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Principle Quantum Number Question

  1. Oct 7, 2009 #1
    1. The problem statement, all variables and given/known data
    The hydrogen spectrum contains a blue line with a wavelength of 434 nm. Photons of blue light are emitted when hydrogens electron drops from the fifth energy level to a lover energy level. What is the lower energy level?


    2. Relevant equations
    En = 13.6 eV/n^2
    E = hc/lambda


    3. The attempt at a solution
    First i'm going to find E.

    E = hc/lambda
    = (6.63 x 10^-34 J s)(3.00 x 10^8 m/s) / (4.34 x 10^-7 m)
    = 4.583 x 10^-19 J

    (1.60 x 10^-19 eV/J) x (4.583 x 10^-19 J)
    = 2.86 eV

    Next i tried 2 different approaches, im not sure either is correct though. I used En = -(13.6eV/n^2) - (-13.6eV/n^2) and i just used En = -(13.6eV/n^2)

    1st approach:
    En = -(13.6eV/n^2) - (-13.6eV/n^2)

    I rearranged this equation to be n^2 = 2(13.6eV)/En5^2 (i'm pretty sure thats incorrect)
    so n^2 = 2(13.6 eV)/2.86(25)
    n = sqrt(0.38041958)
    n = 0.62

    2nd approach:
    En = -(13.6eV/n^2)
    Now i rearranged this equation to be n^2 = En + 13.6eV
    = 2.86eV + 13.6eV
    n = sqrt16.46 eV
    = 4.06

    So im pretty sure my problem lies within rearranging the equations. Any suggestions? thanks
     
  2. jcsd
  3. Oct 7, 2009 #2
    I also just realized i can rearrange En = -(13.6eV/n^2) to be n^2 = 13.6eV/En

    so using that equation it would be n^2 = 13.6eV/2.86eV
    n = sqrt4.75
    = 2.2

    this one makes alot more sense, seeing how the units cancel
     
  4. Oct 7, 2009 #3
    it looks like your problem is with the relationship between the energy of the photon and the change in energy levels.

    [tex]E_{photon}= E_{f}- E_{i}[/tex]

    where the initial state is the 5th energy level, E of the photon you figured out was 2.86, and the final state is what we're solving for.

    so,
    [tex]E_{n}= E_{photon} + E_{5}[/tex]

    and

    [tex] \frac{13.6}{n^{2}} = 2.86 + \frac{13.6}{5^{2}} [/tex]

    now its just algebra to find out n
     
  5. Oct 7, 2009 #4
    thanks, i thought it was that, but i might've messed up the rearrangement, im not very strong when it comes to that... i got n^2 = 13.6eV - 0.544eV/2.86eV (the 0.544 is 13.6/5^2)
    I worked it out and got an answer of 2.14
     
  6. Oct 7, 2009 #5
    i still think you're doing something a bit off here. the .544 and 2.86 should be summed, not divided. and then divide 13.6 by the new sum for your answer. it should give you n^2=(approx.)4, and n=2

    remember that n is a discrete energy level, which are integers, and if your decimal is off by more than a tenth or two from the whole, something might be awry in your calculations.
     
    Last edited: Oct 7, 2009
  7. Oct 7, 2009 #6
    but when i do that i get a decimal, not a whole number. and its 0.544
     
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