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

(N/A) For the Balmer series,the lower energy level is $n_{1} = 2$. The Rydberg formula for the wave number $(\bar{\nu})$ is given by:
$\bar{\nu} = R_{H} \left( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right)$
where $R_{H} = 1.097 \times 10^{7} \ m^{-1}$.
Since the wave number $(\bar{\nu})$ is inversely proportional to the wavelength $(\lambda)$,the longest wavelength corresponds to the smallest wave number.
For the Balmer series,the transition with the smallest wave number occurs from the nearest higher energy level,i.e.,$n_{2} = 3$ to $n_{1} = 2$.
Substituting the values:
$\bar{\nu} = (1.097 \times 10^{7} \ m^{-1}) \left( \frac{1}{2^{2}} - \frac{1}{3^{2}} \right)$
$\bar{\nu} = (1.097 \times 10^{7} \ m^{-1}) \left( \frac{1}{4} - \frac{1}{9} \right)$
$\bar{\nu} = (1.097 \times 10^{7} \ m^{-1}) \left( \frac{9 - 4}{36} \right)$
$\bar{\nu} = (1.097 \times 10^{7} \ m^{-1}) \left( \frac{5}{36} \right)$
$\bar{\nu} \approx 1.5236 \times 10^{6} \ m^{-1}$