Question

What is the wavelength of the emitted light when excited electrons in the hydrogen atoms make transition from the n = 3 in the Balmer series?

Answer 1

λ = 656.1 nm

Explanation:

The Balmer series consists of all transitions that end at the (`n = 2`) energy level.

The Rydberg equation gives the wavelength `λ` for the transitions:

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

where

`R =` the Rydberg constant (`1.0974 × 10^7color(white)(l) "m"^"-1"`) and
`n_1` and `n_2` are the numbers of the energy levels such that `n_1 < n_2`

For a transition from (`n = 3`) to (`n = 2`),

`1/λ = (1.0974 × 10^7color(white)(l) "m"^"-1")(1/4 - 1/9) = (1.0974 × 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") = 6.561 × 10^"-7"color(white)(l)"m" = "656.1 nm"`

This wavelength corresponds to the red line in the hydrogen spectrum.