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What happens to 'unabsorbed' photons?

  1. Apr 24, 2013 #1
    Hi guys,

    This is probably a basic and therefore stupid question, but hey, I'm new.

    When a photon hits an atom, it's energy is 'absorbed' and the electron (if we use hydrogen as an example) jumps up an orbit. It then naturally wants to loose this energy to settle back into it's ground state and subsequently emits a new photon.

    My question is two fold...

    If the energy absorbed needs to be of a certain discreet amount, then what happens to those photons that aren't of a specific energy? They can't pass through or we would get light pass through all solid objects, but if they're not absorbed and reflected by the process above, what happens to them?

    Alternatively, if ALL photons are absorbed (to use a simple term), then given that the newly emitted photon is of a certain quanta of energy (that of the difference between charge shells) then what happens to the difference in energy if the original photon did not match that of the required energy level of a shell jump?

    It seems to me (from me ignorant newbie stand point) that the first part of my question is most relevant as I would be more prone to believe that only the photons that are able to produce an electron jump (i.e. those of specific discreet energy) would be absorbed. So what then happens to the 99.9 % of photons that naturally collide with an atom and can't be absorbed and re-emitted?

    Forgive me if this is a stupid question. I'm sure this info is all over the internet. Just happens I started here.
     
  2. jcsd
  3. Apr 24, 2013 #2

    ZapperZ

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    This is where things are confusing, because you are mixing two very different scenarios without realizing it.

    You started by describing the scenario of light being absorbed by atoms. Fine. But later, you moved to to solids. Please note that the physics of atoms, individual atoms, is very different than the physics of solids. If you learn nothing else from this, that is something that you should be aware of.

    A solid has a lot more going for it. For one thing, there is something called "phonons", i.e. the vibrational modes of atoms that have come together to form the solid. This is a collective phenomenon, and it does not exist with individual atoms. This vibrational modes can be the ones responsible for many properties of the solid, including optical behavior.

    I suggest you check the FAQ sub-forum in the General Physics forum, especially the thread on the question of the speed of photon/light in a medium. That should give you a start on getting an idea on optical behavior of light in solids.

    Zz.
     
  4. Apr 24, 2013 #3
    Thanks ZZ,

    That's a great start, I'll definitely check it you.
     
  5. Apr 24, 2013 #4

    sophiecentaur

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    There is a big difference between the way solids and single atoms behave. We all learn about the hydrogen atom absorbing single frequencies but even with hydrogen when under great pressure, the spectral absorption lines start to spread so there is a range of frequencies absorbed. This is because of the Pauli Exclusion Principle, which says that two electrons (Fermions) cannot have exactly the same Quantum Numbers. When they are very close together, you need more quantum numbers to describe the available energy levels (all the atoms affect each other too) and the familiar energy levels spread into narrow energy bands.
    In condensed matter, the broadened lines become whole bands and so EM waves of any frequency will interact (be absorbed). Solids like glass are an exception, of course, because photons of optical wavelength are not absorbed. But IR and UV are.
     
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