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Quantum Numbers and the use of the Rydberg equation to find a value for n

Submitted by breaghaam on Sun, 10/02/2011 - 20:34
An electron in an excited state of a hydrogen atom emits two photons in succession, the first at 3738 nm and the second at 94.92 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. What were the principal quantum numbers of the initial and intermediate excited states involved?

Two separate applications of the Rydberg equation. Apply this first to the second emission as n1 = 1 for this. You can calculate n2 for this emission. This value might be 2, 3 etc. Then whatever the value, this will be n1 in the second calculation and you can then find n2 for this emission. The value will be at least 3 or 4 or higher etc. gives the formula.