Cerenkov light - a practical calculation

I need to calculate the approximate total amount of energy radiated, via Cerenkov, by a muon as it traverses $10\:\rm{cm}$ of quartz glass. Unfortunately I've spent so much time fiddling with electronics in the lab recently that I've completely forgotten how to do maths.

The formula to use is the Frankâ€“Tamm Equation:

$$\frac{dE}{dx} = \frac{q^2}{4\pi} \int_{v>c/n(\omega)} \mu(\omega)\omega\biggl(1-\frac{c^2}{v^2n^2(\omega)}\biggr)d\omega$$

Where $\mu(\omega)$ and $n(\omega)$ are the permeability and refractive index of the medium being traversed by the charged particle.

For a $5\:\rm{keV}$ muon travelling through quartz glass I have looked up the following numbers (but I'm really not sure over what frequency range to consider the last two):

$$x=0.1\:\rm{m},$$ $$q=1.6\times10^{-19}\:\rm{C},$$ $$v\approx c,$$ $$n(\omega)=1.54-1.46 (\lambda=200-700\:\rm{nm})\approx1.5,$$ $$\mu(\omega)\approx1NA^2.$$

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You are going to need a better equation for refractive index... you need something which captures the deep-uv resonance and the region at even higher frequencies where n decreases below 1, probably around 100 nm. – user2963 Feb 20 '12 at 14:51
Hi zephr. Of course your right if the integral were carried out over the full range of frequencies. In practice however, I'll be using something like this[1] to detect ring patterns and it is sensitive only to wavelengths in the range 300-650nm. [1]: sales.hamamatsu.com/index.php?id=13199716&language=2&; – qftme Feb 20 '12 at 16:06
ok, but you may still need to carry out the full integral, because v will be changing due to the energy loss across all frequencies. Since most of the radiation is in the hard UV range, this may be significant - I can't tell you for sure though. – user2963 Feb 20 '12 at 16:09
Isn't it okay to assume that the muon would be travelling at $c$ throughout it's passage through the glass? It is only 10cm afterall. For a $5keV$ muon: $1-v/c=2.4\times10^{-5}$. Either way, my attempt at the calculation gave $4\times10^{-47}J$ which I really don't think is correct. Basically I'm still stuck :( – qftme Feb 20 '12 at 17:46
@zephyr, could you perhaps type up your calculation into an answer? Allbeit approximate, I think it will be sufficient for my purposes and will therefore duly accept it. – qftme Feb 22 '12 at 11:13