Question

Using the Bohr model, how do you calculate the frequency of light in the hydrogen atom with the energy transition goes from n=4 to n=1?

Answer 1

We use the Rydberg formula to get the wavelength, and then we convert this to a frequency of `3.08 × 10^15color(white)(l) "s"^"-1"`.

Explanation:

The Rydberg formula is

`color(blue)(bar(ul(|color(white)(a/a)1/λ = R_"H"(1/n_"f"^2 - 1/n_"i"^2)color(white)(a/a)|)))" "`

where

`λ` is the wavelength of the radiation,
`R_"H"` is the Rydberg constant (`1.097 × 10^7color(white)(l) "m"^"-1"`)
`n_"i"` and `n_"f"` are the initial and final energy levels

Thus,

`1/λ = 1.097 × 10^7color(white)(l) "m"^"-1"(1/4^2 -1/1^2) = "-1.028 × 10"^7color(white)(l) "m"^"-1"`

`λ = -1/("1.028 × 10"^7color(white)(l) "m"^"-1") = "-9.723 × 10"^"-8"color(white)(l) "m"`

The negative sign shows that energy is emitted.

Frequency and wavelength are related by the formula

`color(blue)(bar(ul(|color(white)(a/a)νλ = c color(white)(a/a)|)))" "`

where

`ν` is the frequency of the light
`c` is the frequency of the light (`2.998 × 10^8color(white)(l) "m·s"^"-1"`)

∴ `ν = c/λ = (2.998 × 10^8color(white)(l) color(red)(cancel(color(black)("m")))·"s"^"-1")/(9.723 × 10^"-8" color(red)(cancel(color(black)("m")))) = 3.08 × 10^15color(white)(l) "s"^"-1"`