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Quantum energy levels

  1. Apr 17, 2008 #1
    hello i have a basic question regarding energy levels of electrons in atoms in bohr's model of the atom. From what I understand if an electron is in an energy level other than the ground state it can either absorb a photon equal to the difference of energy between its energy level and a higher level, or emit a photon with an energy equal to the difference between its energy and a lower energy level.
    What i don't understand is if lets say for a hydrogen atom the electron is at level n=2. In order for it to move up to a higher level n=3 it needs to absorb a photon of energy about 1.9eV. If a photon with an energy of 2.1eV was passing through would it be absorbed partially? moving the electron up a level then exiting with an energy of .2eV? or would it simply pass through not being absorbed? and if so why?
    Also is this the same for electrons? for example the Franck Hertz experiment seems to suggest that electrons can give up some of their kinetic energy through inelastic collisions to electrons in an atom to bump them up an energy level. and then exit with their remaining KE. what makes them be able to give only a partial amount of their energy and what makes the photons on able to give all or nothing of theirs? Thanks for your help and sorry if i am completely wrong on this
  2. jcsd
  3. Apr 18, 2008 #2
    An electron MUST absorb the right amount of energy to go to the next level, so like you said, to move to n=3, the value must be 1.9 eV not 2.0 or 2.1, otherwise nothing will happen. So its the exact value or nothing. In anwer to your second question asking if the electron will partially absorb the energy in a photon, It will not ! Electrons in certain energy levels can only absorb the exact values of energy needed to jump to a higher level.I hope this helped ;).
  4. Apr 18, 2008 #3


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    Don't forget that in practice hydrogen atoms get thermally bounced around. So the hydrogen atom, unless strictly at absolute zero, is not in a pure energy eigenstate. This means thermal motion,and recoil and radiative damping and experimental resolution are important -- this can be seen in plots of hydrogen emission frequencies; the photons coming from an H atom have a line-width rather than a single frequency. The same is true for absorption, with the exception of measurement noise.

    Radiative damping was considered by Weisskopf and Wigner back in 1930 -- they looked at the QM analogue of energy loss of a radiating classical oscillator, which precludes monochromatic emission. The effect is subtle, but is important in understanding the fine details of atomic spectra seen in experiments. Close is good enough. (Old fashioned as I am, I'll cite Condon and Shortley, The Theory of Atomic Spectra, first published 1935. In its day, this book was one of the key texts on QM.)

    But, with a Google or two you can find more modern treatments of the fact that monochromatic spectra are a first approximation to reality.

    In electrical engineering, one talks about bandwidth,same is true in QM.
    Reilly Atkinson
  5. Apr 18, 2008 #4
    hey thanks for your reply. However i am still not convinced about the electron...for example the following post: www(dot)physicsforums(dot)com/showthread(dot)php?t=184144&highlight=energy+levels+electron (sorry it wouldn't let me post a link to this topic even tho it is part of this forum?)
    implies that the electron can give up a portion of its energy to an electron in the atom?

    also the Franck-Hertz experiment was performed by using a stream of electrons and passing them through a tube of mercury vapor. The electrons leave a heated cathode and are attracted toward a grid by an adjustable accelerating potential [tex]Vo[/tex]. The electrons that pass thru the grid will reach the anode provided that they have enough energy to overcome the small retarding potential [tex]\Delta[/tex]V creating a current. When an electron is accelerated by a potential difference [tex]Vo[/tex] it acquires an energy of [tex]Voe[/tex]. As long as this energy is less than the first excitation energy of the mercury atom, only elastic collisions are possible, and there is no way for the electrons to lose any energy in the vapor. So until [tex]Voe[/tex] reaches the first excitation energy, the anode current increases steadily as [tex]Vo[/tex] is increased. Once [tex]Voe[/tex] reaches the excitation energy, some electrons can excite the mercury atoms and lose most of their energy as a result. When these electrons reach the grid, they have insufficeint energy to overcome the retarding potential [tex]\Delta[/tex]V and cannot reach the anode. So when [tex]Voe[/tex] reached the first excitation energy there is a drop in the current. The fact that the drop is observed when [tex]Vo = 4.9V[/tex] tells us that the first excitation energy of mercury is 4.9eV. When [tex]Vo[/tex] increased beyond 4.9V, the current increases again, but when [tex]Voe[/tex] reaches twice the excitation energy some electrons can undergo two inelastic collisions making the current drop again. This information on the Franck-Hertz experiment was taken from the book "Modern Physics for Scientists and Engineers" by Taylor, Zafiratos,and Dubson.

    That right there states that an electron can excite an atom if it has an energy equal to or greater than the difference between any two of the energy levels of the atom. That is an electron can give part of its energy to an electron in an atom and it doesn't have to be all or nothing. But for a photon to excite an atom, its energy must be relatively equal to the difference between any two of the atom's energy levels. If its energy is too big or small the photon is not absorbed. So for example lets say to excite an atom from n=1 to n=2 requires 10eV. If a photon is "fired" at the atom with energy of 12eV it will not be absorbed and will exit with 12eV. But if an electron is fired at the atom with an energy of 12eV it will excite the atom to the n=2 level and will exit with 2eV. Why is this so? does it have to do something with the photon not having any mass?

    sorry this is long and drawn out ...any information on this would be greatly appreciated.
  6. Apr 19, 2008 #5
  7. Apr 21, 2008 #6
    It's not so much the case that the photon needs to have the exact ENERGY as it needs to have the exact FREQUENCY. To drive an electron from one state to a higher state, the atom must pass through the mixed state. In the mixed state, the charge density oscillates at exactly the transition frequency. If the light does not have this exact frequency, it won't drive the oscillation.

    I don't know about the mechanism for driving the transition via an incident electron, but I'm guessing it's not so frequency dependent as the light-driven transition.
  8. Apr 21, 2008 #7


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    The difference between atomic spectra and Frank-Hertz is in the masses involved. Hydrogen is light, and can get bounced around in a gas. Mg, on the other hand is very much more massive than the 5eV needed for absorption to take place so there's no recoil, nor much bouncing around. The bigger the mass, the more precise a photon's energy--or frequency- must be to initiate an atomic transition.
    Reilly Atkinson
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