Question

Compute the shortest and the longest wavelength in the Lyman series of hydrogen atom.

Answer 1

Data: R = 1.097 × 10⁷ m⁻¹

`1/λ = R (1/(n^2) - 1/(m^2))`

For the Lyman series, n = 1. For the short wavelength limit (λ_L∞), m = ∞.

∴ `1/(λ_(L∞)) = R (1/1 - 0) = R`

∴ The short wavelength limit of the Lyman series,

`λ_(L∞) = 1/1.097 xx 10^-7`

= 0.9110 × 10⁷ m

= 911 Å

For the longest wavelength line (λ_Lα) of the Lyman series, m = 2.

∴ `1/(λ_(Lα)) = R (1/1 - 1/4)`

= `(3R)/4`

= `(3(1.097 xx 10^7))/4`

∴ The wavelength of the first Lyman line,

`1/(λ_(Lα)) = 4/3.291 xx 10^-7`

= 1.215 ×  10⁷ m

= 1215 Å