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.

Energy of electron in first excited state in Hydrogen atom is -3.4eV.

1.	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.
Answer» SOLUTION :`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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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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