My previous inquiry into how batteries work to produce power lead me to a new question:
How does the power produced in a battery propagate along a copper wire conductor when the 1.5 volt battery potential yields a visible light photon that is known NOT to propagate in a metal?
The following auxiliary questions arise and are answered by the above
-1- What is "DC" current? ……… DC is a reservoir of electrons transitioning from the higher energy state to the lower and supplying the "bump" a photon supplies to the load.
-2- Why is high voltage more efficient use of power for long distance transmission? Lower "current" is equal to saying "lower rate of electron transition which means less interaction with the elements of the lattice. Each interaction carries a probability of bleeding off some of the energy as heat.
-3- what is voltage and current as classically used terminology? Voltage is the same as classically described. Current is not really a flow of charge past a point but rather the rate of energy level transitions at that point
The probability of transition from valence band to conduction band goes down with increasing energy gap. Why does this affect conduction? See the next diagram. The circles are base states of electrons. The figure eights are the first excited state of an electron. These states are used in a representative fashion to depict relative energy levels and should not be interpreted too literally.
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Observations
- A conductor or semiconductor sets up a wave guide for the propagation of a photon
- In a semiconductor the photon must be the result of a 0.7 volt electron volt potential drop more or less to exceed waveguide cutoff frequency
- In a conductor there is no forbidden band due to the conduction and valence band overlapping. Thus waveguide cutoff frequency is more or less 0 hertz
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