Calculate the wavelength of the first two lines in Balmer series of hydrogen atom.

Answer: \(H_\alpha\): \(\lambda \approx 6563\) Å \((6.563\times10^{-7}\) m); \(H_\beta\): \(\lambda \approx 4862\) Å \((4.862\times10^{-7}\) m)

For the Balmer series, \(\dfrac{1}{\lambda} = R_H\left(\dfrac{1}{2^2} - \dfrac{1}{n^2}\right)\), \(n = 3, 4, 5, \ldots\), with \(R_H = 1.097\times10^7\) m\(^{-1}\).

First line (n = 3 → 2): \(\dfrac{1}{\lambda_1} = R_H\left(\dfrac{1}{4} - \dfrac{1}{9}\right) = \dfrac{5R_H}{36}\)

\(\lambda_1 = \dfrac{36}{5\times1.097\times10^7} = \dfrac{36}{5.485\times10^7} = 6.563\times10^{-7}\) m \(= 6563\) Å.

Second line (n = 4 → 2): \(\dfrac{1}{\lambda_2} = R_H\left(\dfrac{1}{4} - \dfrac{1}{16}\right) = \dfrac{3R_H}{16}\)

\(\lambda_2 = \dfrac{16}{3\times1.097\times10^7} = \dfrac{16}{3.291\times10^7} = 4.862\times10^{-7}\) m \(= 4862\) Å.