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The energy of a hydrogen atom in the ground state is `-13.6 eV`. The eneergy of a `He^(+)` ion in the first excited state will beA. `-13.6eV`B. `-27.2 eV`C. `-54.4eV`D. `-6.8 eV` |
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Answer» Correct Answer - A `E_(He^(+))=-(Z^(2))/(n^(2))xx13.6eV` for `He^(+)` ion, Z=2, and for first excited state, n=3. `:. E_(He^(+))=-(2^(2))/(2^(2))xx13.6eV=-13.6 eV` |
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