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
