1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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.

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


    [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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook