“…As shown in Figure b, with increasing the BaO doping amount from 0 to 0.175, the XRD peak for the BiCuSeO δ ceramic block is slightly shifted from 30.23 to 29.87°. This result indicates that the lattice constant is increased because Ba 2+ (1.42 Å) with a larger ionic radius replaces Bi 3+ (1.03 Å). , BiCuSeO δ is made up of (Cu 2 Se 2 ) 2– conductive layers and (Bi 2 O 2 ) 2+ insulating layers stacked alternatively along the c axis, and these layers are modified as [(Bi 1– x Ba x ) 2 O 2 ] (2–2 x )+ and [Cu 2 Se 2 ] (2–2 x )– due to substitution, respectively. , However, compared with the Bi 0.825 Ba 0.175 CuSeO δ sample, the diffraction peaks for the Bi 0.8 Ba 0.2 CuSeO δ sample shift to a higher angle, which is ascribed to the solid solubility limit of Ba in the BiCuSeO δ matrix. When the doping content exceeds 0.175, the excess BaO reacts with Se to form the BaSeO 3 phase.…”