Atomic Spectra, transmission of photons

[Brewer]

Homework Statement

A molecule with angular momentum L and moment of inertia I has a rotational energy that can be written as [tex]E=\frac{L^2}{2I}[/tex]. Assuming that angular momentum is quantized according to Bohr's rule [tex]L = n\hbar[/tex], find the wavelength of the photons emitted in the n=2 to n=1 tranistion of the [tex]H_2[/tex] molecule. This molecule has moment of inertia [tex]I = \frac{1}{2}mr_0^2[/tex], where m=938[tex]MeV/c^2[/tex] and [tex]r_0[/tex] = 0.074nm.

Homework Equations

I assumed only the ones given in the question.
Possibly the Bohr Model

The Attempt at a Solution

By using the rotational energy alone as given above I got a [tex]\delta E[/tex] = 2.30x{tex]10^{13}[/tex] J, which in turn lead to an emitted wavelength of something of the order 48. I can see that this is clearly wrong.

Do I need to use the Bohr model equation for this as well somehow? The rotational energy given by the numbers in the question is tiny, thus giving a huge rotational energy. But the energies given from the Bohr model are quite small, so simple addition of them is pointless, as the Bohr model won't have much (if any) effect on the overall energy will it?

 

[Dick]

It looks like you have everything you need. Your energy is clearly way off. I would suggest you put the numbers back in again - or show some intermediate steps.

 

[Brewer]

Looks like I forgot the squared term in the numerator of the fraction.

When I remember this, it looks like the wavelength I come out with is of the order [tex]10^{-5}[/tex]. This looks better doesn't it?

 

[f(x)]

Looks like I forgot the squared term in the numerator of the fraction.

When I remember this, it looks like the wavelength I come out with is of the order [tex]10^{-5}[/tex]. This looks better doesn't it?

No idea abt molecules but for Bohr species its generally of the order[tex]10^{-7}[/tex] m

 

[Dick]

Looks like I forgot the squared term in the numerator of the fraction.

When I remember this, it looks like the wavelength I come out with is of the order [tex]10^{-5}[/tex]. This looks better doesn't it?

This is a molecular rotational transition, much longer wavelength than Bohr. 10^(-5) is about right.