# Convertion between MHz and Ångström

1. Jun 29, 2011

### Hjensen

I'm sure this is a rather simple calculation, but I just can't seem to get it right. During a recent lecture a professor gave an example concerning hyperfine-structure splitting. This is the essence of it:

We're considering a Lithium-7 isotope, which exhibits a fine-structure splitting (due to electron spin) and a hyperfine-structure splitting (nuclei spin). The resonant line of the isotope has two fine-structure components at 6707.76Å and 6707.91Å respectively. Each of these is further split into two major hyperfine-structure components, which are separated by 803MHz.

He then calculated the resolution (roughly 45000) and the minimum width required of a diffraction grating with 1800 rulings per millimeter to resolve the fine-structure lines (25mm). Here comes my first question: For this to work, wouldn't we have to assume that the grating can be used at first order only? Is this the case, and if so, why?

Next he did the same for hyperfine-structure components - this is the part I'm really having problems with. Since we have the difference in MHz, we need to convert this to Ångströms. At the lecture he said, at this wavelength, that 803MHz corresponds to 0.012Å. At the time I figured it was a simple calculation involving the usual $f\lambda =v$. However, after many tries I still cannot see how he gets to that result. Any ideas?

Also, in such an experiment, how would one need to illuminate the grating, and where would the spectrum be observed? There are obvious constraints on the angles involved, of course, since $d(\sin\alpha\pm\sin\beta )=N\lambda$. As far as I can see, this will only work for N=1 and +. So the angles must be on the same side from the normal to the grating.

2. Jun 29, 2011

### phyzguy

As far as converting between frequency and wavelength, since the line widths are very small compared to the base wavelengths, we can approximate the lineiwdths as differentials. Starting with:
$$\nu = \frac{c}{\lambda}$$ Now differentiate both sides and solve for d-lambda:
$$d\nu = -\frac{c}{\lambda^2}d\lambda; d\lambda = -d\nu \frac{\lambda^2}{c}$$
If I plug in the numbers (d-nu = 803 MHz, lambda = 6708 Angstroms), I in fact get d-lambda = .012 Angstrom.

On your other questions, I am less sure. Generally the first order is used because the higher orders are a lot less light-efficient, so your professor may have implicitly assumed the first order. I don't really understand your last question about the angles. Are you talking a reflection grating or a transmission grating? Usually transmission gratings are used slightly off-normal, and the specular reflection (n=0) is dumped into a light absorber and the first order is analyzed for the spectrum.

Last edited: Jun 29, 2011
3. Jun 29, 2011

### Hjensen

Thanks a lot - I get it now.

Meanwhile, I've spotted another thing in my notes: It says that the 0.012Å also corresponds to $0.0268cm^{-1}$. How can this be? This just looks like there should be a factor for the purpose of division? Is this the case, or is it something a bit more sophisticated this time?

4. Jun 29, 2011

### phyzguy

Well, wavenumber is just the reciprocal of wavelength. So, proceeding as before (calling the wavenumber n):
$$n = \frac{1}{\lambda}; \nu = \frac{c}{\lambda} = c * n; d\nu = c * dn$$

So if d-nu is 803 MHz, dn is just 803 Mhz / c = .0267 cm^-1.