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!

Balmer Series for ##Li^{2+}##

  1. Oct 12, 2016 #1
    1. The problem statement, all variables and given/known data
    The emission wavelengths of hydrogen-like atoms are related to nuclear charge. How do they scale as a function of Z? What are the longest and shortest wavelengths in the Balmer series for ##Li^{2+}##?

    2. Relevant equations
    ##E_n = -\frac{R}{n^2}## (1)

    ##a_0 = \frac{\hbar^2}{Zme^2}## (2)

    From (2), ##\hbar^2 = a_0Zme^2## (3)

    ##\alpha = \frac{Ze^2}{\hbar c}## (4)

    ##R = \frac{1}{2} mc^2 \alpha^2## (5)

    3. The attempt at a solution
    I started by trying to find an expression for the energy levels of hydrogen-like atoms. Substituting (5) into (1) gives
    ##E_n = -\frac{mc^2\alpha^2}{2n^2}##
    Substitute in (4):
    ##E_n = -\frac{mZ^2e^4}{2\hbar^2 n^2}##
    Sub in (3):
    ##E_n = -\frac{mZ^2e^4}{2Za_0me^2 n^2}##

    Simplifying, ##E_n = -\frac{Ze^2}{2a_0n^2}##. This can be related to wavlength via ##E_n = \frac{hc}{\lambda}##, so emission and absorption wavelengths of hydrogenic atoms are related to nuclear charge Z by the function

    ##\lambda = -\frac{2hca_0n^2}{Ze^2}##

    But I have no idea how to answer the bit about the Balmer series because n=2 and Z=3, so based on my equation isn't there only going to be one answer? How do I get minimum/maximum wavelengths?
     
    Last edited: Oct 12, 2016
  2. jcsd
  3. Oct 12, 2016 #2
    In the Balmer series the final state is an ##n=2## state. The initial state may be anything from ##n=3## to ##n=\infty##. You need to find the energy difference between these states.
     
  4. Oct 12, 2016 #3
    So I use the equation for ##E_n## I found and do ##E_3-E_2## for the maximum wavelength? This gives a negative answer though, can I just multiply that answer by ##-1##? Or do ##E_2-E_3##?
     
  5. Oct 12, 2016 #4
    ##E_3-E_2## is positive from your formula since you got a minus sign in it. That is ##-1/9 > -1/4##.
    I don't know if all your other constants are correct but you can always compare your answer to the Rydberg formula https://en.wikipedia.org/wiki/Rydberg_formula
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Balmer Series for ##Li^{2+}##
  1. Balmer series (Replies: 3)

  2. Balmer series (Replies: 12)

  3. Balmer Series (Replies: 1)

  4. Balmer Series (Replies: 3)

Loading...