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The ground state energy of hydrogen atom...

The ground state energy of hydrogen atom is `-13.6 eV`. The energy of second excited state of `He^(+)` ion in eV is

A

`-27.2`

B

`-3.4`

C

`-54.4`

D

`-6.04`

Text Solution

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The correct Answer is:
D
Second excited state is n = 3,`" "E=-13.6xx(z^(2))/(n^(2))eV=-13.6xx((2)^(2))/((3)^(2))=-13.6xx(4)/(9)=-6.04eV`
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