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A brief question on quantum thermodinamics.

  1. Apr 11, 2015 #1
    Hi, I am new to this forum so I dont really know if this question already exists..

    My question is: When an electron absorbs a foton and climbs to the next energy gap and then returns again, as the energy is quantum energy, how is possible that the foton reemited by the electron has the same energy as the original foton, without any energy lost as the classical thermodinamics would suggest? Is there any energy lost during this transformation?

    // I am 17 years old, be gentle :D

    Thanks!!
     
  2. jcsd
  3. Apr 11, 2015 #2
    I think there are two answers to this.

    1: classical thermodynamics depends on many (as is ##10^{23}##) particles interacting. Thermodynamic laws are statistical laws that apply in those situations, so in the realm of QM where you are dealing with a single atom, the statistical rules don't apply.

    2: The reemitted photon does not need to have the same energy. In an atom with lots of energy levels that are close together the incoming photon could excite multiple electrons and then two lower energy photons could be reemitted. In atoms that are part of a solid, the energy could get transferred into vibrations in the solid and a lower energy photon could be reemitted.
     
  4. Apr 11, 2015 #3

    Drakkith

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    Staff: Mentor

    No energy is lost because the energy given up by the electron as it falls to a lower energy state must be equal to the amount of energy required to excite it into that state to begin with. Now, before the electron can drop into a lower energy state and emit a photon, it is possible that something else interacts with it and takes the energy instead. For example, the atom with the excited electron can collide with another atom and the electron can give up this energy to the other atom without emitting a photon.
     
  5. Apr 11, 2015 #4
    I've been wondering for a while, is there any experiment which has observed interference after a free, or losely coupled, atom has absorbed and emitted a photon?

    Or are we left with only more complex effects to confirm this this type of mechanism?
     
    Last edited: Apr 11, 2015
  6. Apr 11, 2015 #5

    Drakkith

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    Staff: Mentor

    What is a "free or loosely coupled" atom in this context?
     
  7. Apr 12, 2015 #6
    An atom which is sufficiently decoupled from a complex system.
     
  8. Apr 12, 2015 #7
    And would that energy transformation be done without any energy lost in the process of transformation?

    Thanks for all the answers so far!!
     
  9. Apr 12, 2015 #8
    What you are reading are already examples of energy being "lost". The expression "lost" refer to energy changing form into something unusable or different from the original (e.g. kinetic energy becomes thermal energy). No energy is ever lost strictly speaking, because energy is conserved.
     
  10. Apr 12, 2015 #9
    "Lost" energy is energy that has been transferred into a system that is too complicated to keep track of the energy. As ddd123 points out, it is not really lost. When you talk about energy being "lost" in something like an inelastic collision, the energy is just being distributed to the (more than ##10^{23}##) atoms involved in the collision. Once numbers like that are involved, it is preferable to deal with statistical measures of energy like temperature. That is when the thermodynamic law apply.

    Every transfer of energy transfers all of the energy without losing any energy.
     
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