Changeset 2277 for Tutorials


Ignore:
Timestamp:
May 18, 2016 1:24:44 PM (6 years ago)
Author:
vondreele
Message:

fix a few typos

File:
1 edited

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  • Tutorials/StackingFaults-I/Stacking Faults-I.htm

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    233250<p class=MsoNormal>In this exercise you will use GSAS-II to simulate the
    234251diffraction patterns from faulted diamond. Diamond most commonly has the
    235 well-known cubic structure with the space group Fd3m and a=3.5668A. The C-atom
    236 is at 1/8,1/8,1/8 and can be viewed as a cubic stacking of ruffled hexagonal
    237 nets along the cubic cell 111 diagonal. </p>
    238 
    239 <p class=MsoNormal><img width=482 height=361 id="Picture 1"
     252well-known cubic structure with the space group Fd3m and a=3.5668A. The C-atom is
     253at 1/8,1/8,1/8 and can be viewed as a cubic stacking of ruffled hexagonal nets
     254along the cubic cell 111 diagonal. </p>
     255
     256<p class=MsoNormal><img width=482 height=361
    240257src="Stacking%20Faults-I_files/image001.jpg"></p>
    241258
    242 <p class=MsoNormal>The structure of lonsdaleite has those layers stacked
     259<p class=MsoNormal>The structure of londsdaleite has those layers stacked
    243260hexagonally and thus a faulted diamond structure may occasionally have
    244261hexagonal stacked layers instead of all cubic ones. From geometric
     
    248265Otherwise, if there are many such faults then the diffraction pattern will show
    249266streaks. To simulate the streaks one must first develop a model for the
    250 hexagonal net of C-atoms and then show how they stack in either the cubic or hexagonal
    251 forms. GSAS-II uses a suite of subroutines from the DIFFaX program (M.M.J.
    252 Treacy, J.M. Newsam &amp; M.W. Deem, (1991), Proc. Roy. Soc. Lond. 433A,
    253 499-520) to calculate the diffraction pattern via a general recursion algorithm
    254 and a randomized set of stacked layers. NB: this calculation can be quite time
    255 consuming particularly if unreasonable demands are made on it. </p>
     267hexagonal net of C-atoms and then show how they stack in either the cubic or
     268hexagonal forms. GSAS-II uses a suite of subroutines from the DIFFaX program
     269(M.M.J. Treacy, J.M. Newsam &amp; M.W. Deem, (1991), Proc. Roy. Soc. Lond.
     270433A, 499-520) to calculate the diffraction pattern via a general recursion
     271algorithm and a randomized set of stacked layers. NB: this calculation can be
     272quite time consuming particularly if unreasonable demands are made on it. </p>
    256273
    257274<p class=MsoNormal>If you have not done so already, start GSAS-II.</p>
     
    266283select it; the data window will display the default for the General tab.</p>
    267284
    268 <p class=MsoNormal><img width=620 height=334 id="Picture 2"
     285<p class=MsoNormal><img width=620 height=334
    269286src="Stacking%20Faults-I_files/image002.gif"></p>
    270287
     
    274291style='font-family:"Calibri",sans-serif'>diamond</span></b>’.</p>
    275292
    276 <p class=MsoNormal><img width=624 height=370 id="Picture 3"
     293<p class=MsoNormal><img width=624 height=370
    277294src="Stacking%20Faults-I_files/image003.gif"></p>
    278295
     
    286303the window will be redrawn.</p>
    287304
    288 <p class=MsoNormal><img width=624 height=464 id="Picture 4"
     305<p class=MsoNormal><img width=624 height=464
    289306src="Stacking%20Faults-I_files/image004.gif"></p>
    290307
    291308<p class=MsoNormal>The a cell parameter for diamond stacking can be assumed to
    292 be a<span style='font-size:12.0pt;font-family:"Times New Roman",serif;
    293 position:relative;top:3.0pt'><img width=27 height=21
    294 src="Stacking%20Faults-I_files/image005.gif"></span> = <b><span
    295 style='font-family:"Calibri",sans-serif'>2.522</span></b> and the c cell
    296 parameter is a<span style='font-size:12.0pt;font-family:"Times New Roman",serif;
    297 position:relative;top:3.0pt'><img width=27 height=21
    298 src="Stacking%20Faults-I_files/image006.gif"></span> = <b><span
    299 style='font-family:"Calibri",sans-serif'>2.059</span></b> since there are 3
    300 layers along the 111 diamond cell diagonal. Enter these in the appropriate
    301 places; the cell volume will be revised.</p>
     309be a<span style='position:relative;top:3pt'><img width=27 height=21
     310src="Stacking%20Faults-I_files/image005.gif">&nbsp;= <b><span style='font-family:
     311"Calibri",sans-serif'>2.522</span></b> and the c cell parameter is a<span
     312style='position:relative;top:3pt'><img width=27 height=21
     313src="Stacking%20Faults-I_files/image006.gif">&nbsp;= <b><span style='font-family:
     314"Calibri",sans-serif'>2.059</span></b> since there are 3 layers along the 111
     315diamond cell diagonal. Enter these in the appropriate places; the cell volume
     316will be revised.</p>
    302317
    303318<p class=MsoNormal>Now we have to describe the two hexagonal nets that will be
     
    306321will be redrawn.</p>
    307322
    308 <p class=MsoNormal><img width=624 height=479 id="Picture 5"
     323<p class=MsoNormal><img width=624 height=479
    309324src="Stacking%20Faults-I_files/image007.gif"></p>
    310325
     
    315330and the window will be redrawn with one line in the layer table.</p>
    316331
    317 <p class=MsoNormal><img width=624 height=479 id="Picture 6"
     332<p class=MsoNormal><img width=624 height=479
    318333src="Stacking%20Faults-I_files/image008.gif"></p>
    319334
     
    326341the window should show the new position.</p>
    327342
    328 <p class=MsoNormal><img width=624 height=479 id="Picture 8"
     343<p class=MsoNormal><img width=624 height=479
    329344src="Stacking%20Faults-I_files/image009.gif"></p>
    330345
     
    333348will appear. </p>
    334349
    335 <p class=MsoNormal><img width=486 height=363 id="Picture 10"
     350<p class=MsoNormal><img width=486 height=363
    336351src="Stacking%20Faults-I_files/image010.jpg"></p>
    337352
     
    344359for the layer symmetry. The window should look like this when done.</p>
    345360
    346 <p class=MsoNormal><img width=624 height=624 id="Picture 11"
     361<p class=MsoNormal><img width=624 height=624
    347362src="Stacking%20Faults-I_files/image011.gif"></p>
    348363
     
    354369reminiscent of how carbon sheets stack in graphite. </p>
    355370
    356 <p class=MsoNormal><img width=479 height=358 id="Picture 12"
     371<p class=MsoNormal><img width=479 height=358
    357372src="Stacking%20Faults-I_files/image012.jpg"></p>
    358373
     
    375390like.</p>
    376391
    377 <p class=MsoNormal><img width=477 height=357 id="Picture 13"
     392<p class=MsoNormal><img width=477 height=357
    378393src="Stacking%20Faults-I_files/image013.jpg"></p>
    379394
     
    384399twin fault will be shown.</p>
    385400
    386 <p class=MsoNormal><img width=483 height=361 id="Picture 14"
     401<p class=MsoNormal><img width=483 height=361
    387402src="Stacking%20Faults-I_files/image014.jpg"></p>
    388403
    389 <p class=MsoNormal>If you enter 1 2 1 2 1 2 1 2 then the structure of
    390 lonsdaleite will be shown; 1 1 1 1 1 1 or 2 2 2 2 2 2 gives the diamond
    391 structure. </p>
     404<p class=MsoNormal>If you enter 1 2 1 2 1 2 1 2 then the structure of londsdaleite
     405will be shown; 1 1 1 1 1 1 or 2 2 2 2 2 2 gives the diamond structure. </p>
    392406
    393407<p class=MsoNormal>Finally we must select transition probabilities; they should
     
    410424options.</p>
    411425
    412 <p class=MsoNormal><img width=242 height=136 id="Picture 15"
     426<p class=MsoNormal><img width=242 height=136
    413427src="Stacking%20Faults-I_files/image017.gif"></p>
    414428
     
    420434The data window will be redrawn</p>
    421435
    422 <p class=MsoNormal><img width=624 height=474 id="Picture 16"
     436<p class=MsoNormal><img width=624 height=474
    423437src="Stacking%20Faults-I_files/image018.gif"></p>
    424438
     
    428442(or <b><span style='font-family:"Calibri",sans-serif'>U</span></b>); I got</p>
    429443
    430 <p class=MsoNormal><img width=484 height=455 id="Picture 17"
     444<p class=MsoNormal><img width=484 height=455
    431445src="Stacking%20Faults-I_files/image019.gif"></p>
    432446
     
    447461some choices.</p>
    448462
    449 <p class=MsoNormal><img width=322 height=268 id="Picture 18"
     463<p class=MsoNormal><img width=322 height=268
    450464src="Stacking%20Faults-I_files/image020.gif"></p>
    451465
     
    453467lab data</span></b>; a new popup will appear</p>
    454468
    455 <p class=MsoNormal><img width=318 height=342 id="Picture 19"
     469<p class=MsoNormal><img width=318 height=342
    456470src="Stacking%20Faults-I_files/image021.gif"></p>
    457471
     
    490504interesting stuff around each peak.</p>
    491505
    492 <p class=MsoNormal><img width=624 height=587 id="Picture 20"
     506<p class=MsoNormal><img width=624 height=587
    493507src="Stacking%20Faults-I_files/image022.gif"></p>
    494508
     
    504518<p class=MsoNormal>A perhaps useful means of exploring the effects of changing
    505519stacking parameters is doing a sequence of simulations varying one parameter
    506 over a range. To try this out do Operations/Sequence simulations from the
    507 Layers menu; a popup will appear</p>
    508 
    509 <p class=MsoNormal><img width=279 height=191 id="Picture 21"
     520over a range. To try this out do <b><span style='font-family:"Calibri",sans-serif'>Operations/Sequence
     521simulations</span></b> from the Layers menu; a popup will appear</p>
     522
     523<p class=MsoNormal><img width=279 height=191
    510524src="Stacking%20Faults-I_files/image023.gif"></p>
    511525
     
    514528parameter</span></b> pulldown choose <b><span style='font-family:"Calibri",sans-serif'>TransP;0;0</span></b>;
    515529this is the layer 1 to layer 1 transition probability. Then change the no.
    516 steps to <b><span style='font-family:"Calibri",sans-serif'>10</span></b> (11
    517 will be calculated). We have the same choices for instrument broadening as
    518 above; use the default. Press <b><span style='font-family:"Calibri",sans-serif'>Ok</span></b>;
     530steps to <b><span style='font-family:"Calibri",sans-serif'>10</span></b> (11 will
     531be calculated). We have the same choices for instrument broadening as above;
     532use the default. Press <b><span style='font-family:"Calibri",sans-serif'>Ok</span></b>;
    519533the next popup allows selection of a powder pattern (e.g. for comparison and
    520534the range for the calculation). Press <b><span style='font-family:"Calibri",sans-serif'>Ok</span></b>
     
    530544result?</span></b> box; a new plot will appear.</p>
    531545
    532 <p class=MsoNormal><img width=624 height=535 id="Picture 22"
     546<p class=MsoNormal><img width=624 height=535
    533547src="Stacking%20Faults-I_files/image024.gif"></p>
    534548
     
    538552I’ve done this for the next plot.</p>
    539553
    540 <p class=MsoNormal><img width=624 height=535 id="Picture 23"
     554<p class=MsoNormal><img width=624 height=535
    541555src="Stacking%20Faults-I_files/image025.gif"></p>
    542556
    543 <p class=MsoNormal>The first blue line is for pure hexagonal lonsdaleite
    544 stacking and the last magenta line is for pure cubic diamond stacking. You can see
    545 how some lines quickly vanish with the introduction of stacking faults while
    546 other persist across the entire sequence. This is a good place to save your
    547 project file; the sequential result will be included.</p>
     557<p class=MsoNormal>The first blue line is for pure hexagonal londsdaleite stacking
     558and the last magenta line is for pure cubic diamond stacking. You can see how
     559some lines quickly vanish with the introduction of stacking faults while other
     560persist across the entire sequence. This is a good place to save your project
     561file; the sequential result will be included.</p>
    548562
    549563<h2>Simulation 4. Modelling clustering in diamond</h2>
     
    551565<p class=MsoNormal>In this simulation we will explore the possibility that the
    552566stacking history affects the probability of a fault. In the case of diamond, a
    553 fault to form lonsdaleite could be followed by similar layers until a lower
     567fault to form londsdaleite could be followed by similar layers until a lower
    554568probability fault converts the structure back to diamond. The crystal then has
    555 blocks of diamond structure interleaved with blocks of lonsdaleite. This is
     569blocks of diamond structure interleaved with blocks of londsdaleite. This is
    556570best done in a new phase so we don’t mess up the above simulations, but most of
    557571the data in the current phase is useful for the cluster model. The easiest way
     
    567581General tab for the new phase will appear.</p>
    568582
    569 <p class=MsoNormal><img width=624 height=336 id="Picture 24"
     583<p class=MsoNormal><img width=624 height=336
    570584src="Stacking%20Faults-I_files/image026.gif"></p>
    571585
     
    573587of the tabs; select it.</p>
    574588
    575 <p class=MsoNormal><img width=624 height=474 id="Picture 26"
     589<p class=MsoNormal><img width=624 height=474
    576590src="Stacking%20Faults-I_files/image027.gif"></p>
    577591
     
    594608<p class=MsoNormal>&nbsp;</p>
    595609
    596 <p class=MsoNormal><img width=624 height=474 id="Picture 28"
     610<p class=MsoNormal><img width=624 height=474
    597611src="Stacking%20Faults-I_files/image028.gif"></p>
    598612
     
    609623<p class=MsoNormal>&nbsp;</p>
    610624
    611 <table class=MsoTableGrid border=1 cellspacing=0 cellpadding=0
    612  style='border-collapse:collapse;border:none'>
     625<table class=MsoNormalTable border=0 cellspacing=0 cellpadding=0
     626 style='border-collapse:collapse'>
    613627 <tr style='height:21.45pt'>
    614628  <td width=72 valign=top style='width:54.05pt;border:solid windowtext 1.0pt;
     
    748762Transition tables should look like</p>
    749763
    750 <p class=MsoNormal><img width=624 height=620 id="Picture 29"
     764<p class=MsoNormal><img width=624 height=620
    751765src="Stacking%20Faults-I_files/image029.gif"></p>
    752766
     
    762776will look like</p>
    763777
    764 <p class=MsoNormal><img width=624 height=535 id="Picture 30"
     778<p class=MsoNormal><img width=624 height=535
    765779src="Stacking%20Faults-I_files/image030.gif"></p>
    766780
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