1.

The energy of a hydrogen atom in the ground state is -13.6 eV. The energy of a He+ ion in the first excited state will be .....

Answer»

`-13.6eV`
`-27.2eV`
`-54.4eV`
`-6.8eV`

Solution :The ENERGY of a hydrogen like atom in nth orbit is given by
`E_(n)=-Z^(2)xx(13.6)/(n^(2))eV`
For first EXCITED STATE of `He^(+),n=2,Z=2`
`:. E_(He^(+))=-(4)/(2^(2))xx13.6=-13.6eV`


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