(Ba <sub>1x </sub > Sr <sub> x </sub>)<sub>4</sub>(Sm<sub>0.4</sub > Nd <sub>0.6</sub>)<sub>28/3</sub > Ti <sub>18</sub > O <sub&g...(Ba <sub>1x </sub > Sr <sub> x </sub>)<sub>4</sub>(Sm<sub>0.4</sub > Nd <sub>0.6</sub>)<sub>28/3</sub > Ti <sub>18</sub > O <sub>54</sub>(x = 0.02, 0.04, 0.06, 0.08, 0.1 ) 稳固的解决方案被准备由常规固态反应过程。它被发现那(Ba <sub>1x </sub > Sr <sub> x </sub>)<sub>4</sub>(Sm<sub>0.4</sub > Nd <sub>0.6</sub>)<sub>28/3</sub > Ti <sub>18</sub > O <sub>54</sub> 陶艺充分由 BaSm <sub>2</sub 组成 > Ti <sub>4</sub > O <sub>12</sub> 和乐队 <sub>2</sub > Ti <sub>5</sub > 为所有作文的 O <sub>14</sub> 阶段。增加的 x 价值(0.02 x 0.1 ) 在((Ba <sub>1x </sub > Sr <sub> x </sub>)<sub>4</sub>(Sm<sub>0.4</sub > Nd <sub>0.6</sub>)<sub>28/3</sub > Ti <sub>18</sub > O <sub>54</sub> 陶艺不能仅仅获得高展开更多
基金Financial supports of the National Natural Science Foundation of China,the Middle-aged and Young Teachers in Colleges and/or Universities in Guangxi Basic Ability Promotion Project of China (Grant No.KY2016YB534) are gratefully acknowledged by the authors
文摘(Ba <sub>1x </sub > Sr <sub> x </sub>)<sub>4</sub>(Sm<sub>0.4</sub > Nd <sub>0.6</sub>)<sub>28/3</sub > Ti <sub>18</sub > O <sub>54</sub>(x = 0.02, 0.04, 0.06, 0.08, 0.1 ) 稳固的解决方案被准备由常规固态反应过程。它被发现那(Ba <sub>1x </sub > Sr <sub> x </sub>)<sub>4</sub>(Sm<sub>0.4</sub > Nd <sub>0.6</sub>)<sub>28/3</sub > Ti <sub>18</sub > O <sub>54</sub> 陶艺充分由 BaSm <sub>2</sub 组成 > Ti <sub>4</sub > O <sub>12</sub> 和乐队 <sub>2</sub > Ti <sub>5</sub > 为所有作文的 O <sub>14</sub> 阶段。增加的 x 价值(0.02 x 0.1 ) 在((Ba <sub>1x </sub > Sr <sub> x </sub>)<sub>4</sub>(Sm<sub>0.4</sub > Nd <sub>0.6</sub>)<sub>28/3</sub > Ti <sub>18</sub > O <sub>54</sub> 陶艺不能仅仅获得高