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In traditional solar cells, the maximum EQE is 100%, representing the generation and collection of one electron for each photon absorbed from sunlight. However, some advanced materials and configurations developed over the past several years have demonstrated the capability of generating and collecting more than one electron from high-energy photons, representing an EQE of over 100%.
This is pretty awesome. I wonder what the theoretical threshold of photons to freed electrons is if 190% EQE produces 80% energy absorption. Like, how much energy does a photon carry and how much energy is required to free the most promiscuous electrons.
171 sats \ 4 replies \ @0fje0 14 Apr
how much energy does a photon carry and how much energy is required to free the most promiscuous electrons.
That struck me as a fun question to ask ppq.ai. So I did. Here's what it says. It may be pure gobbledygook, of course, but the 25 sats spent on it may just provide a few clues to a proper answer:
The energy a photon carries is given by the equation:
[ E = h \nu ]
where:
  • ( E ) is the energy of the photon
  • ( h ) is Planck's constant ((6.626 \times 10^{-34}) Joule >seconds)
  • ( \nu ) (nu) is the frequency of the photon
To determine how much energy is required to free the most readily liberated electron from an atom, we refer to the "work function" of the material. The work function is the minimum energy needed to remove an electron from the surface of a solid (typically a metal). This energy varies depending on the specific material.
For example, the work function for cesium, which has one of the lowest work functions of any metal, is about ( 2.14 ) electron volts (eV). Other materials can have higher work functions; for instance, platinum has a work function of approximately ( 5.65 ) eV.
To illustrate, if we're dealing with a photon intended to eject an electron from cesium, you would need a photon with an energy of at least ( 2.14 ) eV. The frequency (\nu) of the photon required could be calculated from the photon energy equation above, rearranging it:
[ \nu = \frac{E}{h} ]
Substituting the values:
[ \nu = \frac{2.14 , \text{eV}}{4.135 \times 10^{-15} , \text{eV} \cdot \text{s}} ]
[ \nu \approx 5.17 \times 10^{14} , \text{Hz} ]
This frequency corresponds to light in the visible spectrum, specifically near the violet end.
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Thank you for your answer! I've just zapped you 21 sats so most of the ppq prompt price is on me!
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0 sats \ 1 reply \ @0fje0 14 Apr
I've just zapped you
Thanks, but that should probably have gone to @k00b instead. He effectively crafted the ppq prompt. I would not have known to ask the question that way.
Just zapped you back for posting the article! As a solar panel user myself, these things interest me directly.
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Thank you 🙏♥️
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By citing cesium, it makes me wonder if people have experimented with radioactive solar cells. I'm so out of my depth lol.
I went to ChatGPT then wikipedia and couldn't really find a direct answer for the theoretical EQE limit. It might not be known.
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I have you read at least a dozen books on the topic to be able to answer your question
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27 sats \ 0 replies \ @k00b 13 Apr
More likely it'd require finding one great article on the topic. Naive questions are the only ones I'm capable of and I'm sure it's an easy answer. Also, to be clear, I wasn't asking or expecting you to answer.
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