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Condition of the CD

Of course I meant a CD that had been "treated" previously.

So now we have a QD that responds to a laser working with a treated CD in the drawer. The QD will emit his nanometers (simply responding to the lasre's input), these won't provoke any further change of the CD (because it has already been changed), BUT this emission will have its share on "charge depletion", won't it. Because there's nobody to tell the QD that it should NOT shell out any photons, this being useless waste of "charge".


Klaus


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