Rotational Energy

For a molecule of $O_2$ 
a) Determine the lowest energy of rotation.
b) What is the energy of the photon that will excite the molecule into its first excite state (starting from energy zero).
c) Compare with $H_2$ molecule. 

For free motion ($V =0$) Schrodinger equation is
translation: $-\frac{\hbar^2}{2m}*\frac{d^2}{d x^2}*ψ=Eψ$  
and rotation: $-\frac{\hbar^2}{2I}*\frac{d^2}{dφ^2}*Φ=EΦ$
From the similarity of the two equations we get for rotational motion the solution:
$Φ(φ)=\sqrt{1/2π}*\exp⁡(imφ)$  with $m=0,+or-1,+or-2,…$   (see particle in a ring solutions)
Back into the initial equation with $Φ(φ)$ and we have
$E=\frac{\hbar^2 m^2}{2I}$

For $O_2$ the distance between the 2 atoms is $d=1.2075*10^{-10} m$ and mass of one $O$ atom is $m=16 amu=16*1.66*10^{-27} kg=2.656*10^{-26} kg$ which gives a moment of inertia of
$I=2mR^2=(md^2)/2=1.936*10^{-46}  kg*m^2$

The first nonzero energy is
$E_1=\frac{\hbar^2}{2I}=\frac{(1.0546*10^{-34}) ^2}{2*1.936*10^{-46}}=2.87*10^{-23} J=0.1795 meV$
The wavelength of the corresponding photon is
$E_1=\frac{h c}{λ}$  that is $λ=\frac{h c}{E_1} =\frac{6.626*10^{-34}*3*10^8}{2.87*10^{-23}}=6.926 mm$
This wavelength corresponds to microwaves.
c)
Imagine you have two spinners, one heavier and one lighter about the same dimension. The lighter spinner (Hydrogen) is easier to place in rotation because its moment of inertia is smaller. Thus the first nonzero energy of the $H_2$ molecule is bigger and so the corresponding wavelength for $H_2$ is smaller.


valentin68

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