datagamma-rayswavesx-rays
The greater the frequency range of a transmission medium, the greater the number of bits per second it can transmit. In other words, the bigger the bandwidth in hertz available, the bigger the bandwidth in bits per second that can be transmitted.
Given that, why we don't use gamma rays, x-rays or ultraviolet to transmit data instead of e.g. visible(light) in optical fiber or microwaves(cell phones)?
Yes, X-ray, UV, and even radio waves are made of photons.
The difference is the Energy (or equivalently, the wavelength). See the picture of the Electromagnetic spectrum . The different nomination comes from the time of the discovery. Your eyes can see the visible part. the radio waves can be observed with antennas, etc. The only difference is the way we observe it. But they are all the same (photons).
Here is a scale of the different wavelength for electromagnetic waves:
The Energy reads $E=mc^2$ for massive particles, which a photon is not. The good relation is: $$ E=pc = \hbar c k = h \nu $$ You have to use the complete Einstein relation: $$ E = \sqrt{p^2c^2 + m^2c^4}$$
And the energy is always finite.
For any photon you can calculate it's energy with the formula $E = h \nu$, where $h \simeq 6.626 \times 10^{-34}\:\mbox{J s}$ is Planck constant and $\nu$ the frequency of the photon.
At the same time you can crudely convert energy to temperature with $E = k_B T$, where $k_B \simeq 1.381 \times 10^{-23}\:\mbox{J/K} = 8.617 \times 10^{-5}\:\mbox{eV/K}$ is Boltzmann constant.
This means that $T = \frac{E}{k_B} = \frac{1 \times 10^6\:\mbox{eV}}{8.617 \times 10^{-5}\:\mbox{eV/K}} \simeq 1.16 \times 10^{10}\:\mbox{K}$, i.e. 10 billions Kelvin (or degrees Celsius, at this point the difference is irrelevant) in one single point. If you distribute that energy over a larger radius, you'll have your "fireball", although at these temperatures speaking of fire may be misleading.
A gamma ray burst could completely wipe out the atmosphere if it happens near enough. Fortunately the sources of such powerful events are also distant enough to not have any direct consequence on us.
Best Answer
For data transmission through optical fibers, what you have to worry about is finding the correct material for transmitting the light over long distances. Consider the following diagram for the attenuation through an optical fiber (from NASA through wikipedia):
You can see that there is a minimum in the losses through the fiber in the IR because both the scattering of light in the fiber and the absorption of light through the fiber are at a minimum here. Since we'd like to transmit information with the least amount of power, this will dictate our choice of radiation energy (wavelength) for fiber optics.
For free space transmission, it wouldn't be good to use "high" energy radiation such as gamma rays and x-rays because they could be dangerous for peoples health. Also, the air can scatter high energy radiation (we see a blue sky because of this scattering of blue and violet light for instance).