Question

What is the frequency of limiting line in Balmer series?

Answer 1

`f = 8.225 × 10^14color(white)(l)"Hz"`

Explanation:

The Balmer series corresponds to all electron transitions from a higher energy level to `n = 2`.

The wavelength is given by the Rydberg formula

`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 and
`n_1` and `n_2` are the energy levels such that `n_2 > n_1`

Since `fλ = c`

we can re-write the equation as

`1/λ = f/c = R(1/n_1^2 -1/n_2^2)`

or

`f = cR(1/n_1^2 -1/n_2^2) = R^'(1/n_1^2 -1/n_2^2)`

where `R^'` is the Rydberg constant expressed in energy units (`3.290 × 10^15 color(white)(l)"Hz"`).

In this problem, `n_1 = 2`, and the frequency of the limiting line is reached
as `n → ∞`.

Thus,

`f = lim_(n → ∞)R^'(1/4 -1/n_2^2) = R^'(0.25 - 0) = 0.25R^'`

`= 0.25 × 3.290 × 10^15 color(white)(l)"Hz" = 8.225 × 10^14color(white)(l)"Hz"`