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Radiated power, what does this even mean?

  1. Jul 24, 2006 #1
    Honestly, Im not even sure exactly what radiated power means. Why is the 18.4mW not the radiated power of the light bulb?

    I thought electrical energy was essentially work, which is a product of power and time, so how could radiated power be anything except 18.4mW? Are they referring to 0.5% * 18.4 mW?
     
  2. jcsd
  3. Jul 24, 2006 #2

    quasar987

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    Let me take a wild guess here.

    A light bulb is a resistance. When current passes through it, collisions of the electron of the current with the atoms the the resistance happen. As a result, the speed of the atoms in the resistance increases. "Electrical energy" is converted into "heat energy", and we are told by the fabricant that this conversion happens at a rate of 18.4 mJ/s.

    I'm not sure how the radiation comes into play here, but it is possible that it is emited when the excited atoms fall back to a state of lower energy. An electromagnetic wave is then produced. So heat is transformed into radiation in this way. And we are told that 5% of the 18.4 J transformed per second end up in this way (as visible light).

    So all that to say "ya, 0.5% * 18.4mW" would be the answer.
     
  4. Jul 25, 2006 #3

    chroot

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    The bulb radiates power both in visible light (5%) and other forms of invisible radiation (mostly infrared).

    - Warren
     
  5. Jul 25, 2006 #4

    quasar987

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    Ok, so 18.4mJ is the amount specifically converted to electromagnetic radiation per second. And 5% of that 18.4mW is light in the visible spectrum.

    chroot, how are the electromagnetic radiations produced? I.e. as a result of what mechanisms?
     
  6. Jul 26, 2006 #5

    Tide

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    Light is produced when charged particles accelerate. In the situation you are describing electrons collide (i.e. accelerate) with ions in the filament thereby radiating.
     
  7. Jul 27, 2006 #6

    quasar987

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    Electrons also collide with each other in the conducting wire. According to my notes, a classical analysis gives an intercollision time of 2 fs for a coper wire at 300K. But we rarely see visible light coming out of coper wires. So what's different in a resistance that makes the electron accelerate more as a result of collisions with the atoms of the resistance?

    And a subquestion if I may: why do we say that electrons emit an electromagnetic wave when they accelerate but not when they travel at constant speed? In other words, what's different btw the "emited radiation" caused by acceleration, and the simple propagating deformation of the electric field caused by a moving charge?
     
  8. Jul 29, 2006 #7

    quasar987

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    Anybody?

    ---
     
  9. Jul 30, 2006 #8
    Second question: A charged particle that is moving at constant speed (assuming no external fields) is not absorbing or emitting energy. It is simple inertial movement.
    The accelerated particle needs energy inputs and outputs. Energy is loss when a particle loses energy and it has to go somewhere so it emits the photon (Conservation of Energy). The photon will then accelerate another particle when it is absorbed elsewhere.

    The simple deformation of an electric field is just an abstract construct and really doesn’t mean anything physical. If there is an external field, the particle will be accelerated and it will be absorbing and emitting photons.

    First question: Heat up a copper wire enough (like the filament in the bulb) and it will emit visible light too. The heat energy knocks electrons out of their orbits and as they drop back they emit visible radiation. There is nothing special about the resistance, it is just that it would take an enormous amount of current to heat the copper that much (and it would melt).

    (I am still learning this stuff so tell me if I am wrong!!! But I believe what I said is correct)
     
    Last edited: Jul 30, 2006
  10. Jul 30, 2006 #9

    quasar987

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    Hi interested_learner, thanks for the reply.

    What do you mean by " The accelerated particle needs energy inputs and outputs."? In a gravitational field, there is no analogue to photons emited by decelerating charges. So there does not need to be an "output" other than potential energy. (imo :smile:)
     
  11. Aug 4, 2006 #10

    Tide

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    Of course, electrons collide with each other but since they are like-charged their radiated fields cancel to lowest order. Much more power is emitted in electron-ion collisions.
     
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