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The energy of an electron in the first B...

The energy of an electron in the first Bohr orbit of H atom is `-13.6 eV` The potential energy value (s) of exxcited state(s) for the electron in the Bohr orbit of hydrogen is//are

A

`-3.4 eV`

B

`-4.2 eV`

C

`-6.8 eV`

D

`+6.8 eV`

Text Solution

Verified by Experts

The correct Answer is:
A

`E_(n) = (-13.6)/(n^(2))Z^(2)eV`
For `n = 2`, for H atom,
`E = (-13.6)/(2^(2)) 1^(2) eV = - 3.4 eV`
Other value cannot be obtained for `n = 3,4,5,6` etc
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