Q:  Compute the energy of a photon emitted when a CO molecule
      drops from J = 1 to J = 0.


            E(J = 1)  =  0.00023825 eV

            E(J = 0)  =  0 eV

        So the energy of the photon is

                  0.00023825 eV



  Q:  What is the wavelength of this photon?

                         1240 eV-nm
             lambda  =  --------------  =  2.601 x 10^(6) nm
                        0.00023825 eV


                                        =  2.601 mm