Poor efficiency but nice wide spectrum.
1,100W over 100ft2 is ~11W/ft2. That's rather low average intensity. They also have to be hung higher like other HID to create a uniform spread which means the irradiance drops even further. You'd want multiple of them in a 10×10 imo.
Fluorescent lights are plasma lights too. They electrically excite gases inside a chamber which induces a phase change about the gasses (4 phases of matter, solid, liquid, gas, plasma; gasses become a plasma) and as the newly formed plasma fluctuates within its state, electrons are jumping back and forth between energy levels and in the instance when they vacate higher orbitals for lower orbitals they expell radiation in a frequency/wavelength which happens to lie within the ~UV/visible range. Fluorescent lights realistically only emit higher frequencies of light (depending on Hg vapor pressure), much and most of it UV light. In order to create visible light out of the high energy UV created from the excited plasma, they powder the inside of the tube with a phosphor layer. This phosphor layer absorbs some of the energy from the UV photons, and as a result the UV photons no longer have enough energy to truly be UV photons. At this point the UV photons have been effectively converted into visible light photons.
"Ya, but within the visible light range, there's lots of colors, so what color photon was created?"
The amount of energy which the phosphor layer absorbed from each photon will determine the final color of the converted photon. As far as colors go, red is the lowest energy photon and blue is the highest energy photon. If the phosphor layer absorbed much of the UV photons initial energy, then the resulting photon would be Red in color, and if the phosphor layer absorbed only a small amount of the UV photons initial energy, then the resulting photon would be closer to Blue in color. How much energy a photon has, determines its wavelength/frequency/color. The phosphor layer absorbs some of the original photons energy, and this % of energy absorbed by the phosphor layer is converted to heat which increases the phosphor layer temp, but the resulting photons that are re-emitted out the other side have been effectively downgraded in energy which ultimately results in the phenomenon of converting the hard to see UV photons into useful visible light photons. This is why old fluorescent, or CFL with visible cracks on the inside of the tube emit greater amounts of UV. The cracks are areas void of phosphor coating, which means no absorption/conversion is happening at these places, and if the total % of cracks increases with time, then so will the UV.
Other induction lights will use different mixtures of gasses to create different spectrums of light rather than relying soley on the phosphor coating to convert the higher energy light into broader lower energy light.
This is also what most LED do. Full spectrum (or any phosphored) LED actually use only 1 or 2 mono color LED of a lower wavelength (typically ~450nm base pump) to which they slap a phosphor film over the top to then convert/downgrade the high energy base pump color(s) into a wider spectrum. Kinda like shining a spot light through certain spots of a stained glass window would create different colors on either side of the glass. White light goes into the stained glass, mono colored light comes out, but in the case of LEDs and fluorescents or lights utilizing a phosphor filter, its the other way around, UV or blue light (ie relatively mono color) color goes into the stained glass/phosphor, white light or wide spectrum visible light comes out.
You can't upgrade photons. At least not yet. So far we can only take energy away. Hypothetically and generally speaking, this means that you'd need to use a high energy base pump to create a wide blend of visible light. If you used a green base pump you'd only be able to create green -> red/IR photons. You couldn't create blue photons with a green base pump because you can't upgrade photons.