What exactly do light-sensitive elements react to?

I had an old (about 1970s) large power transistor (I think it was a field-effect one, but I am not completely sure...). Knowing it is supposed to be light-sensitive I took the top of the case off, and it has a small EMF between the collector and emitter when exposed to light. But the EMF from directly under  electroluminescent lamps in the school corridor is about 10 times smaller than from lying on a sun-lit windowsill. So light-intensity can't be the only thing the EMF depends on... But what is the other thing? 
PS It can't be UV, because well... we have glass in the windows at school, although it risks to get broken, as there is a football cup at the moment.

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The spectral response of a silicon p-n junction is expected to look something like the graphs below.  The way the junction responds is by producing current (A), and the way its being forced is with light power (W), so the vertical axis "response" is given in units of A/W (amperes/watt).  If you're wondering where the ideal shaped graph came from, I think that's just based on the assumption of perfect quantum efficiency, one photon/second producing one electron/second. That plus a sharp cut-off at the wavelength where the photon energy equals the bandgap energy.  BTW, if you drew the graph as a function of frequency, or photon energy, the line would slope the other way.  Also the low response at low wavelength (high frequency) reflects the fact the incomming photons have much more energy than that needed move just one electron across the junction, but it's still just 1electron per photon, so energy is being lost/wasted at high photon energies.

You might be wondering how this response for diode current, how this translates into a collector-emitter voltage seen across your transistor?  And I don't know the answer to this.  I dont' know what's happening with your transistor.  I know the base-to-emitter on an NPN transistor is a PN junction, is a diode. But it is not clear how a base-to-emitter photocurrent translates into a collector-emitter voltage without the transistor being hooked up to something else.

Anyway, the graphs came from here and here:
seandogue6 years ago
Photons have energy. when the photons strike a semiconductor, they impart some or all of that energy to the outer valence shells (electron orbit. If enough energy is released, the electron becomes decoupled from its host and become motile.
rickharris6 years ago
They react to the energy in the photons that strike them. Different wavelengths have different energy levels.
Re-design6 years ago
They are sensitive to intensity changes AND wave length. Not all light is the same wave length and some elements may be almost totally blind to one wavelength while very excited to another wave length.
gruffalo child (author)  Re-design6 years ago
But the lamps in the corridor are nearly the same 'white-light' as the sun, and glass isolates most other wavelengths (UV)
You sure about that spectrum ? Check with your favourite spectroscope. Glass DOES pass some UV, not a lot, but a little bit.

The whole theory of photoelectricity was a major challenge for physics in the late 19th - early 20th century. Quantum theory was BORN from the study of the photoelectric effect.

You'll have to study solid state physics to gain a deep insight into what is going on, and that's beyond 'ibles really.

KelseyM is our resident Real Physicist.....and I am sure he will be along shortly....