Is it possible to produce electricity from all wavelengths of electromagnetic spectrum beside visible light ?Like using gamma rays or x-rays .
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$\begingroup$ You mean by the photoelectric effect? Absolutely. $\endgroup$– rurouniwallaceCommented Apr 15, 2013 at 23:10
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$\begingroup$ Currently No, develop one, and go public. $\endgroup$– OptionpartyCommented Apr 16, 2013 at 0:30
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$\begingroup$ Without specifying whether this shall be achieved by a single apparatus or more, or what range of efficency is to be reached, the question lacks some incentive to think about it. $\endgroup$– GeorgCommented Apr 16, 2013 at 11:53
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3 Answers
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No, this is not possible with current (and feasible) technology. There are cutoffs at both low and high energies which make collecting energy difficult.
Low Energies:
Electricity is generated in photovoltaic ('solar') cells when an electron is knocked from a 'valance band' into a 'conduction band'. This process requires a critical threshold of energy per photon called the 'work function' (same ideas as for the photoelectric effect), which depends on the nature of the material - but is necessarily a non-zero number.
High Energies:
Photons with enough energy to ionize material (i.e. above the far UV) knock electrons clear off - making them difficult to collect, unless you add energy to the system, like in a photo-multiplier or Geiger-Muller tube.
This leaves infrared, optical, and near UV light -- all of which are currently used in solar energy production.
As @CrazyBuddy points out, these are also the only bands which contain significant energy as the peak of the solar spectrum is in the optical (and the atmosphere absorbs lots of energy far outside the optical).
answered Apr 16, 2013 at 5:01
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Currently, we're using EM waves in the visible & UV region of spectrum because, it's freely available via solar radiation. Yes, we can. But, the photoelectric current depends only on the intensity of incident radiation, irrespective of frequency provided the frequency is above the threshold frequency.
Because, current depends only on the rate of flow of electrons. Hence - the higher the intensity (more photons), more electrons are knocked off. If you're increasing the frequency, the electrons have more kinetic energy.
But, there isn't a source for X & $\gamma$-rays. It's quite expensive and potentially hazardous too...
answered Apr 16, 2013 at 4:33
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$\begingroup$ Your stipulation "provided the frequency is above the threshold frequency" is a huge one. This is the primary reason why using the full EM spectrum is hard. $\endgroup$ Commented Apr 16, 2013 at 7:27
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$\begingroup$ Hi @BrandonEnright: Well, Do I have to specify it in my answer? BTW, Thanks for your revision ;-) $\endgroup$ Commented Apr 16, 2013 at 12:45
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This is in addition to the high & low energies zhermes mentioned.
At radio frequencies we can extract energy from the radiation with a tuned circuit and a rectifier. Think of a crystal radio set: it extracts enough energy from the transmitted signal to produce sound in a headphone.
It has been suggested this could be used to beam energy down to earth from a solar power satellite. See for example this Wikipedia article.
