The XRD pattern of pure and Pb2+doped Ba2Be2B2O7is presented in Fig. 1, which is in agreement with Ref. [11]. This indicated that Pb2+could be doped into Ba2Be2B2O7 compound instead of Ba up to mol fraction of 0.07 without formation of any secondary phase. The two possible sites available for incorporating Pb2+in Ba2Be2B2O7 lattice are either the Ba+2 sites or the Be2+sites. The Pb2+(1.19˚A for C.N = 6) ion is a much larger ionic radius, compared with that of Be2+(0.27˚A for C.N = 4) ion. So, Pb2+ion is expected to occupy the Ba2+sites and not Be2+sites according to the ionic size considerations.