Calc K-Shell X-Ray Series Wavelengths for Zirconium Atom

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SUMMARY

The discussion focuses on calculating the K-shell X-ray series wavelengths for the zirconium atom, specifically λ(Kα) = 79.8832 pm, λ(Kβ) = 67.401142 pm, and λ(Kγ) = 63.90653 pm. The series limit was incorrectly calculated using the formula 1/λ = Z²R(1/(nl)² - 1/(nu)²) with Z = 40, nl = 1, and nu = infinity, leading to an erroneous result of 56.9543545 pm. A participant suggested that the calculation may have been flawed due to incorrect input, proposing that 3/4 of 79.88 pm is approximately 59.91 pm as a potential correction.

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  • Understanding of X-ray spectroscopy and K-shell transitions
  • Familiarity with the Rydberg formula for spectral lines
  • Basic knowledge of atomic structure, specifically for the zirconium atom (Z = 40)
  • Proficiency in using scientific calculators for precise calculations
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  • Review the Rydberg formula for calculating wavelengths in atomic spectra
  • Study the principles of X-ray emission and absorption in elements
  • Explore advanced topics in quantum mechanics related to electron transitions
  • Practice calculating wavelengths for other elements in the periodic table
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Students and professionals in physics, particularly those specializing in atomic physics, X-ray spectroscopy, and materials science, will benefit from this discussion.

freefallin38
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Calculate the three longest wavelengths and the series limit for the K-shell x-ray series of the zirconium atom.
(a) λ(Kα) = 79.8832 pm
(b) λ(Kβ) = 67.401142 pm
(c) λ(Kγ) = 63.90653 pm
(d) The series limit is _____ pm

So I have solved parts a-c. For part d, the series limit, I tried solving using the following formula:
1/lambda=Z^2*R(1/(nl)^2-1/(nu)^2), where nl=1, nu=infinity, Z=40 so it reduced to:
Lambda=1/(Z^2R)= 56.9543545pm, but this answer is wrong. Can anyone see where I'm going going wrong?
 
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At a guess, I'd say you may have punched something into your calculator wrong for part (d). Isn't 3/4 of 79.88 more like 59.91 ?
 

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