Q:

How long would a light emitting diode be able to power itself?

If I connect a solar panel to a light emitting diode and point the diode towards the solar panel, shining some light over it, some energy will be given back to the emitting diode.
This will stablish a cycle: the light emitting diode emits light, captured by the solar panel and used to power the light emitting diode, continuing the process.
Eventually, though, this cycle will end. How can we find how long would the cycle go on? Stating differently, how long would the light emitting diode power itself?

A:

Basically nothing happens. You might expect some exponential decay in the amount of light remaining in your circuit, but the decay is almost instantaneous.
First, if you're seeing it, a lot of light is escaping from this circuit. You could improve this by enclosing your circuit in a mirrored cavity, and using some kind of sensor or a pinhole to observe. The light would still escape very quickly - assume each time the light crosses the diameter ($d$) of the cavity it has a chance to escape via the pinhole or absorption by the sensor, proportional to the fraction of area ($A_p/A_c$) it covers. The amount of light in the cavity then decreases exponentially and we can solve 
$$(1-A_p/A_c) = e^{-t_d/\tau}$$
where $t_d = d/c$, for the time constant 
$$\tau = -\frac{d}{c \ln (1-A_p/A_c)}$$
For a 1m cavity with a 0.1mm wide pinhole, this is about 1.3 seconds... I was actually not expecting a human time scale there! But in the real world, with imperfect mirrors ($10^{-5}$ losses each reflection) the time constant drops to about 300 microseconds.
Second, consider the actual circuit in the context of this mirrored cavity. The LEDs and solar panels both are not particularly efficient, about 20%, and so not much better than holes in our mirrored cavity wall in terms of how much light escapes the circuit. (into the form of heat, rather than leaving the system entirely) If the solar panels are just 10% of the internal surface area, the decay time constant is only a few nanoseconds.

