Question
Download Solution PDFThe ratio of the longest wavelength to the shortest wavelength \((\frac{\lambda_L}{\lambda_S})\) in Paschen series of hydrogen spectrum is:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Hydrogen spectrum:
- The hydrogen spectrum is an important piece of evidence to show the quantized electronic structure of an atom.
- The hydrogen atoms of the molecule dissociate as soon as an electric discharge is passed through a gaseous hydrogen molecule.
- It results in the emission of electromagnetic radiation initiated by the energetically excited hydrogen atoms.
- The hydrogen emission spectrum comprises radiation of discrete frequencies.
- First three spectral series in the hydrogen spectrum: Lyman, Balmer, and Paschen series
- The formula for the wavelength, \(\frac{1}{λ}=R(\frac{1}{p^2}-\frac{1}{n^2})\) where, p = higher energy level, n = lowest energy level, λ = wavelength
- Here R = 109677 is called the Rydberg constant for the hydrogen atom
Calculation:
The longest wavelength in the Paschen series was obtained when p = 3, n = 4
For the longest wavelength,
\(\frac{1}{λ_l}=R(\frac{1}{p^2}-\frac{1}{n^2})\)
\(\frac{1}{λ_l}=R(\frac{1}{3^2}-\frac{1}{4^2})\)
\(\frac{1}{λ_l}=R(\frac{1}{9}-\frac{1}{16})\)
\(λ _l=\frac{144}{7R}\)
The shortest wavelength in the Paschen series was obtained when p = 3, n = ∞
For the shortest wavelength,
\(\frac{1}{λ_s}=R(\frac{1}{p^2}-\frac{1}{n^2})\)
\(\frac{1}{λ_s}=R(\frac{1}{3^2}-\frac{1}{∞^2})\)
\(λ _s=\frac{9}{R}\)
Ratio, \(\frac{λ_l}{λ _s}=\frac{144/7R}{9/R}\)
\(\frac{λ_l}{λ _s}=\frac{16}{7}\)
Last updated on Jun 19, 2025
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