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Energy of electron in first excited stat...

Energy of electron in first excited state in Hydrogen atom is `-3.4eV`. Find K.E. and P.E. of electron in the ground state.

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`because` Energy of electron in frist excited state in hydrogen atom, `E=-3.4eV`
`therefore" K.E. of electron K"=-E=+3.4eV`
`"and P.E. of electron U"=2E=2xx(-3.4)=-6.8eV`
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