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  • Mar 30, 2011, 03:37 PM
    pop000
    Excited state
    given ion like hydrogen that it electron are in the Excited state n.
    the ionization energy of this ion from the Excited state n is 7.65eV.

    irradiation in light with frequency of 6.65x10^14 sec^-1 making to jump to Excited state n+1.

    how I can calculate what is the Excited state n and who is the ion of this possibility:He+ , Li2+


    tell me if I correct here: the E=Hf is 6.65x10^14 sec^-1

    thanks for help. And sorry for my bad english :)
  • Mar 31, 2011, 07:42 AM
    Unknown008

    The ionisation energy for the ion is 7.65 eV.

    The ionisation energy for the next excited state (n+1) is given by E = hf.

    In your answer you the answer as the frequency? :confused:

    I get E = 3.99 x 10^-13 J = 2.49 eV

    I don't know if there is a relationship (formula relating) between n and the energy, but if not, there should be a table when you have some value of:

    n = 7.65 eV
    n + 1 = 2.49 eV

    And from there, deduce which of He^+ or Li^2+ is correct.
  • Apr 1, 2011, 04:27 AM
    pop000
    I will try to ask in a little better english.

    Given a ion Hydrogen-like that it's electron is in an excited state n.
    The ionization energy of the ion from the excited state n is 7.65eV.
    Radiating the electron in light frequency of 6.65x10^14sec-1 transfers it to excited state n+1.

    I need to calculate what is the excited state n, and which is ion: He+ or Li2+ ?

    If we know that radiating the electron in light frequency of 6.65x10^14sec-1 transfers it to excited state n+1.
    Does it not means that 6.65x10^14sec-1 = E=hf?

    I need to use the Bohr model.

    Thank you very much!
  • Apr 1, 2011, 07:32 AM
    Unknown008

    6.65x10^14sec-1 = E = hf

    Energy is not frequency...
  • Apr 3, 2011, 11:21 AM
    pop000
    Comment on Unknown008's post
    yes that correct.
    by the way did you used with the formula -13.6*Z^2/n^2 ?
    if not can you give me the formula that you used to calculate it ?
    thank you.
  • Apr 4, 2011, 12:10 AM
    Unknown008

    Wait... I did a mistake earlier >.< I think there was some number on my calculator from a previous work which got in the calculation.

    I used the simple E = hf



    Now, using:





    Meaning that the ion gained 2.75 eV to get on to the n+1 state.

    The ionisation energy from the n+1 state is thus = 7.65 - 2.75 = 4.90 eV

    Okay, so the ionisation energy for:

    n = 7.65 eV
    n + 1 = 4.90 eV

    Now, for this formula you gave, this is the first time I see it and I did some research.

    Okay, so let's try He^+

    Z = 2





    If En and En+1 give satisfactory values of n, then He^+ is the one.

    If not, we try with Li^+

    Z = 3
  • Apr 6, 2011, 11:15 AM
    pop000
    So many thanks :)
    Again help me a lot

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