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In hydrogen atom, energy of first excit...

In hydrogen atom, energy of first excited state is `- 3.4 eV`. Then, `KE` of the same orbit of hydrogen atom is.

A

`+3.4 eV`

B

`+6.8 eV`

C

`-13.6 eV`

D

`+13.6 eV`

Text Solution

Verified by Experts

The correct Answer is:
A
K.E. of an electron in nth orbit.
`= -E_(n) = (-3.4 eV) = 3.4 eV`
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