Skip to main content
Chemistry LibreTexts

Solutions 11

  • Page ID
  • Q11.1

    Calculate the wavelength of the highest energy Balmer emission line of hydrogen.

    The equation for the emission wavelength of hydrogen is: 

    \[\frac{1}{\lambda} = R_H(\frac{1}{n_f}^2 - \frac{1}{n_i}^2)\]

    where \(R_H = 109737.32  cm^{-1}\), \(n_f = 2\) and, \(n_i = infinity\). 

    \[\frac{1}{\lambda} = 109737.32 cm^{-1}(.25 - 0)\]

    \[\lambda = 3.65x10^{-5} cm = 365 nm\]


    Calculate the wavelength of an ultraviolet transition in the Lyman emission series of Hydrogen gas from the n = 2 level and to the n= 1 level.

    \[\frac{1}{\lambda} = R_H(\frac{1}{n_f}^2 - \frac{1}{n_i}^2)\]

    \[\frac{1}{\lambda} = 109737.32 cm^{-1}(1 - .25)\]

    \[\lambda = 1.22x10^{-5} cm = 122 nm\]


    Calculate the energy of a photon absorbed for an electron to jump from a n=3 and n=4 state of atomic hydrogen? Calculate the energy of a photon emitted for an electron to jump from a n=4 and n=3 state of atomic hydrogen?

    a) \[\frac{1}{\lambda} = R_H(\frac{1}{n_f}^2 - \frac{1}{n_i}^2)\]

    \[\frac{1}{\lambda} = 109737.32cm^{-1}(\frac{1}{9} - \frac{1}{16})\]

    \[\frac{1}{\lambda} = 5334.45cm^{-1}\]

    \[\lambda = 1.875x10^{-6}m\]

    \[E = \frac{hc}{\lambda} = 1.06x10^{-19} or .66 eV\]

    b) Same answer as a.



    Calculate the wavenumber of the wavelength of the light emitted from the \(n=8\) to \(n=6\) transition.

    \[\nu = \frac{1}{\lambda} = R_H(\frac{1}{n_f}^2 - \frac{1}{n_i}^2)\]

    \[\nu = \frac{1}{\lambda} = 109737.32cm^{-1}(\frac{1}{36} - \frac{1}{64}) = 1333.86 cm^{-1}\]



    Explain what the expected emission spectrum would be for atomic hydrogen if continuum mechanics (i.e., classical mechanics) were applicable for describing it.

    The emission spectrum of atomic hydrogen without quantization would be a continuous broadband spectrum.