Fisika Zat Padat Pertemuan Pertama : Ceramic

  

Fisika Zat Padat

Pertemuan Pertama :

Ceramic

  

Fisika Zat Padat

Pertemuan Pertama :

Ceramic

  

Pengaruh jari-jari ion

Pengaruh jari-jari ion

kecil pada pergerakan

kecil pada pergerakan

  

(pacing) ion-ion besar

(pacing) ion-ion besar

KOORDINASI ION-ION SEBAGAI FUNGSI RASIO JARI-JARI IONIK

  Bilangan Disposisi ion-ion sekitar ion Rasio jari-jari (r/R) koordinasi pusat Linear 0.155 - 0

  2 Corners of triangle 0.225 – 0.155

  3 Corners of tetrahedron 0.414 – 0.225

  4 Corners of octahedron 0.732 – 0.414

  6 Corners of cube 1.0 – 0.732

  8

  GEOMETRI KOORDINASI TRIANGULAR GEOMETRI KOORDINASI TRIANGULAR Geometri koordinasi triangular ion-ion kecil (jari-jari = r) dan ion-ion besar (jari-jari = R) Geometri koordinasi triangular ion-ion kecil (jari-jari = r) dan ion-ion besar (jari-jari = R) Bukti : rasio jari-jari (r/R) Bukti : rasio jari-jari (r/R) Jadi rasio jari-jari (r/R) koordinasi triangular = 0,155 – 0,225. Jadi rasio jari-jari (r/R) koordinasi triangular = 0,155 – 0,225.

  Pekerjaan rumah 1 : Buktikanlah rasio jari-jari (r/R) pada :

  

a) Koordinasi tetrahedron (CN = 4) sebesar 0,225 – 0,414 ?

  

b) Koordinasi oktahedron (CN = 6) sebesar 0,414 – 0,732 ?

  c) Koordinasi kubus (CN = 8) sebesar 0,732 – 1,000? Pekerjaan rumah 1 : Buktikanlah rasio jari-jari (r/R) pada :

  

a) Koordinasi tetrahedron (CN = 4) sebesar 0,225 – 0,414 ?

  

b) Koordinasi oktahedron (CN = 6) sebesar 0,414 – 0,732 ?

  c) Koordinasi kubus (CN = 8) sebesar 0,732 – 1,000?

  JARI-JARI IONIK (Å) JARI-JARI IONIK (Å)

  0.62 Ba +2

  0.69 Ti +4

  0.65 Au +1

  1.37 Cr +3

  0.64 Mg +2

  0.69 U +6

  0.82 Au +3

  0.85 Cr +6

  0.45 Mn +4

  0.53 V +5

  0.53 B +3

  0.20 Cs +1

  1.70 N +3

  0.16 W +6

  1.35 Cu +1

  0.62 Co +3

  0.96 Na +1

  1.02 Y +3

  0.95 Be +2

  0.43 Cu +2

  0.72 Nd +3

  1.04 Zr +4

  0.81 Bi +3

  0.93 Dy +3

  0.92 Ni +2

  0.72 CO ₃ -2

  1.31 Bi +5

  0.74 Er +3

  0.97 O +6

  0.10 OH -1

  0.63 Li +1

  0.68 At +7

  Ion Valensi Jari- jari Ion Valensi Jari- jari Ion Valensi Jari- jari Ion Valensi Jari- jari Ac +3

  0.62 Rh +3

  1.15 Br +7

  0.39 F -7

  0.08 Pb +2

  1.24 Ag +1

  1.22 C +4

  0.18 F -1

  1.34 Pt +4

  0.65 Ag +2

  0.89 Ca +2

  1.00 Fe +2

  0.77 Rb +1

  1.48 Al +3

  0.51 Cd +2

  0.99 Ga +3

  0.69 Am +3

  1.33 Ta +5

  1.03 Ce +3

  1.13 Ge +4

  0.50 S -2

  1.83 Am +4

  0.91 Cl +7

  0.27 Hg +3

  1.11 Sb +5

  0.62 As +3

  0.58 Cl -1

  1.81 I -1

  2.18 Sn +2

  0.93 As +5

  0.47 Co +2

  0.76 K +1

  1.33

  Pekerjaan rumah 2: Predict the coordination number (bilangan

koordinasi) and crystal structure for the ions solids

CsCl and NaCl? Use the following ionic radii for the prediction : Cs + = 0,170 nm; Na + = 0,102 nm; Cl - = 0,181 nm

  

Pekerjaan rumah 2:

Predict the coordination number (bilangan

koordinasi) and crystal structure for the ions solids

CsCl and NaCl? Use the following ionic radii for the prediction : Cs + = 0,170 nm; Na + = 0,102 nm; Cl - = 0,181 nm

  INTRODUCTION : FERROELECTRIC

  INTRODUCTION : FERROELECTRIC

MATERIAL APPLICATION

MATERIAL APPLICATION

BST/PZT, IR SENSOR, SATELLITE

FERROELECTRIC/PYROELECTRIC MATERIAL

  

Ferroelectric material perovskite structure (PZT/BST)

(a) Up Polarization (b) Down Polarization

Ferroelectric material perovskite structure (PZT/BST) (a) Up Polarization (b) Down Polarization (a)

  (b)

  1 and 0 = binary

   Use at room temperature,

   Wide range of response frequency,

   Quick response in comparison with other temperature sensor,

   High quality materials for pyrosensor is unnecessary.

   Use at room temperature,

   Wide range of response frequency,

   Quick response in comparison with other temperature sensor,

   High quality materials for pyrosensor is unnecessary.

  Merit of Ba x Sr 1-x TiO

  3 and PbZr x Ti 1-x O

  3 based pyrosensor compared to other infrared sensor materials, such as semiconductors are : Merit of Ba x Sr 1-x TiO

  3 and PbZr x Ti 1-x O

  3 based pyrosensor compared to other infrared sensor materials, such as semiconductors are :

  Lattice constants of Ba x

Sr

  1-x TiO

  3 and PbZr x Ti 1-x O

  3 ferroelectric ceramic using XRD method Lattice constants of Ba x Sr 1-x TiO

  3 and PbZr x Ti 1-x O

  3 ferroelectric ceramic using XRD method

  Methodology PTZT ferroelectric ceramic Methodology PTZT ferroelectric ceramic

  5.3020 gram 6.6980 gram 0.0600, 0.1200, 0.1296, lead titanate lead zirconate 0.1800 gram tantalum oxide (PbTiO , 99 %] [PbZrO , 99,7 %] [Ta O , 99,9 %] _{3 3 2 5} A pellet pressing by the presser 12 grams of Mixing by at 31.43 MPa for PTZT powder bowl of

  15 minutes and sintering by _o polishing for furnace at 850 C for 10 hours.

  6 hours.

  PTZT thin

  Analyzing and

  Preparation DC

  12 grams

  films of

  characterizing :

  Unbalanced Magnetron Pt (200)/

  of PTZT the structure (XRD Diano

  Sputtering (UBMS) SiO /Si(100) ₂ bulk.

  type 2100E), morphology

  method substrate

  surface and films thickness (SEM JEOL Stop type JSM-35C).

  

Flow diagram on research

  

Spectrum XRD PTZT

Spectrum XRD PTZT

  Analytic method for cubic Analytic method for cubic

  (1) 2 d sin , 

   

  2

  2

  2 1 h k l  

  (2) 

  2

  2 d a

  where : d = interplanar spacing; a, b, c = the lattice constants;

  h, k, l = the planes indices;  = wave length (Cu element = 1.54056 Å);  = angle of diffraction. Perhitungan lengkap dari konstanta kisi dari target PZT didadah 1.08 % Ta ₂ O ₅

  (PTZT) menggunakan metode analitik struktur kubik Puncak

  2 (deg.)

   (deg.)

  Sin ² 

  (Sin ² )/2

  (Sin ² )/3

  (Sin ² )/4

  (Sin ² )/5 1 21.85 10.925 0.03588 0.01794 0.01196 0.00897 0.00718

  2 31.10 15.550 0.07180 0.03589 0.02393 0.01795 0.01436 3 38.35 19.150 0.10751 0.05375 0.03593 0.02688 0.0215 4 44.55 22.275 0.14354 0.07177 0.04785 0.03589 0.02871 5 50.15 25.075 0.17944 0.08972 0.05981 0.04486 0.03589 6 55.30 27.650 0.21516 0.10758 0.07172 0.05379 0.04303 7 64.40 32.200 0.28370 0.14185 0.09326 0.07093 0.05674 8 68.35 34.175 0.31525 0.15763 0.10508 0.07881 0.06305 9 73.10 36.550 0.59622 0.17717 0.11811 0.08858 0.07087

  Puncak 2

  2 31.10 15.550 0.01197 2.00099 2 1 1 0

  8 68.35 34.175 0.05254 8.78626 9 3 0 0

  7 64.40 32.200 0.04663 7.79756 8 2 2 0

  6 55.30 27.650 0.03586 5.99663 6 2 1 1

  5 50.15 25.075 0.02991 5.00111 5 2 1 0

  4 44.55 22.275 0.02392 4.00063 4 2 0 0

  3 38.30 19.150 0.01792 2.99629 3 1 1 1

  1 21.85 10.925 0.00598 1.00000 1 1 0 0

  (deg.) 

  s h k l

  )/A

  2

  )/6 (Sin

  2

  (deg.) (Sin

  9 73.10 36.550 0.05906 9.87568 10 3 1 0 untuk memperoleh nilai konstanta kisi dengan

menggunakan hubungan . Jadi nilai konstanta kisi

struktur kubuk untuk material PZT doping 1,08 %

Ta O ) diperoleh sebesar :

  2

  5 a =  /(2A) = 1,54056/(20,03589) = 4,066 Å. Cu Cohen method for tetragonal Cohen method for tetragonal

  2 ² sin C B A , 

         

  (1)    

  2 d sin ,    _{2 2}   (3) sin C B A , ,

      

          _{2 2}

  2

  2

  2  1 h k l

   sin C B A ,

      

    

      

     .

  2

  2

  2 (2) d a c

  where : d = interplanar spacing; a, b, c = the lattice constants;

  h, k, l = the planes indices;  = wave length (Cu element = _{2 2 2}

• + k ;

  ;  = l 1.54056 Å);  = angle of diffraction,  = h _{2 2 2 2 2}

  /(4c /(4a ); A, B, C =  = 10 sin 2; A = D/10; B =  ); C =  numeric number. We carry out A, B, C using Cramer method. No Puncak h k l 2

     ²

  25 6. 2 1 1 55,4 27,70

  Perhitungan lengkap dari konstanta kisi dari target PZT didadah 1 % Ta ₂ O ₅

   320

  81 9. 3 1 0 73,6 36,80 10 100

  9

  64 8. 3 0 0 70,2 35,10

  8

  1 7. 2 2 0 65,2 32,60

  25

  5

  5

   1. 1 0 0 21,6 10,80

  16 5. 2 1 0 50,2 25,10

  4

  1 4. 2 0 0 44,7 22,35

  4

  2

  4 3. 1 1 1 38,8 19,40

  2

  1 2. 1 1 0 31,5 15,75

  1

  (PTZT) menggunakan metode Cohen (struktur tetragonal)

  No 

  2

    Sin

  2

  2 Sin

  2

   

  

  2

  1. 0,1355157 0,0351118 1,355157 1,836450 2. 0,2730048 0,0736799 2,730048 7,453159 3.

  1 2 0,3926323 0,1103310 3,926323 15,41602 4. 0,4947641 0,1446003 4,947641 24,47915 5. 0,5902596 0,1799452 5,902596 34,84064 6.

  1 5 0,6775535 0,2160781 6,775535 45,90787 7. 0,8240600 0,2902740 8,240600 67,90748 8. 0,8852566 0,3306310 8,852566 78,36793 9. 0,9202833 0,3588293 9,202833 84,69214

  

  2

  7 360,9008 No      sin

  2

    sin

  2

    sin

  2

   1. 1,355157 0,0351118 0,0475819 2. 5,460095 0,1473598 0,2011497 3. 3,926323 7,852647 0,2206620 0,1103310 0,4331953 4. 19,79056 0,5784011 0,7154302 5. 29,51298 0,8997258 1,0621435 6. 6,775535 33,87767 1,0803906 0,2160781 1,4640449 7. 65,92480 2,3221917 2,3920314 8. 79,67310 2,9756794 2,9269332 9. 92,02833 3,5882927 3,3022459

   10,701858 335,4753 11,847815 0,3264091 12,5447559

  11,847815 320 C

  7 B 335, 475300 A     

  0,3264091

7 C 2 B 10, 701858 A .

       12,5447559 335, 475300 C 10,701858 B 360,900800 A

     

  Untuk memperoleh bilangan numerik A, B, C dipecahkan dengan metode Cramer, maka diperoleh bilangan numerik

  C = 0,0035726585 dan B = 0,035501007.

  Selanjutnya diperoleh konstanta kisi PZT doping 1 % Ta O (PTZT)

  2

  5

  untuk :

  a = Cu/(2C) = 1,54056/(20,035726585) = 4,075 Å;

c = Cu/(2B) = 1,54056/(20,035501007) = 4,088 Å; c/a ratio = 1,003 No Puncak h k l 2

     ²

   1. 1 0 0

  21.85 2. 1 1 0 31,10 3. 1 1 1 38,35 4. 2 0 0 44,55 5. 2 1 0 50,15 6. 2 1 1 55,30 7. 2 2 0 64,40 8. 3 0 0 68,35 9. 3 1 0 73,10

  

  Pekerjaan Rumah ke -2

  Perhitungan lengkap dari konstanta kisi dari target PZT didadah 1,08 % Ta ₂ O ₅ (PTZT) menggunakan metode Cohen (struktur tetragonal).

  2

  2

  2

  2

   Sin Sin

     2   No 1.

  2.

  3.

  4.

  5.

  6.

  7.

  8.

  9.

  

  2

  2

  2

       sin   sin   sin  No 1.

  2.

  3.

  4.

  5.

  6.

  7.

  8.

  9.

  