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Questions
Compute the shortest and the longest wavelength in the Lyman series of hydrogen atom.
Find the shortest and longest wavelengths in the Lyman series of hydrogen atom.
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Solution
Given: R = 1.097 × 107 m−1
`1/λ = R_H (1/(n^2) - 1/(m^2))`
For the Lyman series, n = 1 and for the shortest wavelength, m = ∞.
∴ `1/(λ) = R (1/1^2 - 1/∞^2) = R`
∴ The short wavelength limit of the Lyman series,
`λ = 1/R`
= `1/(1.097 xx 10^7)`
= 0.9110 × 10−7 m
= 911 Å
For the longest wavelength in the Lyman series, n = 1 and m = 2.
∴ `1/(λ) = R (1/1^2 - 1/2^2)`
= `R (1/1 - 1/4)`
= `R ((4 - 1)/4)`
= `(3R)/4`
= `(3(1.097 xx 10^7))/4`
∴ The wavelength of the first Lyman line,
`1/λ = 4/3.291 xx 10^-7`
= 1.215 × 10−7 m
= 1215 Å
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