Rotational energy levels vs l quantum number

  • Thread starter svayl
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  • #1
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Main Question or Discussion Point

H2 has a moment of inertia equal to 4.603 x 10-48 kg m2.

1) Calculate its bond length.
2) For the first 3 rotational energy levels, find the
a) l quantum number
b) ml quantum number
c) the degeneracy of each rotational level
d) energy eigenvalues
e) the magnitude of l


Ok so I calculate bond length using the I=mu(reduced mass) * r^2

I get confused when it comes to l. Are the first three rotational energy levels equal to 1,2,3? or 0,1,2? So would l be equivalent to these energy levels and the m will be +/- l?

Where would I find the energy eigenvalues and magnitude of l?

Thanks in advance!
 

Answers and Replies

  • #2
DrClaude
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The notation is strange, as ##l## is usually used for the orbital angular momentum of electrons, and it ##J## that is used for molecular rotation.

Are the first three rotational energy levels equal to 1,2,3? or 0,1,2?
Rotational angular momentum (like orbital angular momentum) starts at 0 (the solution to the angular differential equation is expressed in terms of the spherical harmonics).

So would l be equivalent to these energy levels and the m will be +/- l?
The ##M_J## quantum number is quantized (!), with value from ##-J## to ##+J## (not just the two extremes, but all intermediate values, separated by 1).

d) energy eigenvalues
The rotational energy of a diatomic molecule is ##E_\mathrm{rot} = B J (J+1)##, with ##B## the rotational constant,
$$
B = \frac{\hbar^2}{2 I}
$$
e) the magnitude of l
The magnitude of quantum orbital angular momentum (including spin) is always ##\sqrt{l(l+1)} \hbar##.
 

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