[例]某电子在两能级间跃迁的能量差为4969×10-19 J,求其波长为多少纳米?其波数为多少? [解]由AE=hv=hc/λ得 λ=hc/△E 6.626×1034×3×1010/4.969 1019 =4×105cm =400nm d=1/λ=1/4×105cm=25000 cI[例] 某电子在两能级间跃迁的能量差为4.969 × 10-19 J，求其波长为多少纳米？其波数为多少？ [解] 由 ΔE = h ν = h c / λ 得 λ = h c / ΔE = 6.626 × 10-34 × 3 × 1010 / 4.969 × 10-19 = 4 × 10-5 cm = 400 nm σ = 1 / λ = 1 / 4 × 10-5 cm = 25000 cm-1