Question

Calculate the wavenumber for the longest wavelength transition in the Balmer series of atomic hydrogen.

Answer 1

For the Balmer series, $n_i = 2$.
Thus, the expression of wave number is given by,
$\bar v = [ \frac {1}{(2)^2}-\frac {1}{n_f^2}] (1.097×10^7 m^{-1})$
Wavenumber is inversely proportional to wavelength of transition. Hence, for the longest wavelength transition, Wavenumber has to be the smallest. For Wavenumber to be minimum, $n_f$ should be minimum.
For the Balmer series, a transition from $n_i = 2$ to $n_f = 3$ is allowed.
Hence, taking $n_f = 3$, we get:
$\bar v = (1.097×10^7)[\frac {1}{2^2} - \frac {1}{3^2}]$

$\bar v = (1.097×10^7)[\frac {1}{4} - \frac {1}{9}]$

$\bar v = (1.097×10^7)[\frac {9-4}{36}]$

$\bar v = (1.097×10^7)(\frac {5}{36})$

$\bar v = 1.5236×10^6 \ m^{-1}$