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Early Quantum Mechanics

  1. Apr 30, 2007 #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 hydrogen's electron drops from the fifth energy level to a lower energy level. What is the lower energy level?

    2. Relevant equations

    1/lambda = R[1/m e2 - 1/n e2]

    3. The attempt at a solution

    1/4.34 x 10 e-7 m = 1.097 x 10 e7 m e-1 [1/m e2 - 1/5 e2]
    1/4.34 x 10 e-7 m [1/m e2 - 1/25] = 1/1.097 x 10 e7 m
    1/[1/m e2 - 1/25] = 4.34 x 10 e-7m/1.097 x 10 e7
    1/m e2 - 1/25 = 1.097 x 10 e7/4.34 x 10 e-7 + 1/25
    n e2 = 4.34 x 10 e-7/1.097 x 10 e7 + 25/1
    =3.96 x 10 e-14 + 25
    n = 5

    Grrrrr! I keep getting n = 5 and I don't know what I'm doing wrong. Can anybody please help me?
  2. jcsd
  3. Apr 30, 2007 #2

    Chi Meson

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    here, the "m" is the shell number you want, and the "n" is the shell it came from, in this case 5.

    Edit: No, that's what you did. Um...

    When you write out "1/m e2" do you mean "1/m^2"? No...

    Your math is difficult to follow. I'm thinking you are making the algebra harder than it needs to be. Your formula is correct and you should get the right results; the problem here is algebra.

    Try solving the problem without cramming in all the numbers first.

    ARe you required to work out the math for this? Does the fact that this photon is visible provide enough info?
    Last edited: Apr 30, 2007
  4. Apr 30, 2007 #3
    I did mean 1/m e1 = 1/m^2

    I guess my algebra is off but I've tried it several times and each time I keep getting 5. Do you have any suggestions?
  5. Apr 30, 2007 #4

    Chi Meson

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    In what you wrote, your second line is all wrong.

    Again, work out the algebraic manipulation using the variables, not the numbers, it's less confusing.
  6. Apr 30, 2007 #5
    Aaaaah, I got it now. Thanks Chi!!!!
  7. Apr 30, 2007 #6

    Chi Meson

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    Just wanted to point out, that if a hydrogen emission is visible (with only one red exception) then it is an emission in the "Balmer series," all of which are from transitions to the second shell.

    Emissions that are UV, are the "Lyman series" which transition to level 1, and IR emissions are "Paschen series" which transition to level 3. The highest energy paschen emission is barely visible red.
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