在氢原子光谱中怎么计算可见光区的四条主要谱线的波长

用公式ΔE = E1 - E2 = hν= hc/λ计算可见光区的四条主要谱线的波长
2025-03-15 12:34:56
推荐回答(1个)
回答1:

解:光子能量ε=hν=hc/λ=6.626×10^(-34)Js×3×(10^8)m/s÷[4340×10^(-10)m]≈4.58×10^(-19)J=4.58×10^(-19)J÷[1.6×10^(-19)J/eV]≈2.86eV。

而氢原子基态能级是Eo=-13.6eV,氢原子的激发态能级能量和基态能量之间存在下列关系:En=Eo/n²,n为氢原子核外电子的主量子数。也就是说:E1=-13.6eV,E2=-3.4eV,E3=-1.51eV,E4=-0.85eV,也就是说ΔE=2.86eV=E5-E2,也就是说这条谱线是从氢原子的第四激发态(第五能级)跃迁到第一激发态(第二能级)的谱线。

★☆★2.86eV是紫色光。

另:为什么氢原子核外电子能量是负的呢?那是因为我们选取的电子的势能平面在无穷远处,也就是说自由态的电子势能不小于零,而束缚态的电子势能都是小于零的!根据库仑定律,电子与质子之间的库仑引力Fc=ke²/r²,其中k=9×(10^9)Nm²/C²,是库仑常数;e=1.6×10^(-19)C,是电子电量;r=5.29166×10^(-11)m,是玻尔半径。所以,库仑引力势能Ep=dFc/dr=-ke²/r,带入数据可求得Ep=-27.2eV。加上电子的动能Ek=0.5mv²,由于电子做圆周运动,库仑引力提供向心力:mv²/r=ke²/r²,所以,mv²=ke²/r,代入动能表达式:Ek=ke²/2r,所以电子总能量E=Ep+Ek=-ke²/2r=-13.6eV。

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