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Electron and wavelength problem

  1. Dec 10, 2015 #1
    1. The problem statement, all variables and given/known data
    Electron on n=6 level emits wavelength of 410.2nm.
    What energy level does it move too?

    2. Relevant equations
    1/L = R ( 1/N^2 - 1/n^2)
    where N is initial and n final
    R=1.096779*10^7 m

    3. The attempt at a solution
    1/410.2 = 10967790 (1/36 - 1/n^2)
    1/4498975152 = (1/36 - 1/n^2)
    -1/36 -1/36
    4498975152 =n^2
    sqrt
    n=67074......
    this number is wrong, it is not an energy level, and not one of the answers, what did I do wrong?
     
  2. jcsd
  3. Dec 10, 2015 #2

    DrClaude

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    Staff: Mentor

    Careful here. N is the lower level and n is the upper level, not necessarily initial and final.

    Put the units in your equation, you will see where it went wrong.
     
  4. Dec 10, 2015 #3
    alright,
    R is in m^-1
    and L is in nm,

    so it will be .0000004102m*10967760m^-1 = .449897515 m/m
    Now I add 36 and take the sqrt, which gives me a number slightly larger than 6, which still is not correct. Did I miss other units?
     
  5. Dec 10, 2015 #4

    DrClaude

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    Staff: Mentor

    Almost. I think it is simpler to make calculation using powers of 10 (less chance of an error):
    410.2×10-9 m × 1.096779×107 m-1 = ?

    You can't do that,
    $$
    \frac{1}{a} + \frac{1}{b} \neq \frac{1}{a+b}
    $$
    You need to add 1/36 to 1 over the number you will get for λ×R.
     
  6. Dec 10, 2015 #5
    I thought since all the terms were under 1 that I could just solve the denominators, like this property

    X(2x+5) = x14
    2x+5= 14

    also, how do you display fractions on this forum like you did?
     
  7. Dec 10, 2015 #6

    DrClaude

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    Staff: Mentor

    That makes use of the properties of the logarithm
    $$
    \begin{align*}
    x^a &= x^b \\
    \log(x^a) &= \log(x^b) \\
    a \log(x) &= b \log(x) \\
    a &= b
    \end{align*}
    $$
    It doesn't work for ##a^{-1} + b^{-1}##.

    I used the Latex capabilities of PF. See https://www.physicsforums.com/help/latexhelp/ for more info.
     
  8. Dec 10, 2015 #7
    Thank you. I did it all out and the value I got it .66666, which still does not seem right
     
  9. Dec 10, 2015 #8

    DrClaude

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    Staff: Mentor

    Indeed. I get a nice value. Could you show your calculation?
     
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