Energy of a photon is given by the expression, E=hf. Is this the frequency of most prominent wave? Due to the localized nature of photon, photon consist of all possible wave length from zero to infinity and consequently all frequency from zero to infinity.
It is given by that expression when its energy is absorbed, but it is not anymore a photon then. When it gets hv, it gets localized and is matter already with mass hv/c^2 and is affected by time, before the absorption wasn't matter, had zero mass and wasn't affected by time and had all the properties of a non local entity.
Ramanand
As per the quantum cannon, if a photon has a well defined four-momentum (h\omega,\hbar \vec k) then it has not a well defined position. If only the frequency is known---like in atomic transitions, when there is no a-priori information on the direction of motion of photons--you will have an indetermination in the components of \vec k of the order 2 h w / c. The corresponding indetermination in position is \lambda/2, where \lambda is the wave length.
Speed of light(c) = wavelength(𝜆) * frequency (f). if f =0, then c = 0. No light ... it's dark. Therefore, f must be between zero and infinity.
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