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Homework Help: Finding wavelength Please help

  1. Jul 27, 2010 #1

    Ush

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    1. The problem statement, all variables and given/known data
    Two of the wavelengths emitted by a hydrogen atom are 9.50×10-8 m and 4.65×10-6 m. What are the m and n values for the first wavelength?

    2. Relevant equations
    E = -13.6eV/n2
    E = -13.6eV/(m2 - n2)
    restriction:m > n
    E = hf
    f = c/λ
    1 eV = 1.6 x 10-19J

    3. The attempt at a solution

    E = -13.6eV/(m2 - n2)
    hc/λ = -13.6eV/(m2 - n2)
    hc/λ(-13.6eV) = 1/(m2 - n2)
    λ(-13.6eV)/hc = m2 - n2
    λ(2.179E-18J)/hc = m2 - n2

    let λ =9.50×10-8

    (9.50×10-8)(2.179E-18J)/hc = m2 - n2
    1 = m2 - n2
    1 = (m-n)(m+n)

    ...now what? o_o
    ...am I doing this right?
    ...if I do the same process with the other wavelength then I get

    51 = mb2 - nb2
    51 = (mb-nb)(mb+nb)

    ...but nothing says m and mb or n / nb are related to each other..

    this looks like something I did in h.s. -I don't remember how I did it =[

    please help
     
  2. jcsd
  3. Jul 27, 2010 #2
    It should be: E = -13.6 ev * (1/n2 - 1/m2)
     
  4. Jul 27, 2010 #3

    kuruman

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    Incorrect. Should be E = -13.6 eV {(1/n2)-(1/m2)}!!!

    1/4 - 1/3 is NOT 1/(4-3) = 1/1 !!!
     
  5. Jul 27, 2010 #4

    Ush

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    what's the difference?
    if m > n then.. ex.. if m = 3 and n = 2 .... 1/3^2 - 1/2^2 = a negative number! a negative number multiplied by -13.6 is a positive number :S
     
  6. Jul 27, 2010 #5

    Ush

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    i mean, if I switch the two, wont I have to switch the restriction anyway? to make E positive? I'm confused =S
     
  7. Jul 27, 2010 #6
    read carefully, what hikaru & kuru are pointing out.
     
  8. Jul 27, 2010 #7

    Ush

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    Oh,

    -13.6eV/(m2 - n2) ≠ E = -13.6 ev * (1/n2 - 1/m2)

    but if m > n, then shouldn't it be
    E = -13.6 ev * (1/m2 - 1/n2)
    in order to get a positive energy? =S

    if I put n before m i'll get a negative energy,
    that doesn't make sense because photons have to have a positive energy!
    I'm almost certain of that =/

    Try 2:

    E = -13.6 ev * (1/m2 - 1/n2)
    hc/λ = -13.6 ev * (1/m2 - 1/n2)
    hc/λ(-13.6eV) = (1/m2 - 1/n2)

    let λ =9.50×10-8

    -0.96 = (1/m2 - 1/n2)
    and now..?
     
  9. Jul 27, 2010 #8

    kuruman

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    Correct, now you see the difference.

    The formula that you are using gives the change in energy of the atom. If it is positive, the atom gains energy, if it is negative it loses energy. And yes, a photon always has positive energy. You should be 100% sure of that. Take the absolute value of the difference as the energy of the photon, which could either be gained or lost by the atom.

    Try different pairs of integers m and n, but be systematic about it.
     
  10. Jul 27, 2010 #9

    Ush

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    using different pairs of integers..
    1/52 - 1/1 = -0.96

    is there no mathematical way of doing this, other then guessing?
     
  11. Jul 27, 2010 #10

    kuruman

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    Think about it. You have one equation and two unknowns.

    Of course, you can fix n and then find the shortest wavelength of a photon (highest energy) that is emitted by a transition to that level. This means that m is infinite, or 1/m2 = 0. In that case, the shortest wavelength to the n= 2 level is 2.29x10-6 m, which is much longer than 9.6x10-8 m. Any n other than 1 will give you even longer wavelengths. Therefore n must be equal to 1. Once you establish that, there is a finite number of m values that will do the job. That is what I mean by "systematic".
     
  12. Jul 27, 2010 #11

    Ush

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    oh..
    so giving us another wavelength in the question is just to confuse us right?
     
  13. Jul 27, 2010 #12

    Ush

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    thank you for your help =)
     
  14. Jul 27, 2010 #13

    kuruman

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    I don't know, and I cannot read the mind of the person who wrote the problem. Anyway, I think you now understand how this works.
     
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