The Balmer series in the hydrogen spectrum corresponds to the transition from $n_1 = 2$ to $n_2 = 3, 4, ...$ This series lies in the visible region. Calculate the wave number of the line associated with the transition in the Balmer series when the electron moves to the $n = 4$ orbit. $(R_H = 109677 \ cm^{-1})$

(N/A) The wave number $\bar{v}$ is given by the Rydberg formula: $\bar{v} = R_H \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right] \ cm^{-1}$.
For the Balmer series,the transition is from $n_2$ to $n_1 = 2$.
Given $n_2 = 4$ and $R_H = 109677 \ cm^{-1}$.
Substituting the values: $\bar{v} = 109677 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \ cm^{-1}$.
$\bar{v} = 109677 \left( \frac{1}{4} - \frac{1}{16} \right) \ cm^{-1}$.
$\bar{v} = 109677 \left( \frac{4-1}{16} \right) \ cm^{-1} = 109677 \times \frac{3}{16} \ cm^{-1}$.
$\bar{v} = 20564.44 \ cm^{-1}$.