- Rietveld 법을 이용한 $(Na_{0.3}Sr_{0.7})(Ti_{0.7}M_{0.3})O_3 (M=Ta, Nb)$ 계에서의 결정구조 해석과 상전이 특성
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- 1995년|32권 5호|pp.582-586 (5 pages)
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- 한국세라믹학회
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The crystal structures of (Na0.3Sr0.7)(Ti0.7M0.3)O3 (M=Ta, Nb) compounds were determined using the Rietveld method. Due to the tilting of a oxygen octahedron, (Na0.3Sr0.7)(Ti0.7Nb0.3)O3 had a superlattice of doubled a, b and c of simple perovskite. The crystal structure of (Na0.3Sr0.7)(Ti0.7M0.3)O3 was tetragonal with a space group 14/mmm. The crystal structure of (Na0.3Sr0.7)(Ti0.7M0.3)O3 was a cubic with space group Pm3m, in which no tilting of oxygen octahedron was observed. The difference in the oxygen tilting of these two materials was due to the larger covalency of Nb-O bond than that of Ta-O bond, which induced a strong $pi$Nb0 bonding in (Na0.3Sr0.7)(Ti0.7M0.3)O3. Therefore, the higher transition temperature of (Na0.3Sr0.7)(Ti0.7M0.3)O3 could be related to the larger tilting of oxygen octahedron.