1.

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`

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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