An electron in an excited state of a hydrogen atom emits two photons in succession, the first at 1817 nm and the second at 97.20 nm, to return to the ground state (n=1). For a given transition, the wavelength of the emitted photon corresponds to the difference in energy between the two energy levels.

The energy emitted or required for electron de-excitation / excitation is given by Planck's equation:

E = hv = hc/λ; where h is the Planck's constant, v is the frequency of the emitted wave, c is the speed of light in a vacuum, λ is the wavelength of the emitted light.

To calculate the wavelength of photons emitted upon return to the ground state directly from the second excited state, we must first calculate the energies of each transition. These work out to be:

E (3 to 2) = [ (6.63 * 10 ^-34) * (3 * 10^8) ] / (1817 * 10^-9) = 1.09 * 10^-19 J

E (2 to 1) = [ (6.63 * 10 ^-34) * (3 * 10^8) ] / (97.2 * 10^-9) = 20.5 * 10^-19 J

E (3 to 2) + E (2 to 1) = 21.59 * 10^-19 J

Calculating the wavelength of the state 3 to 1 jump:

λ = [ (6.63 * 10 ^-34) * (3 * 10^8) ] / 21.59 * 10^-19

λ = 92.1 nm