Information on the Photoelectric Effect: https://en.wikipedia.org/wiki/Photoelectric_effect
Summary: Some metals release electrons when struck by a certain frequency of photon.

What does this mean in the context of this question?

Assume that you have found the right metal that will release photons when struck by infrared light. It would seem that you would be able to have completely passive cooling within a house, by setting up panels of this materials to absorb heat (and perhaps generate energy, as a bonus).

What material would you likely need? Would it be cost-effective? Is this even possible for consumer use?

  • $\begingroup$ This is the basis of solar panel technology, though you wouldn't have a cooling effect if the photon is absorbed anymore than a black wall in the sun would be cool to the touch. $\endgroup$
    – Neil
    Mar 2, 2016 at 16:03

1 Answer 1


Although you don't make this clear in your question, I'm guessing you propose that the infrared radiation from the walls etc in a house could be converted to electricity by the solar panel. The end result would be that you'd be removing thermal energy from the house and this would cool it.

The problem is that you need the temperature of the light to be hotter than the temperature of the solar panel otherwise thermal excitation of electrons swamps the effect of the light. By the temperature of the light we mean the temperature of the black body with the same peak wavelength as the light. In this case the light is emitted by the walls, floor, etc so the temperature of the light will be the temperature of the walls, floor, etc and of course the solar panels. So you won't get any electricity out.

The reason we get electricity from solar panels in sunlight is because the temperature of sunlight is the same as the temperature at the surface of the Sun - about 5,700 degrees. That's a lot hotter than the solar panel!


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