Question

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

Hint:

As the transition moves to higher \( n_2 \) values, the wavelengths of spectral lines in the Balmer series decrease.

Answer 1

The wavelength \( \lambda \) of a spectral line in the hydrogen atom can be determined using the Rydberg formula: \[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] where \( R_H = 1.097 \times 10^7 \, {m}^{-1} \) is the Rydberg constant, and \( n_1 \) and \( n_2 \) are the principal quantum numbers, with \( n_1<n_2 \). For the first line in the Balmer series (\( n_1 = 2, n_2 = 3 \)): \[ \frac{1}{\lambda_1} = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R_H \left( \frac{1}{4} - \frac{1}{9} \right) \] \[ \lambda_1 = \frac{1}{R_H \left( \frac{5}{36} \right)} = 6.56 \times 10^{-7} \, {m} = 656 \, {nm} \] For the second line (\( n_1 = 2, n_2 = 4 \)): \[ \frac{1}{\lambda_2} = R_H \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R_H \left( \frac{1}{4} - \frac{1}{16} \right) \] \[ \lambda_2 = \frac{1}{R_H \left( \frac{3}{16} \right)} = 4.86 \times 10^{-7} \, {m} = 486 \, {nm} \]