Practice portal › Atoms › Hydrogen Spectrum and Transitions
Asked in RS Academy GUJCET booklet · Spectral series and wavelengths
Idea: infrared photons carry less energy than ultraviolet ones, so the transition we want must release less energy than 4→3.
For a hydrogen-like atom the energy released is Δ E=13.6Z²(1/(n_f²)-1/(nᵢ²)) eV, so compare the brackets.
4→3: 1/9-1/(16)=7/(144)≈0.049.
2→1: 0.75; 3→2: 0.139; 4→2: 0.188 — all larger than 0.049, so all ultraviolet.
5→4: 1/(16)-1/(25)=9/(400)=0.0225, less than half the energy of 4→3.
Only 5→4 releases less energy than the ultraviolet line, so it is the transition that can give infrared light.
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