Question

(a) Find the photon energy and wavelength for the series limit (shortest wavelength) in the Balmer series (n_f = 2) for the hydrogen atom.
___________eV
_________ nm

(b) Calculate the wavelength for the three longest wavelengths in this series.
longest
____________nm
second longest
____________nm
third longest
____________nm

Answer 1

(a)We know that,

E = -13.6 / n² eV

For n = 2

E = -13.6/4 = -3.4 eV

corresponding wavelength = = 1240 eV nm / E = 1240 eV nm / 3.4 = 364.71 nm

Hence, E = -3.4 eV and = 364.71 nm.

(b) we have Ef = E2 = -3.4 eV

E = E3 - E2 = -13.6/9 eV- (-3.4) eV = 1.89 eV

1 = 1240 eV nm / 1.89 = 656.1 nm

E = E4 - E2 = -13.6/16 eV- (-3.4) eV = 2.55 eV

2= 1240 eV nm / 2.55 = 486.3 nm

E = E5 - E2 = -13.6/25 eV- (-3.4) eV = 2.856 eV

3= 1240 eV nm / 2.856 = 434.2 nm

Hence, Longest = 1=  651.1 nm ; second longest = 2 = 486.3 nm and third longest = 3 = 434.2 nm.