f1 sin 2∏ ( hx1 + ky1 + lz1 )] I ἁ f1 [ cos 2∏ x 0 ]2 I ἁ f1 [1] I ἁ f1 If replaced with one that is twice the scattering factor, then the new intensity is going to be calculated as: I = [ f1 cos2∏ ( hx1 + ky1 + lz1 ) + f1 sin 2∏ ( hx1 + ky1 + lz1 )] Since f1 = ½ f2 Therefore it follows that F2 = 2 f1 And I ἁ f1 Substituting f1 = ½ f2 The equation becomes I ἁ ½ f2 2I ἁ f2 Hence, as the intensity factor is doubled, the intensity also doubles. Solution 2. The instantaneous intensity is given by Yci = ( K /ᵘ ) .∑ [ Ljk . ǀFjkǀ2 . ᵩ ijk . Pjk . (Ck / Vk2 )] + ybi The terms which are considered important in the equation include the following:- Vk= volume of unit cell ᵩ ijk = Braggs angle…this is the diffraction angle that is measured experimentally Pjk = this is referred to as the Primitive cubic. It is the measure that explains more about the diffraction condition as found in the primitive unit cells. Some of these cells include the P cubic and the P tetragonal. Here, the cell referred to is at a unit cell corner Fjk = this is a reference to a Bravais lattice. It is called the orthorhombic. Solution 3. a. ) Bragg’s law xy = yz = d sin θ xyz = 2d sin θ xyz = nλ nλ = 2d sin θ and the Primitive cubic example nλ = 2d sin θ 1 h 2 + k 2 + l2 --- = d2 a2 Lambd 1.5406 angstroms a 3.1648 angstroms 2Theta h k l h2+k2+l2 sqrt( h2+k2+l2) d 2d sin θ θ 2 θ 58.276 2 0 0 4 2 1.5824 3.1648 0.4668 27.83 55.66 73.198 2 1 1 6 2.4494897 1.271 2.542 0.6065 37.34 74.68 87.024 2 2 0 8 2.8284271 1.1189 2.2378 0.6664 41.79 83.58 100.651 3 1 0 11 3.3166247 0.9542 1.9084 0.6072 37.39 74.78 141.928 2 2 2 12 3.4641016 0.9136 1.8272 0.6431 61.64 123.28 131.184 3 2 1 14 3.7416573 0.8458 1.6916 0.9107 65.6 131.2 154.603 4 0 0 16 4 0.7912 1.5824 0.5736 35 70 b. ) The h , k, l values for each line are: 2 θ h k l 58.276 2 0 0 73.198 2 1 1 87.024 2 2 0 100.651 3 1 0 141.928 2 2 2 131.184 3 2 1 154.603 4 0 0 c. ) In checking the lattice type, the following steps are carried out:- Since the value of 1/ (d2) is given by : 1 h 2 + k 2 + l2 --- = d2 a2 and a is given by = λ ( √ ( h2 + k2 + l2 ) ) 2 sin θ Reflection h k l ( h2+k2+l2= even) 2 θ 1 2 0 0 4 58.276 2 2 1 1 6 73.198 3 2 2 0 8 87.024 5 2 2 2 12 141.928 6 3 2 1 14 131.184 7 4 0 0 16 154.603 We will plug in the values for the crystal system a = 1.5406 (√ ( h2 + k2 + l2 ) ) 2 sin θ = 1.5406 (√ ( 4 ) ) 2 sin 154.603 = 3.59 Therefore, since 3.59 is close to the value of a, it follows that the lattice is of type BCC. Assignment 2 Solution 1. From the Bragg’s law nλ = 2d sin θ 1 h 2 + k 2 + l2 --- = d2 a2 a = λ ( √ ( h2 + k2 + l2 ) ) 2 sin θ a = 1.54060 (√ ( h2 + k2 + l2 ) ) 2 sin θ We will plug in the values of θ for each crystal system and compare the crystal parameter. The first crystal to be considered is the one with the values of the h, k, and l. Therefore, a1 = 1.54060 {√ ( 4+0+4 ) } = -2.375025204 2 sin 11.76 a2 = 1.54060 {√ ( 1+0+16 ) } = 11.75316153 2 sin 12.84 a3 = 1.54060 {√ ( 1+1 +0 ) } = -1.280776369 2 sin 16.73 a4 = 1.54060 {√ ( 0+0+36 ) } = -4.931421179 2 sin 17.64 a5 = 1.54060 {√ ( 1+1+9 ) } = 31.79293816 2 sin 18.93 a6 = 1.54060 {√ ( 4+0+4 ) } = 3.229597452 2 sin 19.59 a7 = 1.54060 {√ ( 0+4+16 ) } = 3.550614007 2 sin 20.67 a8 = 1.54060 {√ ( 1+1+36 ) } = 19.48717335 2 sin 21.75 a9 = 1.54060 {√ ( 4+1+1 ) } = -2.093487216 2 sin 24.01 a10 = 1.54060 {√ ( 4+1+1 ) } = 2.797156275 2 sin 26.11 a11 = 1.54060 {√ ( 0+1+64 ) } = 6.746791085 2 sin 27.11 a12 = 1.54060 {√ (4 +1+16 ) } = -7.048238349 2 sin 28.79 a13 = 1.54060 {√ ( 9+0+0 ) } = -2.31829047 2 sin 29.93 # h k l h2+k2+l2 sqrt( h2+k2+l2) 2 θ θ Sin θ 2sin θ a 1 0 1 2 5 2.236068 23.51 11.76 -0.7252315 -1.45046 -2.375025204 2 1 0 4 17 4.123106 25.68 12.84 0.27022757 0.540455 11.75316153 3 1 1 0 2 1.414214 33.45 16.73 -0.8505534 -1.70111 -1.280776369 4 0 0 6 36 6 35.27 17.64 -0.9372146 -1.87443 -4.931421179 5 1 1 3 11 3.316625 37.86 18.93 0.08035734 0.160715 31.79293816 6 2 0 2 8 2.828427 39.18 19.59 0.67461578 1.349232 3.229597452 7 0 2 4 20 4.472136 41.33 20.67 0.9702227 1.940445 3.550614007 8 1 1 6 38 6.164414 43.49 21.75 0.24367044 0.487341 19.48717335 9 2 1 1 6 2.44949 48.02 24.01 -0.9012914 -1.80258 -2.093487216 10 1 2 2 9 3 52.21 26.11 0.82616049 1.652321 2.797156275 11 0 1 8 65 8.062258 54.21 27.11 0.92049051 1.840981 6.746791085 12 2 1 4 20 4.472136 57.57 28.79 -0.4887585 -0.97752 -7.048238349 13 3 0 0 9 3 59.85 29.93 -0.9968121 -1.99362 -2.31829047 Fig shoing the X-ray diffraction pattern. The comparison for the a=4.75870, b= 4.75870, and c= 12.99290 values is done with the calculated a value on the table to check for a corresponding value. It is seen that the value of a which is for the specimen α –Al203 corresponds closely with the observation 7 with a (h, k, l) value of (0, 2, 4) which is 3.550614007. The value of b which is for the specimen LaAl103 corresponds closely with the observation 7 with a (h, k, l) value of (0, 2, 4) which is 3.550614007. The value of c which is for the specimen PrAl103 corresponds closely with the observation 2 with a (h, k, l) value of (1, 0, 4 ) which is 11.75316153. Therefore it follows that, since the value a2 for the second observation is closely related to the real value of a with the figure of merit being taken into consideration. Hence it shows that the X-ray powder was contaminated with PrAl103 more than the other specimens α –Al203 and LaAl103. Solution 2. Through the use of data mining techniques, especially those that are geared to highlight some more information on the process identification phase and the other two phases; the minor and the trace phases can be tracked and then recorded for analysis purposes. These data mining attempts help in combining a variety of several other data clues so that the identification of an element can be achieved. The method employed in this experiment is generic. It has the potential to identify and quantitate of even minor and trace phases involving t e diffraction data. This is because, by itself, it is less likely for one to get enough explanation on a data collected but when combined with other observations; one can get a satisfactory analysis. In going about the experiment, various data sets were collected in the ICCD instrumental Clinics. This instrumental clinic was useful in the collection of the specimen. The instruments that were employed in the identification process such as the optical detectors and defractometers were regulated so as to achieve the various inputs needed. For the data mining, there was a variety of data that was used and some used for referencing. These include the Search/match algorithm, the ICDD powder diffraction file ( Pdf +2011 edition), and the Inorganic crystal Structure Database ICSD. The focus in the method was dependent on the sample of the data that was available for the experiment. There were different categories of information which were used in the experiment analysis. These include the information database derived from other experimental analysis. Their aim in this experiment was to make sure the experiment is done on the established standards. This is to help in achieving the goal with less interference or foreseen mishandlings. Also, it included other relevant information that could be used in the derivation or calculation from analysis of patterns. These could include the indexed Candidate unit cell for the unknown or the location and their various intensities arising from the unidentified peaks of the phase analysis. In this experiment, the X-ray radiation is used. This is because of its unique property that allows it to travel through the solid material (Brant et al 1999). This property of the electromagnetic nature is characterized by its shorter wavelengths therefore being able to penetrate through the solids. References Brant, William E., and Clyde A. Helms 1999 Fundamentals of Diagnostic Radiology. 2nd ed. Baltimore: Williams & Wilkins, Read More