Question

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.

Answer 1

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