作者用快速淬火方法制备了(Fe_(1-x)Cr_x)_(80)B_(20)(x=0,0.05,0.1,0.2,0.3)的非晶态薄带.用磁天平测量了升温和降温的磁化强度与温度的关系.温度范围为77～1200K,升温速率为10K/分,磁场强度为12000Oe.得到非晶态合金的比饱和磁化强度σ从x=0时的198emu/g减少到x=0.3时的33cmu/g,推导出Fe原子具有2.05玻尔磁子,Cr原子以4.01玻尔磁子与Fe原子磁矩成反平行.实验结果说明B的电子转移到Fe原 The authors prepared amorphous ribbons of (Fe_ (1-x) Cr_x) _ (80) B_ (20) (x = 0,0.05,0.1,0.2,0.3) by rapid quenching method. And the relationship between the magnetization and the temperature at which the temperature is lowered, the temperature range is 77-1200 K, the heating rate is 10 K / min and the magnetic field strength is 12000 Oe, and the specific saturation magnetization σ of the obtained amorphous alloy decreases from 198 emu / g at x = 0 To 33cmu / g at x = 0.3, it is deduced that the Fe atom has a 2.05 Bohr magnet, and the Cr atom has an antiparallelism with the Fe atom Mohr’s moment at 4.01 and the Cr atom shows that the B electron transfer to the Fe