Danny <email@example.com> wrote in message
Quantum mechanics applies to all known systems, however it is not
needed really unless high accuracy is desired in Most situations. The
exceptions are atomic and similiarly dimensioned problems. Although one
could use QM for radio waves, it isnt really needed unless the utmost in
accuracy is desired. The quantum-like nature of the atomic systems becomes
blurred as the dimensions of the system becomes much larger so that one is
talking about large aggregates of systems. The macroscopic properties that
are measurable are statistical Averages of the quantum features of a system.
Hope that helps a lil.
As you may know, the absorbtion of wavelengths by a atom or molecule is
frequency (v) dependent according to E=hv where h is Plancks Constant and
and E is the energy. That being the case I would expect that the transition
occur rather sharply to very sharply. It could be a bit more distrubuted
dependent upon the exact conditions and constiuency of these molocules that
are involved in the detection of the light. A medical eye doctor would have
the name of these ' devices ' [ cones ?.
The photoelectrons are the electrons which are emitted by a atom or
molecule when the frequency ( E=hv ) is sufficiently high to do so. The atom
has to overcome the ionization energy of the atom to be displaced thusly.
Any additional energy could appear as kinetic energy in the electron. So,
no, these electrons never see the power supply if I understand your
question correctly. You might try and rephrase it. Basically the
photoelectric equation is Nothing more then a restatement of the
Conservation of Energy.