The spacing of the rotational energy levels for the hydrogen molecule H₂ is most nearly
A. 10−9 eV
B. 10−3 eV
C. 10 eV
D. 10 MeV
E. 100 MeV
(GR9277 #90)
Solution:
Rotational kinetic energy: E = L²∕2I
Angular momentum: L² = l(l+1)ħ² with l = 0,1,2,⋯
Moment Inertia: I = mr²
For hydrogen atom:
r = 0.529 × 10−10 m
m =10−27 kg (mass of proton)
→ I = 10−27(0.529×10−10)2 ≈ 10−47 kgm2
ħ = h∕2π = 6.63×10−34∕ 2(3.14) ≈ 10−34
For l = 0 → L = 0 → E = 0
For l = 1 → L = 2ħ² → E = ħ²∕I
The spacing of the rotational energy levels:
∆E = ħ²∕I − 0 = (10−34)2∕ 10−47 = 10−68+47 = 10−21 J
Convert to eV: 1 eV = 1.6 × 10−19 J
→ ∆E = 10−21∕ 1.6 × 10−19 ≈ 10−3 eV
Answer: B
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