Question

Calculate the wavelength of the first line in lyman series of the hydrogen spectrum (R = 109677 cm-1) how to do this?

Answer 1

`121.6 \text{nm}`

Explanation:

`1/lambda = \text{R}(1/(n_1)^2 - 1/(n_2)^2) * \text{Z}^2`

where,

R = Rydbergs constant (Also written is `\text{R}_\text{H}`)
Z = atomic number

Since the question is asking for `1^(st)` line of Lyman series therefore

`n_1 = 1`
`n_2 = 2`

since the electron is de-exited from `1(\text{st})` exited state (i.e `\text{n} = 2`) to ground state (i.e `text{n} = 1`) for first line of Lyman series.

Therefore plugging in the values

`1/lambda = \text{R}(1/(1)^2 - 1/(2)^2) * 1^2`

Since the atomic number of Hydrogen is 1.

By doing the math, we get the wavelength as

`lambda = 4/3*912 dot \text{A}`

since `1/\text{R} = 912 dot \text{A}`

therefore

`lambda = 1216 dot \text{A}`
or

`lambda = 121.6 \text{nm}`

Answer 2

`λ = "121.569 nm"`

Explanation:

The first emission line in the Lyman series corresponds to the electron dropping from `n = 2` to `n = 1`.

(Adapted from Tes)

The wavelength is given by the Rydberg formula

`color(blue)(bar(ul(|color(white)(a/a) 1/λ = -R(1/n_f^2 -1/n_i^2)color(white)(a/a)|)))" "`

where

`R =` the Rydberg constant (`"109 677 cm"^"-1"`) and
`n_i` and `n_f` are the initial and final energy levels

For a positive wavelength, we set the initial as `n = 1` and final as `n = 2` for an absorption instead.

`1/λ = -"109 677 cm"^"-1" × (1/2^2 -1/1^2)`

`= "109 677" × 10^7color(white)(l) "m"^"-1" (1/4-1/1)`

`= "109 677 cm"^"-1" × (1-4)/(4×1)`

`= -"109 677 cm"^"-1" × (-3/4) = "82 257.8 cm"^"-1"`

`λ = 1/("82 257.8 cm"^"-1") = "1.215 69" × 10^"-5"color(white)(l) "cm"`

`= "1.215 69" × 10^"-7"color(white)(l) "m"`

`= ul"121.569 nm"`