Photoelectric effect violating entropy?

AI Thread Summary
Photovoltaic cells convert high-energy photons from thermal radiation into electric potential, raising questions about entropy in an isolated system. The discussion highlights that introducing a photovoltaic cell into a gas container allows for energy conversion, which may seem to reduce entropy. However, the efficiency of the photoelectric effect at finite temperatures could counter this effect, suggesting an equilibrium between the gas and the cell. The conversation emphasizes the importance of understanding system boundaries and the implications of the second law of thermodynamics. Ultimately, the placement of the capacitor within the system could change the dynamics of energy transfer and entropy.
Smacal1072
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Hey all,

I posted this a week or so ago, but never really got closure on this issue:

Photovoltaic cells utilize the photoelectric effect to produce a voltage whenever an incident photon's energy is higher than the band gap of the silicon used for the cell.

Consider an isolated system, say a container of gas. If the temperature of the gas is above 0 K, the gas particles produce thermal radiation in the form of photons. The energy of these photons are distributed according to Planck's law. Regardless of the temperature of this gas, there will always be some high-energy photons radiated by the gas. Since we're assuming the system is isolated, none of these photons leave the system.

Let's say we throw in a photovoltaic cell into this container, which is connected to an ideal capacitor with infinite capacitance. Let's say the capacitor is outside of the system. Photons radiated by the gas with energy above that of the band gap generate a tiny voltage in the cell by the photoelectric effect, which charges the capacitor. Over time, the hot gas cools, since some of it's higher-energy radiation is being converted to electric potential (similar to evaporation of water).

Assuming all of this is OK so far, the problem is: How is this possible? Isn't this reducing the entropy of the system, by directly converting thermal energy of the system into electric potential? Something seems wrong.

I appreciate your help!
 
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The first thing I'd check is the nature of the photoelectric effect in materials at finite (i.e., nonzero) temperature. If the efficiency of the photoelectric effect decreases with increasing temperature, you may have your answer right there. We would obviously expect an equilibrium between the gas and the solid, with no energy being transferred.

You might check, for example, "Photoelectric Effect" (Chapter 5), Handbook of Physics, ed. by E. U. Condon and Hugh Odishaw. Also see Robert T. Ross, "Thermodynamic Limitations on the Conversion of Radiant Energy into Work," J. Chem. Phys. 45 (1) (1966) and Ross and Ta-Lee Hsiao's "Limits on the yield of photochemical solar energy conversion," J. Appl. Phys. 48(11) (1977) which addresses the analogous question of photovoltaics. (Note: I haven't studied these resources; I'm suggesting them because they might help you find the answer for yourself.)
 
Smacal1072 said:
Assuming all of this is OK so far, the problem is: How is this possible? Isn't this reducing the entropy of the system, by directly converting thermal energy of the system into electric potential? Something seems wrong.
You drew a system boundary and then penetrated it, so it isn't a closed system anymore!
 
Smacal1072 said:
Assuming all of this is OK so far, the problem is: How is this possible? Isn't this reducing the entropy of the system, by directly converting thermal energy of the system into electric potential? Something seems wrong.
You drew a system boundary and then penetrated it, so it isn't a closed system anymore! The second law says, "The entropy of an isolated (ie, closed) system..."

So what happens if you locate the capacitor inside the system...?
 
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