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Consider a hydrogen-like atom whose energy in nth excited state is given by `E_(n) = (13.6 Z^(2))/(n^(2))` When this excited makes a transition from excited state to ground state , most energetic photons have energy `E_(max) = 52.224 eV`. and least energetic photons have energy `E_(max) = 1.224 eV` Find the atomic number of atom and the intial state or excitation.A. 2B. 4C. 5D. None of these |
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Answer» Correct Answer - A Maximum energy is liberated for transition `E_(n)rarr1` and minimum energy for `E_(n)rarrE_(n-1)` Hence `(E_(1))/(n^(2))-E_(1)=52.224eV " " ....(i)` and `(E_(1))/(n^(2))-(E_(1))/((n-1))=1.224eV" " ....(ii)` Solving Eqs. (i) and (ii), we get `E_(1)=-54.4eV and n=5` But `E_(1)=-(13.6Z^(2))/(1^(2))` `:. -54.4=(13.)/(1^(2))Z^(2)rArrZ=2` |
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