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 Frank–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