Pend Fisika Zat Padat (3) close packing

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Pend Fisika Zat Padat (3) close packing

  1. 1. Pertemuan 3CLOSE PACKING STRUCTURE<br />IwanSugihartono, M.Si<br />JurusanFisika, FMIPA<br />UniversitasNegeri Jakarta<br />1<br />
  2. 2. Crystals<br />06/01/2011<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />2<br /><ul><li>Crystal structure basics
  3. 3. unit cells
  4. 4. symmetry
  5. 5. lattices
  6. 6. Diffraction
  7. 7. how and why - derivation
  8. 8. Some important crystal structures and properties
  9. 9. close packed structures
  10. 10. octahedral and tetrahedral holes
  11. 11. basic structures
  12. 12. ferroelectricity</li></li></ul><li>Objectives<br />By the end of this section you should:<br /><ul><li>understand the concept of close packing
  13. 13. know the difference between hexagonal and cubic close packing
  14. 14. know the different types of interstitial sites in a close packed structure
  15. 15. recognise and demonstrate that cubic close packing is equivalent to a face centred cubic unit cell</li></ul>06/01/2011<br />3<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
  16. 16. Packing<br />Can pack with irregular shapes<br />06/01/2011<br />4<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
  17. 17. Two main stacking sequences:<br />If we start with one cp layer, two possible ways of adding a second layer (can have one or other, but not a mixture) :<br />06/01/2011<br />5<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
  18. 18. Two main stacking sequences:<br />If we start with one cp layer, two possible ways of adding a second layer (can have one or other, but not a mixture) :<br />06/01/2011<br />6<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
  19. 19. Let’s assume the second layer is B (red). What about the third layer?<br />Two possibilities:<br />(1) Can have A position again (blue). This leads to the regular sequence …ABABABA…..<br />Hexagonal close packing (hcp)<br />(2) Can have layer in C position, followed by the same repeat, to give …ABCABCABC…<br />Cubic close packing (ccp)<br />06/01/2011<br />7<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
  20. 20. Hexagonal close packed<br />Cubic close packed<br />06/01/2011<br />8<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
  21. 21. No matter what type of packing, the coordination number of each equal size sphere is always 12<br />We will see that other coordination numbers are possible for non-equal size spheres<br />06/01/2011<br />9<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
  22. 22. Metals usually have one of three structure types:<br />ccp (=fcc, see next slide), <br />hcp or <br />bcc (body centred cubic)<br />The reasons why a particular metal prefers a particular structure are still not well understood<br />06/01/2011<br />10<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
  23. 23. ccp = fcc ?<br />Build up ccp layers (ABC… packing)<br />Add construction lines - can see fcc unit cell<br />c.p layers are oriented perpendicular to the body diagonal of the cube<br />06/01/2011<br />11<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
  24. 24. Hexagonal close packed structures (hcp)<br />hcp<br />bcc<br /><br />06/01/2011<br />12<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
  25. 25. Recurring themes...<br />Foot and mouth virus<br />Body centred cubic<br />06/01/2011<br />13<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
  26. 26. Packing Fraction<br />We (briefly) mentioned energy considerations in relation to close packing (low energy configuration)<br />Rough estimate - C, N, O occupy 20Å3<br />Can use this value to estimate unit cell contents<br />Useful to examine the efficiency of packing - take c.c.p. (f.c.c.) as example<br />06/01/2011<br />14<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
  27. 27. So the face of the unit cell looks like:<br />Calculate unit cell side in terms of r:<br />2a2 = (4r)2<br />a = 2r 2<br />Volume = (162) r3<br />Face centred cubic - so number of atoms per unit cell =corners + face centres = (8  1/8) + (6  1/2) = 4<br />06/01/2011<br />15<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
  28. 28. Packing fraction<br />The fraction of space which is occupied by atoms is called the “packing fraction”, , for the structure<br />For cubic close packing:<br />The spheres have been packed together as closely as possible, resulting in a packing fraction of 0.74<br />06/01/2011<br />16<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
  29. 29. Group exercise:<br />Calculate the packing fraction for a primitive unit cell<br />A = 2 r<br />06/01/2011<br />17<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
  30. 30. 06/01/2011<br />18<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
  31. 31. 06/01/2011<br />19<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
  32. 32. MencariFraksi Packing<br />Jumlah atom efektifdalam unit cell = 12(1/6)+2(1/2)+3=6<br />06/01/2011<br />20<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
  33. 33. Primitive<br />06/01/2011<br />21<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
  34. 34. Close packing<br />Cubic close packing = f.c.c. has =0.74<br />Calculation (not done here) shows h.c.p. also has =0.74 - equally efficient close packing<br />Primitive is much lower: Lots of space left over!<br />A calculation (try for next time) shows that body centred cubic is in between the two values.<br />THINK ABOUT THIS! Look at the pictures - the above values should make some physical sense!<br />06/01/2011<br />22<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
  35. 35. Hitunglahefisiensi packing dankerapatandariNaClbiladiberikan data sebagaiberikut:<br />Jari-jari ion Na = 0,98 A<br />Jari-jari ion Cl = 1,81 A<br />Massa atom Na = 22,99 amu<br />Massa atom Cl = 35,45 amu<br />?<br />Solusinya:<br />Parameter kisi, a = 2 (Jari-jari ion (Na + Cl)) = 5.58 A<br />Fraksi Packing:<br /> = Volume ion yang adadalamsebuah unit cell Volume unit cellnya<br /> = 4 (4/3) phi (r3Na + r3Cl) / a3 = 66,3 %<br />Density:<br /> = Massa unit cell / Volumenya<br /> = 2234 kg m-3<br />1 amu = 1,66 x 10-27 kg<br />06/01/2011<br />23<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
  36. 36. Summary<br /><ul><li>By understanding the basic geometry of a cube and use of Pythagoras’ theorem, we can calculate the bond lengths in a fcc structure
  37. 37. As a consequence, we can calculate the radius of the interstitial sites
  38. 38. we can calculate the packing efficiency for different packed structures
  39. 39. h.c.p and c.c.p are equally efficient packing schemes</li></ul>06/01/2011<br />24<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
  40. 40. THANK YOU<br />06/01/2011<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />25<br />

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