Question

An electron in a hydrogen atom absorbs a photon of energy and jumps from the ground state to the `n = 3` level. Then it returns in two steps to the ground state. What are the wavelengths of the photons involved?

Answer 1

The atom absorbs a photon of wavelength 102.6 nm and emits two photons of wavelengths 656.3 nm and 121.5 nm.

Explanation:

(Adapted from Tes)

The electron first absorbs a photon and is excited to `n = 3` (a).

Then it emits one photon when it drops to `n=2` (b) and another when it drops from `n=2` to `n=1` (c).

(a) Excitation to `n=2`

The wavelength is given by the Rydberg formula

`color(blue)(bar(ul(|color(white)(a/a) 1/λ = R(1/n_2^2 -1/n_1^2)color(white)(a/a)|)))" "`

where

`R =` the Rydberg constant (`1.097 × 10^7color(white)(l) "m"^"-1"`)
`n_1 =` the upper energy level
`n_2 =` the lower energy level

For this part,

`n_1 = 3`
`n_2 = 1`

`1/λ = 1.097 × 10^7color(white)(l) "m"^"-1" × (1/1^2 -1/3^2) = 1.097 × 10^7color(white)(l) "m"^"-1" (1/1-1/9)`

`= 1.097 × 10^7color(white)(l) "m"^"-1" × (9-1)/(9×1) = 1.097 × 10^7color(white)(l) "m"^"-1" × 8/9 = 9.751 × 10^6 color(white)(l)"m"^"-1"`

`λ = 1/(9.751 × 10^6 color(white)(l)"m"^"-1") = 1.026 × 10^"-7"color(white)(l) "m" = "102.6 nm"`

(b) Dropping from `n=3` to `n = 2`

`1/λ = 1.097 × 10^7color(white)(l) "m"^"-1" × (1/2^2 -1/3^2) = 1.097 × 10^7color(white)(l) "m"^"-1" (1/4-1/9)`

`= 1.097 × 10^7color(white)(l) "m"^"-1" × (9-4)/(9×4) = 1.097 × 10^7color(white)(l) "m"^"-1" × 5/36 = 1.524 × 10^6 color(white)(l)"m"^"-1"`

`λ = 1/(1.524 × 10^6 color(white)(l)"m"^"-1") = 6.563 × 10^"-7"color(white)(l) "m" = "656.3 nm"`

(c) Dropping from `n=2` to `n = 1`

`1/λ = 1.097 × 10^7color(white)(l) "m"^"-1" × (1/1^2 -1/2^2) = 1.097 × 10^7color(white)(l) "m"^"-1" (1/1-1/4)`

`= 1.097 × 10^7color(white)(l) "m"^"-1" × (4-1)/(4×1) = 1.097 × 10^7color(white)(l) "m"^"-1" × 3/4 = 8.228 × 10^6 color(white)(l)"m"^"-1"`

`λ = 1/(8.228 × 10^6 color(white)(l)"m"^"-1") = 1.215 × 10^"-7"color(white)(l) "m" = "121.5 nm"`