InterviewSolution
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InterviewSolution
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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. |
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Answer» Maximum energy is liberated for transition `E_(n)` to `E_(n-1)` and Maximum energy for `E_(n) to E_(n-1)` Hence , `E^(1)/n^(2)-E=52.224eV`…(1) and `E_(1)/n^(2) - E_(1)/(n-1)^(1) = 1.224eV`.....(2) Solving above equation simultaneously, we `E_(1)=- 54.4eV` and `n=5` Now `E_(1)=-(13.6Z^(2))/1^(2)=-54.4eV n=5` energy state . |
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