# What is the frequency of limiting line in Balmer series?

Sep 5, 2017

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

$f = c R \left(\frac{1}{n} _ {1}^{2} - \frac{1}{n} _ {2}^{2}\right) = {R}^{'} \left(\frac{1}{n} _ {1}^{2} - \frac{1}{n} _ {2}^{2}\right)$

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"