Uncovering the Shortest Wavelength of Be3+ Lines

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SUMMARY

The shortest wavelength of the triply ionized beryllium (Be3+) line spectrum can be calculated using the Rydberg formula: 1/λ = Rz²(1/n₁² - 1/n₂²). In this case, the atomic number (Z) for beryllium is 4, and the shortest wavelength for the doubly ionized lithium (Li2+) is 1979.8 nm. By applying the Rydberg constant and the appropriate quantum numbers, one can derive the shortest wavelength for Be3+ in the same series.

PREREQUISITES
  • Understanding of the Rydberg formula for spectral lines
  • Knowledge of atomic structure and quantum numbers
  • Familiarity with line spectra and ionization states
  • Basic principles of spectroscopy
NEXT STEPS
  • Calculate the shortest wavelength for Be3+ using the Rydberg formula
  • Explore the differences in spectral lines between Li2+ and Be3+
  • Study the implications of ionization states on spectral emissions
  • Investigate the applications of spectroscopy in identifying elements
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Students in physics and chemistry, particularly those studying atomic spectra, quantum mechanics, and spectroscopy techniques.

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Homework Statement



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 1979.8 nm. For the same series of lines in the beryllium spectrum, what is the shortest wavelength?

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The Attempt at a Solution



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Hehe... give it a shot man. Tell you what, the equation you need is:

\frac{1}{\lambda}=Rz^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})

where R is the rydberg constant, z is the atomic number, and n2 is the shell it jumps from to n1 (the final shell)
 

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