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Electron energy states and photon emission?

  1. Aug 17, 2014 #1
    Are lower energy electron orbitals always closer to nucleus than higher energy orbitals? Is this energy proportional to the inverse square law and Coulomb's law?

    When an electron jumps down to a lower energy orbital, is potential energy not just converted to kinetic energy, and so where does the energy to emit a photon comes from?
     
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  3. Aug 17, 2014 #2

    jtbell

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    The orbitals are probability distributions. For different quantum numbers, they overlap:

    http://hyperphysics.phy-astr.gsu.edu/hbase/hydwf.html

    Because orbitals are "spread out" in space, it's not meaningful to give them a definite kinetic energy or potential energy, only a definite total energy. The photon emitted in a transition gets its energy from the difference between the total energies of the two orbitals involved.
     
  4. Aug 17, 2014 #3

    WannabeNewton

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    As given this is not a well-defined question because energy levels are not surfaces of constant radius in space: an electron in an energy level has an entire spectrum of possible positions it can be found at that are not confined to a single radius. That being said we can calculate the most likely position that an electron will be found in when it is in a certain energy level. I will simply state the result here and refer you to "Quantum Mechanics: Concepts and Applications"-Zettili problem 6.6 p.375 2nd ed. for the calculation. For electrons with energy level ##n## and total angular momentum ##l = n-1## (which is the maximum possible total angular momentum they can have) the most likely radial position is ##r_n = n^2 a_0## where ##a_0## is the Bohr radius. So what you asked can be recast in terms of ##r_n## in which case we can say that the most likely radial position of the electron is larger for higher energy levels.

    No the energy is ##E_n = -\frac{e^2}{2a_0}\frac{1}{n^2}## where as above ##n## are the energy levels.

    What is true is that ##E_n = -\frac{e^2}{2 r_n} = \frac{1}{2}\langle V \rangle_n = -\langle T \rangle_n## where ##\langle V \rangle _n ## is the average of the Coulomb potential in a given stationary state ##|n \rangle## and ##\langle T \rangle_n## is the same for the kinetic energy (this is basically the Virial theorem). However it makes no sense to say ##E_n## itself is exactly the Coulomb potential because stationary electrons do not have a well-defined position (they do have a well-defined angular momentum) and even classically this isn't true because the energy of a central motion includes both kinetic and potential; indeed ##E_n## includes both kinetic energy and the energy of the Coulomb interaction but in terms of operators with fluctuations in value due to the inherent probabilistic nature of QM.

    We can't describe what happens to the electron, or its energy, in between transitions. We can only say that there is some transition probability of it going from one stationary state to another with a "jump" from one fixed energy to another. This energy is not solely potential energy, it is the total energy of the electron; it includes both kinetic and potential energy contributions.

    The energy for spontaneous emission comes, very loosely speaking, from vacuum fluctuations in the ground state energy of the electromagnetic field.
     
  5. Aug 17, 2014 #4
    Is "most likely" the same thing as "average" distance?


    Is there any equation about relationship between angular momentum and "most likely" distance per energy level?


    Quantization and zero-point energy? Is that related to "Dirac sea"? Is there any interpretation of it that is more mechanical and less statistical than QM?
     
  6. Aug 17, 2014 #5

    jtbell

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    No. "Most likely" corresponds to the peaks of the distributions shown on the page that I linked to: the distance with the largest probability. "Average" corresponds to sampling a large number of distances randomly from the distribution and calculating their average in the usual way.

    If you're acquainted with the statistical concepts of mean, median and mode as described here:

    http://math.about.com/od/statistics/a/MeanMedian.htm

    "most likely" corresponds to the mode, and "average" corresponds to the mean.

    I thought this is simply what "triggers" spontaneous emission, not the source of the outgoing photon's energy (which equals the difference in energies between the two orbitals).

    For a crude analogy, imagine a ball sitting on top of a hill. When it rolls down to the bottom of the hill, it gains kinetic energy equal to the difference in potential energy between top and bottom; but in order to start it rolling in the first place, you have to give it a small push (which involves a small amount of energy).
     
  7. Aug 17, 2014 #6

    WannabeNewton

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    It can be calculated in principle but unlike for the case of ##l = n-1## the calculation isn't exactly straightforward. I'll see if I can find the relation in any of the books I own.

    Not really. It's just the statement that even if the electromagnetic field is in its vacuum state it still has quantum mechanical fluctuations in field strength (not energy, which I incorrectly stated earlier-see below).

    Not that I know of. EDIT: Actually this might be of interest: http://www2.famaf.unc.edu.ar/~vmarconi/moderna1/emision_estimulada_AJP.pdf

    Yes you're right. What I stated earlier is technically incorrect since the vacuum fluctuations are of field strength and not of energy as the vacuum state is an eigenstate of the Hamiltonian, which agrees with what you've said.
     
    Last edited: Aug 17, 2014
  8. Aug 17, 2014 #7
    I understand there is mode number for a collection of discrete numbers, like energy levels are quantized, for some reason. But I thought the distance is a function of continuous rather than discrete distribution. I mean while electrons are staying in one and the same energy state, is their distance from nucleus constant or it has a probability to be smoothly closer to or further away from this most probable distance?


    I was indeed thinking about spontaneous emission when I asked the question, to keep it as simple as possible. Is "spontaneous" the same thing as "random"?
     
  9. Aug 17, 2014 #8

    Dale

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    While staying in a definite energy state the electrons have no definite distance from the nucleus. If you measure the distance* then you will find some random value with a probability distribution given by the wavefunction squared in the usual way. That probability distribution is a continuous distribution, so it can assume essentially any value.

    *You will also change the state so that it no longer has a definite energy.
     
    Last edited: Aug 17, 2014
  10. Aug 17, 2014 #9

    WannabeNewton

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    No a spontaneous process is one that happens "by itself" i.e. it isn't forced to happen. The canonical example is heat flowing from a hot object to a cold object. The heat flow is spontaneous. That being said, spontaneous emission specifically is an inherently probabilistic process.
     
  11. Aug 17, 2014 #10
  12. Aug 17, 2014 #11
    http://en.wikipedia.org/wiki/QED_vacuum

    I find it is related to Van der Waals forces, Casimir effect and vacuum permittivity and permeability. In what units is this vacuum energy measured and how it relates to vacuum permittivity and permeability?
     
  13. Aug 17, 2014 #12
    It's caused, but it is not known when exactly the jump will occur? Is total energy of the electron just before the jump greater than new energy after the jump, or equal?


    Does it mean that electrons get "excited" by some influx of energy even in the case of spontaneous emission? So in the case of stimulated emission this extra energy comes from the outside in the form of photons, and for the spontaneous emission the extra energy comes from within spatial constraints of the atom itself?
     
  14. Aug 19, 2014 #13

    WannabeNewton

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    Vacuum energy is measured in units of energy, nothing surprising there. The actual choice of units is quite arbitrary.

    It is Joules in SI but length in geometrical units.

    It's probabilistic so you cannot predict when exactly it will occur. What you can calculate is the transition probability per unit time for the process to occur. This is most simply done using the Einstein coefficients by considering the electron as being immersed in a gas of photons but you can also use Fermi's golden rule in the full formalism of QFT.

    For a spontaneous emission? It has to be less by conservation of energy. The emitted photon carries away the difference in energy between the initial and final states in the transition. The selection rules for the Hydrogen atom will tell you what possible final states are available for transition. Usually one works in the dipole approximation.

    Well no. The electrons don't get an influx of energy in the case of spontaneous emission. What happens is the vacuum state of the EM field still has fluctuations in field strength. The electron interacts with these field strength fluctuations of the EM vacuum and radiates as a result.
     
  15. Aug 19, 2014 #14
    Yes, spontaneous. Do they never spontaneously jump up to higher energy orbitals?

    With gravity without air, does falling object preserve its total energy, or it has lower total energy on lower altitudes like those electrons?
     
  16. Aug 20, 2014 #15

    WannabeNewton

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    You can't spontaneously jump up an energy level. There have to be photons incident on the electron in order for it to jump and by definition there are no photons in the EM vacuum. An object freely falling under gravity does have constant energy but so does an electron in a stationary state of the hydrogen atom if we ignore the surrounding photon gas. If we include the photon gas then the total energy of the electron plus photon gas is conserved. In the gravity scenario if I shot a ball up in the air towards the object falling down such that the ball is much larger than the object then the collision will be inelastic but the energy of the combined system will still be constant in the gravitational field.
     
  17. Aug 20, 2014 #16
    How can we say then any emission was spontaneous if we know the electron had to have absorbed a photon to be in the higher energy orbital to begin with?

    Do electrons jump up to higher energy levels due to plain atom kinetic collision, or is temperature and pressure always followed by photons which actually mediate energy transfer to electrons?

    What prevents an electron to go further down closer to nucleus to even lower energy levels?
     
  18. Aug 20, 2014 #17

    WannabeNewton

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    In spontaneous emission there is no photon absorption. There is only photon emission and the electron drops to a lower energy level as a result.

    I think you have the wrong conceptual picture in mind. The electron doesn't literally jump a level and the photon doesn't literally collide with the electron. These are words used to represent more abstract things. In reality, the photon scatters off the electron and scattering has a very precise meaning in QFT as an interaction amplitude through some field, the electromagnetic field in this case.

    It can keep going until it is in the ground state, which is stable.
     
  19. Aug 20, 2014 #18
    But how can we tell if some emission is spontaneous if we don't know whether the electron absorbed a photon three days ago and is the reason why it was on a higher energy level just before we measured its photon emission?


    Yes, but aren't there energy levels even below the ground state? Why electrons don't prefer even lower energy levels? What's so special about ground level energy?
     
  20. Aug 20, 2014 #19

    Dale

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    The ground state is the lowest energy, by definition. There is no state with lower energy.
     
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