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Ionisation Energy of Hydrogen

  1. Nov 21, 2009 #1
    Hi, i've a question that shows me a diagram for wavelength, there's 4 of them

    11.1 x 106
    10.5 x 106
    9.7 x 106
    8.4 x 106

    then it asked me to determine the ionisation energy of hydrogen in KJ mol-1 by using the above spectrum.

    From what i know
    E = hf
    f = c/lambda
    lambda being the wavelength

    The equation should be using the largest frequency so i should pick the shortest wavelength right? which is 8.4 x 106

    But the calculation behind uses the value of 11.1 x 106 . Can anyone explain?
     
  2. jcsd
  3. Nov 21, 2009 #2

    drizzle

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    You know that the electron of a hydrogen atom has 13.6 eV less energy than a motionless electron infinitely far from the nucleus. Thus, the http://en.wikipedia.org/wiki/Ionization_energy" [Broken]needed to set this electron free is supposed to equal 13.6 eV or more but not less. So, among the wavelengths you've stated, which one would provide the sufficient energy needed to set it free?
     
    Last edited by a moderator: May 4, 2017
  4. Nov 21, 2009 #3

    chemisttree

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    What are the units for wavelength?
     
  5. Nov 21, 2009 #4

    drizzle

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    I think these numbers are to the power of -6!
     
  6. Nov 22, 2009 #5

    Redbelly98

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    Good question. crays, can you tell us what units are used for wavelength in the problem statement.

    This does not answer the question about units, but if it's true than we need crays to clarify whether the exponents are +6 or -6.
     
  7. Nov 22, 2009 #6

    drizzle

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    I wasn't answering chemisttree's Q, but I think it's meters in this case, probably.
     
  8. Nov 23, 2009 #7
    So sorry for the late reply, it is written

    wave number, v (x 106 m-1) by m-1 i assume it is wave length.

    From my book it says that frequency is equals to c/lambda but not proportional to. Clarification please. cause if its wavelength, it should be h x 1/wavelength x c.
     
  9. Nov 23, 2009 #8

    Redbelly98

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    Wave number is 1/λ, i.e. the reciprocal of the wavelength. That is why the units are m-1 instead of m.

    The frequency is simply c·(1/λ), or c-times-wavenumber. Hence the largest frequency goes with the largest wavenumber, 11.1 x 106 m-1.

    EDIT:
    The frequency is c/λ. The photon energy is hc/λ = h x frequency.

    If you examine the units in h, c, and λ, you'll find the expressions in the above paragraph work out to s-1 for frequency and J for energy, just as they should.
     
  10. Nov 23, 2009 #9
    Thanks alot for clarification and the help :)
     
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