In their lowest-energy configuration, the electrons in a hydrogen molecule occupy an energy level of n = 1, where n refers to the electrons’ principal quantum number. When a hydrogen molecule absorbs energy, however, either in the form of heat or light, the electrons transfer to higher energy levels. The electrons cannot sustain this “excited” state, however, and eventually relax back to a lower level. Under the law of conservation of energy, the extra energy the electrons absorbed cannot simply disappear when the electrons relax to a lower energy level. The energy differences between the levels determines the energy of the radiation emitted during the relaxation process. Four transitions, in particular, emit radiation in the visible portion of the electromagnetic spectrum, the wavelength range of which spans approximately 400 to 700 nanometers. Scientists refer to these four transitions — all of which involve a transition to the n = 2 state — as the Balmer series, and their wavelengths are calculated from the equation (1 / lambda) = R * (1/4 – 1/n^2), where lambda represents the wavelength of the light emitted during the transition, R represents the Rydberg constant, or 1.097 x 10^7 reciprocal meters, or 1/m, and n represents the quantum number of the orbital from which the excited electron relaxed.
1. Calculate the term (1/4 – 1/n^2) for n = 3. In this case, 3^2 = 9 and the term becomes (1/4 – 1/9), or (0.250 – 0.111) = 0.139.
2. Multiply the result from Step 1 by 1.097 x 10^7. For n = 3, 0.139 * (1.097 x 10^7) = 1.52 x 10^6 1/m.
3. Calculate the wavelength, lambda, by taking the reciprocal of the result from Step 2. Continuing the previous example, lambda = 1 / (1.52 x 10^6) = 6.56 x 10^-7 meters.
4. Convert the result from Step 3 from meters to nanometers by multiplying by 1 x 10^9. In this case, (6.56 x 10^-7) * (1 x 10^9) = 656 nanometers, or 656 nm.
5. Repeat Steps 1 through 4 for n = 4, n = 5 and n = 6. Doing so should result in calculated wavelengths of 486, 434 and 410 nm, respectively.