# Rotational energy of H2 molecule

1. Apr 21, 2010

### Oddbio

1. The problem statement, all variables and given/known data
Consider the H2 molecule. The two nuclei (protons) have spin 1/2 and can therefore be in a total spin S=0 or an S=1 state.
What is the orbital angular momentum of the two-nucleon system in the lowest energy state for the two values of the total spin?

2. Relevant equations
$$H_{rot}=\frac{\vec{L}^{2}}{2I}$$

For diatomic molecules:
$$E_{rot}=\frac{\hbar^{2}l(l+1)}{2I}$$

3. The attempt at a solution
Well it's asking for the lowest energy state, which would be n=1, so l=0. Therefore the answer would be zero.
But that can't be right.. so is "l" in this case a different "l" than the one used in the hydrogen atom for example?
Or perhaps the value of "n" is irrelevant to this question so then "l" could basically have any value you want.
In that case the answer would be plugging in l=1.
It only asks for the orbital angular momentum though, so that would just be with this equation:
$$\vec{L}^{2}=l(l+1)\hbar$$
right?
If so, then my answer would be sqrt(2*h-bar) is that it?

I doubt that's correct though because then why would it ask for the energy of both spins, because the way I'm doing it would give the same answer for the spin singlet and the spin triplet.
I must be missing something, can anyone give me a little advice? Thanks.

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