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A singly excited gas of hydrogen-like ions has a ground-state…
Question
A singly excited gas of hydrogen-like ions has a ground-state ionization energy of 122.4 eV. (Take the hydrogen ground-state binding energy as 13.6 eV.) For the same ion, find the energy in eV of the photon emitted in the transition from n = 3 to n = 1.
✓ Verified answer: 108.8checked by our engine — not a guess
Step-by-step solution
For a hydrogen-like ion the ground-state ionization energy is 13.6*Z^2.
Given 13.6*Z^2 = 122.4, so Z^2 = 9, Z = 3 (Li2+).
The energy levels are E_n = -13.6*Z^2/n^2.
The n=3 -> n=1 photon energy is 13.6*Z^2*(1/1 - 1/9) = 122.4*(8/9) = 108.8 eV.
Final answer108.8
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