The reason behind that a gas of such hypothetical atoms emits a continuous spectrum rather than the observed line spectrum.

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Chapter 29, Problem 1CQ

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The reason behind that a gas of such hypothetical atoms emits a continuous spectrum rather than the observed line spectrum.

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The light observed that is emitted by a hydrogen atom is explained by a simple model of its structure with one proton in its nucleus and an electron bound to it, but only with internal energies of the atom satisfying EH=−RH/n2EH=−RH/n2 where RHRH is the Rydberg constant and nn is an integer such as 1, 2, 3 ... and so on. When a hydrogen atom in an excited state emits light, the photon carries away energy and the atom goes into a lower energy state. Be careful about units. The Rydberg constant in eV is 13.605693009 eV That would be multiplied by the charge on the electron 1.602× 10-19 C to give 2.18× 10-18 J A photon with this energy would have a frequency f such that E=hf. Its wavelength would be λ = c/f = hc/E. Sometimes it is handy to measure the Rydberg constant in units of 1/length for this reason. You may see it given as 109737 cm-1 if you search the web, so be aware that's not joules. The following questions are intended to help you understand the connection between…

Q) A hydrogen atom emits radiation as a result of an electron transition to a lower energy level. Determine the highest frequency possible due to this transition if the atom emits a series of lines that lie in the visible part of the spectrum. Then, if the electron ends up in n = 1 level, prove that the atom emits a series of lines of wavelength that are not in the visible part of the spectrum.