4.解:J从0→1跃迁时 △v 由:△=2B→B =10.30cm1=10.30×102m h h B →I =2.718×10kg 8丌2Ic 8丌2Bc × 0 1×35 =163×10-27K nma(1+35)×6.02×1023 |2.718×10 -47 I=m2→r=.= V163×1027 129.3pm1 2 1 10.30 10.30 10 2 ~ 2 ~ − − = =    = B  B = cm m   4 7 2 2 2 2.718 10 . 8 8 kg m Bc h I Ic h B − =  = =    4. 解：J从0→1跃迁时 由： K g N m N m N m N m H Cl H Cl 2 7 2 3 0 0 0 0 1.63 10 (1 35) 6.02 10 1 35 − =  +    = +   = pm I I r r 129.3 1.63 10 2.718 10 2 7 4 7 2 =   =  = = − −  