Line Spectra of Li2+ & Be3+: Shortest Wavelength

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SUMMARY

The discussion focuses on the line spectra of doubly ionized lithium (Li2+) and triply ionized beryllium (Be3+), specifically identifying the shortest wavelength emitted by each. The shortest wavelength for Li2+ is established at 1222.2 nm. The formula used for calculating the wavelengths is 1/λ = 1.097 x 107 (Z2) (1/nf2 - 1/ni2). The participant successfully determined the shortest wavelength for Be3+ by applying the known wavelength of Li2+ as the final state (nf) in their calculations.

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  • Understanding of atomic structure and ionization states
  • Familiarity with the Rydberg formula for spectral lines
  • Knowledge of quantum numbers (ni and nf)
  • Basic principles of spectroscopy
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  • Explore the line spectra of other ionized elements for comparative analysis
  • Learn about quantum numbers and their significance in electron transitions
  • Investigate the implications of ionization energy on spectral lines
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Students of physics and chemistry, researchers in atomic spectroscopy, and educators teaching concepts of atomic structure and spectral analysis will benefit from this discussion.

StudentofPhysics
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1. Doubly ionized lithium Li2+ (Z = 3) and triply ionized beryllium Be3+ (Z = 4) each emit a line spectrum. For a certain series of lines in the lithium spectrum, the shortest wavelength is 1222.2 nm. For the same series of lines in the beryllium spectrum, what is the shortest wavelength?



2. 1/ lamba = 1.097 x 10^7 (z^2) (1/nf^2 - 1/ni^2)



3. My only question is what I use for the n's? Since its the smallest I assume ni = 0. What about nf?
 
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Assuming an equation like (2) is valid, does the answer depend on the particular values of n1 and n2?
 
ok, i misunderstood what nf should be.

I figured it out now by solving for it by the first given wavelength then applying that number as the nf for the second.

Thank you.
 

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