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 [itex]f\lambda =v[/itex]. 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 [itex]d(\sin\alpha\pm\sin\beta )=N\lambda[/itex]. 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.