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

wave number = R_H [1/n1^2 - 1 / n2^2]

                      = 1.0973 x 10^7 [1/n1^2 - 1 / n2^2]

for lymen :

n1 = 1 , n2 = 2

n1 = 1 , n2 = infinity

wave number = 1.0973 x 10^7 [1/1 - 1 / 2^2] = 8.23 x 10^6 m-1             ---------------- least energy

wave number = 1.0973 x 10^7 [1/1 - 1 / infinity] = 1.0973 x 10^7 m-1 --------------- most energy

Balmer :

n1 = 2 , n2 = 3

n1 = 2 , n2 = infinity

wave number = 1.0973 x 10^7 [1/2^2 - 1 / 3^2] = 1.524 x 10^6 m-1    --------------- least energy

wave number = 1.0973 x 10^7 [1/2^2 - 1 / infinity] = 2.743 x 10^6 m-1 --------------- most energy

Paschen series :

n1 = 3 , n2 = 4

n1 = 3 , n2 = infinity

wave number = 1.0973 x 10^7 [1/3^2 - 1 / 4^2] = 5.334 x 10^6 m-1       --------------- most energy

wave number = 1.0973 x 10^7 [1/3^2 - 1 / infinity] = 1.219 x 10^6 m-1 --------------- least energy