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Hydrogen spectrum question

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  1. Mar 11, 2014 #1
    1. The problem statement, all variables and given/known data
    Hello,

    I have a question regarding hydrogen spectral emission.

    A hydrogen source is viewed with a grating spectrometer, one spectral lines occurs at 20.5 degrees in the 1st order. What angle will this line appear in 2nd order (viewed through same spectrometer)

    a) cannot find this angle without ruling spacing of grating
    b) cannot find this angle without wavelength of spectral line
    c) it IS possible to find this angle with the given info



    Please help!



    2. Relevant equations

    d(sinθ)=mλ

    3. The attempt at a solution

    When I think about this, do we not need the ruling spacing to find the wavelength? From there, can we not find the 2nd order angle? The problem is, there is no option for A and B. And that's assuming I'm on the right track.
     
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  3. Mar 11, 2014 #2

    ehild

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    There is the equation d(sinθ)=mλ, and given that θ=20.5 degrees when m=1. Can you find λ/d?
    The question is the angle when m=2. Do you need anything else but λ/d?


    ehild
     
  4. Mar 11, 2014 #3

    d is the space between the grating, correct? And λ is the wavelength. I don't have either. Just an angle for m=1. Still confused, but thank you for the reply.
     
  5. Mar 11, 2014 #4

    BvU

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    Why this confusion? You don't have either, but you do have the ratio ! And what do you really need ? Aha!
     
  6. Mar 11, 2014 #5
    So in my case the angle appearing at the 2nd order will be 41.0 degrees? Is that correct?
     
  7. Mar 11, 2014 #6

    BvU

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    No. Just try to imagine where the fifth order would end up!

    No that's corny. You have an equation for this ##\theta##. Solve it.
     
  8. Mar 11, 2014 #7
    Yes, it would be over 90°, makes no sense. Which leads me to believe I need grating. So would my answer be

    a) cannot find this angle without ruling spacing of grating

    ?
     
  9. Mar 11, 2014 #8

    BvU

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    No. It's just that if ##\sin \theta = q##, that doesn't mean that ## \sin 2\theta = 2q##.
    You have q from the 20.5 ##^\circ##, now find ##\theta_2##!
     
  10. Mar 11, 2014 #9
    sinθ = q
    sin20.5 = q
    0.350207 = q

    0.350207*2 = 0.700415

    sin^-1(0.700415) = 44.5°

    Is that correct?

    By the way, thank you for all you're help, BvU. I appreciate you're time.
     
  11. Mar 11, 2014 #10

    BvU

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    I had the same answer.... So either it's right or we are both wrong.
     
  12. Mar 12, 2014 #11

    BvU

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    Thanks for all the thanks, but, really, once is more than adequate...
     
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