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Wave Properties of Electrons and their Atomic Orbitals

  1. Sep 12, 2009 #1
    I understand that electrons are found inside Atoms within specific orbitals, or quantized distances from the nucleus. There are a maximum number of electrons found within each orbit, etc. etc.

    I recently read that the reason for this quantized nature of orbitals has to do with the wave nature of electrons, or more specifically, the sine waveform representing the momentum of atomic electron(s). The author explained that (and I paraphrase here) an electron's sine wave, which represents a specific momentum, has a specific amplitude. And based on this amplitude, the wave can only exist within a specific circumference that allows the peaks and valleys of the wave to overlap, therefore reinforcing the wave and allowing it to persist.

    This all makes sense. What I fail to understand is why the electron has to have a specific energy level. In other words, what keeps the electron from capturing a very low energetic photon that gives it just slightly more momentum, resulting in a slightly different amplitude, and a slightly higher orbit. Can you not have an infinite number of wave amplitudes within a given circumference that would allow for overlapping and reinforcing? :confused:

    I would appreciate it if someone could explain this in simple terms that a non-Physicst like me could understand. And if anyone has a link to a website that clearly explains this concept (preferrably with diagrams), I would also appreciate that greatly. :smile:

    Thanks in advance. Daisey
     
  2. jcsd
  3. Sep 13, 2009 #2
    you must think to the electron as a wave. The allowed orbits in an atom are those with an integer number of wavelength, that is periodic orbits. The the electron cannot absorb an arbitrary small momentum from a photon, otherwise it would be in a non periodic orbits and the wave would have destructive interference. The electron can only absorb a momentum amount such that it passes from a periodic orbit to another periodic orbit.
     
  4. Sep 13, 2009 #3
    Yes. I understand that. I am trying to visualize this in my mind. Take an electron with a given amplitude X which is in an atomic periodic orbit Z. If that wave with amplitude X were to take a slightly higher atomic orbit (say, Z+1), I completely understand why there would be destructive interference at that slightly higher orbit for that wave with amplitude X. The peaks and valleys of the wave would not overlap. There would be destructive interference. It also makes sense to me that at that same higher orbit (Z+1) there should be conceptually a wave with some slightly different amplitude that could exist at that orbit where the wave peaks and valleys would overlap in phase. Is this not the case? :confused:

    It's possible that maybe I am not understanding something inherent about waves. Given a wave of amplitude X. Could there also not exist a wave that has a very slightly different amplitude? X plus or minus some very small factor?

    Daisey
     
  5. Sep 13, 2009 #4
    The amplitude in this example doesn't play any particular role. Forget the amplitude and think to the wavelength (or the frequency if you like). The first N=1 orbit is the orbit of all the possible configuration of the wave with length equal to a wavelength of N=2. The first N=2 is that with length equal two times the wavelength N=2, and so on. The wavelength of N is fixed by the Coulomb potential along the orbit N. It determines the momentum and thus the energy of the electron in that orbit. To satisfy the condition that peaks and valley always overlap the allowed wavelength, and thus energies, are only particular solutions. They are indeed quantized. Note however that for the Coulomb potential the wavelengths are not a multiple of the wavelength of the fundamental orbit N=1, as for instance in a harmonic potential.
     
  6. Sep 13, 2009 #5

    jtbell

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    Remember that the QM "wave function" is actually related to the probability distribution of finding the particle at various locations. Its amplitude is determined by the requirement that the total probability of the particle being found somewhere must be 1. We call this normalization of the wave function.

    Also, don't think of something like wrapping a vibrating string around into a circle around the nucleus. The wave function is a three-dimensional one, sort of like the one that gives the pressure of sound at various locations inside a closed spherical cavity. The atom's energy levels correspond to the resonance modes of the sound in the cavity. (I say "sort of" because the QM wave doesn't have a fixed boundary in this case (an atom doesn't have "walls") and it's a complex function not a real function.
     
  7. Sep 13, 2009 #6
    Got it. Sorry. Again, I keep confusing my terms. Amplitude = possibility. Wavelength = energy.

    I understand what you are saying, but I am still confused as to why, inside an atom, there are only a given number of possible orbits where waves can potentially be in phase. Take orbits N=1 and N=2. Why can there not exist a wave with a particular wavelength that can exist (be in phase) in between these two orbits? I realize the energy (or wavelength, or frequency) has to be different for the wave that exists at N=1 and N=2. Why can there not be a wave with a wavelength (with different energy) that is in phase at N=1.5?
     
  8. Sep 13, 2009 #7
    This might be my problem. I am trying to picture this concept in my mind, and a this is kinda of how I am picturing it. :frown:
     
  9. Sep 13, 2009 #8
    Those orbits (with a certain energy En) are stationary (or quasi-stationary) ones. In fact one can prepare an atom in a QM superposition of different orbitals with any given average energy E ≠ En. With time all excited states in this superposition will radiate (their amplitudes will decay) so after a while the atom will return in its lowest state n=1 by radiating some photons (think of cooling down a cup of coffee via vaporisation).
     
    Last edited: Sep 13, 2009
  10. Sep 13, 2009 #9
    I try to illustrate this using simple example. As jtbell says, consider just for a moment a string of a guitar (fixed string at the boundaries) or in a circle (periodic boundary condition). In this case only particular frequencies v_n are allowed, that is the ones that satisfy the condition at the boundary. In this particular case v_N = N v, where the fundamental frequency is fixed by the inverse of the string length. But, as jtbell says, we must consider a three dimensional string, something like a vibrating cavity.
    An easy example is the quantum harmonic oscillator. The harmonic oscillator has alway the same periodicity T (Galileo's ischronism) In the quantum harmonic oscillator, as in a guitar string, the allowed frequencies are v_N = v N with v=1/T is the frequency of the oscillator. This just means that the energy is quantized E_N = h v_N = h N v (+ the vacuum energy).
    Substituting the harmonic potential with the Coulomb potential you get the analogous quantization of the energy/frequency, but now the orbits at different energy has different periods. The N-th orbital is just the three dimensional wave solution with the N-th allowed frequency level of this discretized spectrum. In general, to obtain the solution with an integer numbers of wavelengths, the periodicity condition that you need is T_N v_N = n (T_N the period of the N-th orbit).
     
  11. Sep 13, 2009 #10
    I think I understand this part. You are saying (referring to my poorly drawn diagram) that for a fixed string (which I think can be represented by N=1 in my diagram), that only particular frequencies are allowed. I just cannot understand why a "fixed string" cannot exist between N=1 and N=2 in my diagram. Sure, the "string" will have to be of a different length. But what's wrong with that?

    The rest of your response (harmonic oscillator, periodicity, discretized spectrum) went way over my head, and I failed to follow it. :blushing:

    I think what is happening here is this cannot be understood in a two-dimensional format, like my mind works. :smile:
     

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  12. Sep 13, 2009 #11
    Daisey, you are right: one can prepare a string in any initial form. Mathematically its state is a superposition of different discrete modes. Now let it vibrate. Let us admit that all harmonics with n > 1 produce a sound but that with n = 1 does not (too low frequency). Then, with time you will find your string in the lowest discrete mode n = 1 which is stationary (no energy losses into the environment). The same is valid for atoms. Read my previous post (#8).
     
  13. Sep 13, 2009 #12
    Are you saying that many possible orbits between N=1 and N=2 in my drawing are possible (excited states), and do exist? And over time, for some reason, all orbits between N=1 and N=2 will decay (lose energy) to N=1 (a non-excited state)?
     
  14. Sep 13, 2009 #13

    http://en.wikipedia.org/wiki/File:Harmonic_partials_on_strings.svg

    the different waves drawn in this picture, from the bottom, are N=1, N=2, N=3 .... (in an harmonic potential or in a guitar string). They correspond to the N=1, N=2 ... of your picture and they are the possible vibration of a guitar string.
    Here an example with N=6 http://www.pas.rochester.edu/~afrank/A105/LectureVI/ewave.jpg

    For a more complete vision see the preview of http://demonstrations.wolfram.com/BohrsOrbits/ where n is the orbital number that we are calling N,

    If N is not an integer the waves at the waves at the boundary are with a fixed value (say zero). This is what I mean when I say that the waves inside the intervals has a non integer number of wave length. For this reasons these non integer solutions are not allowed.

    http://en.wikipedia.org/wiki/File:Drum_vibration_mode01.gif

    http://en.wikipedia.org/wiki/File:Drum_vibration_mode21.gif

    here there is a similar example in 2D, the first picture is with N=1 and the second N=2. The orbital is the 3D analogous but in a Coulomb potential. In this case the frequency spectrum is E_N =R/N^2 (R is the Rydberg constant), which differers from the harmonic one E_N =N h v.

    Probably my english is not so good.

    (that's fun http://www.falstad.com/loadedstring/ )
     
  15. Sep 13, 2009 #14
    Yes, this is correct. Take, for example, an atom (Hydrogen for simplicity) in its ground state n=1 (E=E1). Then hit it with an electron with the kinetic energy K < E3-E1. As a result of scattering you will obtain an atom in a superposition of first two states: ψ = A1ψ1 + A2ψ2. The third and higher orbitals will not be present in this superposition because the projectile had not sufficient energy.

    The state ψ will have an average energy E = A12E1+A22E2. The amplitude A2 will decay exponentially with time: A2=A2(0)*exp(-t/τ) due to (a weak) interaction with the quantized electromagnetic field (photon field). With time t → ∞ you will find the atom in the state ψ1 and a photon radiated.

    Another example: you hit the atom with a resonance photon and obtain a pure ψ2. The with time you will obtain a superposition similar to the previous one where A2 is decaying with time and A1 is growing from zero to unity. At any intermediary time the electron state will not be the second or the first orbital but something intermediary, changing with time from ψ2 to ψ1. If the state ψ2 has a big half-life time τ, the superposition will change slowly so any intermediary orbit is possible.

    Classically it corresponds to slowly falling from a higher to a lower orbit with losing the energy into radiation.
     
    Last edited: Sep 13, 2009
  16. Sep 14, 2009 #15
    Well, Bob. I have a better understanding (thanks to you and Halcyon-on) regarding what is happening inside the atom with these excited electrons. But I still don't understand what makes the N=1 state the lowest possible energetic state. What about N=0.5? Surely there must be a lower possible orbit, for the same reason that there is no possible smallest positive number in math - one can always add another '0' to the right of the decimal point. What would keep the proxy wave of the electron from taking on a VERY SLIGHTLY smaller wavelength, and assuming a VERY SLIGHTLY lower nuclear orbit?
     
  17. Sep 14, 2009 #16

    alxm

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    Science Advisor

    Well, n is just a number. You can number them any way you want :) But there's still always a lowest state.

    Well, there's a lot of ways to explain this.

    One way to explain it is that you've simply got two opposing 'forces' at work here. One is the attraction of the nucleus felt by the electron - making it 'want' to be in the area near the nucleus. The other is that constraining the electrons motion to a small area of space increases its energy.

    So given those two opposing forces, there has to be some optimal 'orbit' which has the lowest possible energy. Any other 'orbit' will be higher in energy, due to one force or the other.

    The fact that such a lowest-energy state, or ground state, must exist can (and has) been proven mathematically. It's called the variational theorem.
     
  18. Sep 14, 2009 #17
    Ok, you understand that the electron “falls down” from higher orbits to lower ones and radiates the excess of energy. In QM it stops at some level. Now, in order to get it closer (on average) to the nucleus you have to supply some energy to the electron. It is due to the wave nature of it. There is an uncertainty principle: the smaller particle/wave localization is, the higher energy it has. The ground state corresponds normally to the minimum of the product ∆p*∆x.

    Take, for example, a string. The lowest state is A*sin(π*x/L), OK? Then prepare the string in the state sin[(π/2)*x/L]. This state is a superposition of the lowest and all highest states so its average energy is higher than E1. With time all excess of energy will be radiated and the system will return to the ground state.

    (One of the ways to get permanently the electron closer to the nucleus is to increase the nucleus charge Z. Then the stationary orbit will be smaller.)
     
  19. Sep 14, 2009 #18
    Bob,

    Interesting. Regarding atoms in Atomic orbits...

    1. When you increase the charge on an atom (ie. the nucleus), do electrons always then move closer to the nucleus? Is not increasing charge the same as gaining energy? I thought a gain in energy always resulted in electrons moving farther from the nucleus.
    2. Electrons when observed (as particles) are found in "shells" around the nucleus. Do these "shells" have to be full? In other words, can the next higher shell contain ANY electrons if the lower shell is not full?
    3. If the lower energy shells HAVE to be full (question #2), how can an electron gain energy and move into a higher orbit without an electron taking it's place in a lower orbit?
    4. Take two identical solid objects composed of atoms (two pieces of Iron, for example). If one of the electrons in each of the atoms in one of these pieces of Iron were to gain energy (take a higher orbit), how would the two pieces of iron appear different (look, feel, etc) to a person observing them?
     
  20. Sep 14, 2009 #19
    1) in this case you are changing the potential and thus the orbits, where there was the orbit N with momentum p_n there will be another orbit with another momentum and, in general the wavelegth associated to this momentum doen't fill a integer number of period on that orbit. To the allowed orbits are in general different E_N = Z R /N^2 (Z is the number of proton in the atom).

    2) The orbitals are distribution of probability so the wave function of the electron, which gives the probability, is not located in a particular region. By the way this orbitls looks in this way (depending on the angular momentum they are not spheres)
    300px-Neon_orbitals.JPG

    3) this is an unstable configuration. Usually one says the the electron is excited. however they always satisfy the Pauli principle: is not possible that the two electron have the same orbit, same angular momentum and spin. When a shell is full no others electrons are allowed.

    4) the metals are characterized by the fact that the last electron of every atom is shared with all the other atom. Since these electrons are free they form an electron sea which is responsible of the metal conductivity, of the particular reflecting aspect of the metals and of others important properties of the metals.
     
  21. Sep 14, 2009 #20
    Okay. I realize we are talking about waves of probability here. But when observed, these waves appear as particles. So regarding an electron that is orbiting a nucleus: For an electron to jump to a higher orbit, it has to gain energy, usually associated with absorbing a photon (I believe). And as stated in earlier posts, this also results in a higher wavelength (which is the wave equivalent of saying that a particle takes a higher orbit). So, are you saying "changing the potential" would result in this electron wanting to absorb a photon? Doesn't an electron have to do this to change orbit?

    So, in my example there is a lower shell that is missing an electron. And the electron that is in a higher state (higher orbit) which should be in a lower orbit is said to be "excited", and this atom is considered unstable.

    Yes, I remember reading about this. But my question was different. In my example of a piece of iron, think of a different electron in each atom (not the valence electron). Let's say that each atom in the piece of iron is in an unstable configuration, as defined in text above. I'm just curious how this piece of iron would be perceived on a macro basis by someone observing that piece of iron. Would it be hot? Would it be vibrating? Would it feel / look the same as a piece of iron in which every atom was in a stable configuration?
     
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