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Light emission and energy of states

  1. Nov 9, 2014 #1
    1. The problem statement, all variables and given/known data

    The emission spectrum of thermally excited sodium atoms practically consists of a single intensive line at 589 nm wavelength. What is the energy difference (in eV units) between the excited and ground states of the sodium atom?

    2. Relevant equations
    E = hc/lambda, we also know that electron electric charge = 1.6 * 10-19

    3. The attempt at a solution
    I know that the energy of a photon emitted from an excitation of an electron is E = hc/lamda and by dividing that result by the electric charge of the electron, i get the eV units, but I do not understand the ground state and excited states.
     
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  3. Nov 10, 2014 #2

    Orodruin

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    The photon is emitted when the atom transits from the excited to the ground state. If the energy difference between the states is E, how much energy must the photon have in order for energy to be conserved?
     
  4. Nov 10, 2014 #3

    andrevdh

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    The energy levels of an atoms are quantized.
    That is the electrons can only occupy certain levels in the atom.
    The electrons fill up from the bottom towards the top levels, but
    when the atom is excited, as by an increase in the temperature
    of the material in this case, some of the electrons can be raised
    to higher energy levels leaving some lower energy levels unoccupied.
    We then get photons being emitted by the atom when the electrons
    return to the ground states - the unoccupied lower energy levels.
     
  5. Nov 10, 2014 #4
    Ok. Thank you. So the atom emits an photon when it drops down from the excited state to a lower state, and absorb an photon when it goes from a lower to a higher energy state. I do know the different states of a hydrogen atom (calculated by the rydebergs formula) but is it possible for use that formula in conjuncton with other atoms like f.ex sodium?
    Thank you.
     
  6. Nov 10, 2014 #5

    andrevdh

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    No. The interactions of the various states come into play and changes the formula quite dramatically.
     
  7. Nov 10, 2014 #6
    Ok. So does every atom have a constant number (which does not have to be calculated) for every energy level?
     
  8. Nov 10, 2014 #7

    andrevdh

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    I would think that the quantum mechanics is so complex that such formulas cannot even be derived and
    thus probably don't even exist.
     
  9. Nov 10, 2014 #8
    Ok, thank you.

    So according to my question, which formula would be the most appropriate to use for calculation of the energy difference in ground and exited state in a sodium atom?

    Thank you
     
  10. Nov 11, 2014 #9

    ehild

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    The energy of the photon emitted is equal to the energy difference between the excited and ground states. You know the wavelength of the emitted photon. What is its energy?
     
  11. Nov 11, 2014 #10
    E = hc/lambda, hence
    E= (6.6*10-34)*(3.0*108) / (5.89*10-7m)

    So this will give me the energy difference of the ground state and excited state as the energy of the photon is equal to the energy difference?
     
  12. Nov 11, 2014 #11

    andrevdh

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    You can use the formula you gave above. lambda is the wavelength of the photon and E is its energy, which is also the
    energy difference between the excited and ground state. This energy became available as the electron fell back to the ground
    state from the excited state of the atom, that is it transited from a higher energy level to a lower energy level in the atom.
     
  13. Nov 11, 2014 #12

    ehild

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    Yes, but in what units? Remember you have to give the energy difference in eV-s.
     
  14. Nov 11, 2014 #13

    andrevdh

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    That gives you the energy in joules since you have worked in S.I. units. Convert it to electron volts.
     
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