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The energy of a hydrogen atom in the gro...

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 be

A

`-13.6 eV `

B

`-27.2 eV `

C

`-54.4eV`

D

`-6.8 eV`

Text Solution

Verified by Experts

The correct Answer is:
A
`E_(n)=-13.6((Z^(2))/(n^(2)))=(-13.6)((4)/(4))=-13.6ev`
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