Question

Calculate the wave number corresponding to the most and least energetic spectral lines in the Lyman, Balmer and Paschen series for the hydrogen atom.

Answer 1

The wave number is calculated using the formula :

Here, R = Rydberg constant = 1.097373 × 10⁷ m⁻¹.

For the Lyman series,

n_f = 1

For the least energetic spectral line in this series, n_i = 2

Putting values in the formula and calculating, we get :

v = 8.22*10^6 m^-1

For the most energetic spectral line in this series, n_i = infinity

Putting values in the formula and calculating, we get :

v = 10.96*10^6 m^-1

For the Balmer series,

n_f = 2

For the least energetic spectral line in this series, n_i = 3

Putting values in the formula and calculating, we get :

v = 1.52*10^6 m^-1

For the most energetic spectral line in this series, n_i = infinity

Putting values in the formula and calculating, we get :

v = 2.74*10^6 m^-1

For the Paschen series,

n_f = 3

For the least energetic spectral line in this series, n_i = 4

Putting values in the formula and calculating, we get :

v = 0.53*10^6 m^-1

For the most energetic spectral line in this series, n_i = infinity

Putting values in the formula and calculating, we get :

v = 1.21*10^6 m^-1