If you had just googled you'd have found your answer here...
The frequency spectrum of Cherenkov radiation by a particle is given by the Frank–Tamm formula:
$d^2E \over dxd\omega$$=$$q^2 \over 4\pi$$\mu(\omega)\omega$$(1-{c^2 \over v^2n^2(\omega)})$
The Frank–Tamm formula describes the amount of energy E emitted from Cherenkov radiation, per unit length traveled x and per frequency $\mu$. $\mu(\omega)$ is the permeability and $n(\omega)$ is the index of refraction of the material the charge particle moves through. q is the electric charge of the particle, v is the speed of the particle, and c is the speed of light in vacuum.
Unlike fluorescence or emission spectra that have characteristic spectral peaks, Cherenkov radiation is continuous. Around the visible spectrum, the relative intensity per unit frequency is approximately proportional to the frequency. That is, higher frequencies (shorter wavelengths) are more intense in Cherenkov radiation. This is why visible Cherenkov radiation is observed to be brilliant blue. In fact, most Cherenkov radiation is in the ultraviolet spectrum—it is only with sufficiently accelerated charges that it even becomes visible; the sensitivity of the human eye peaks at green, and is very low in the violet portion of the spectrum.
As in sonic booms and bow shocks, the angle of the shock cone is directly related to the velocity of the disruption. The Cherenkov angle is zero at the threshold velocity for the emission of Cherenkov radiation. The angle takes on a maximum as the particle speed approaches the speed of light. Hence, observed angles of incidence can be used to compute the direction and speed of a Cherenkov radiation-producing charge.
But mind that all of the above applies to particles moving in a medium at a velocity higher than the speed of light in that medium.
