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InterviewSolution
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The groud state energy of hydrogen atom is `-13.6 eV`. When its electron is in first excited state, its exciation energy isA. 3.4 eVB. 6.8 eVC. 10.2 eVD. zero |
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Answer» Correct Answer - C Given ground state energy of hydrogen atom `E_(1)=-13.6eV` Energy of electron in first excited state (ie, n=2) `E_(2)=-(13.6)/((2)^(2))eV` Therefore, excitation energy `DeltaE=E_(2)-E_(1)` `=-(13.6)/(4)-(13.6)` `=-3.4+13.6=10.2eV` |
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