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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 KE and PE of electron in the ground state.

A

3.4 eV, 6.8 eV

B

`-6.8 eV, 3.4 eV`

C

`+3.4 eV, (-6.8 eV)`

D

`-3.4 eV, (+6.8 eV)`

Text Solution

Verified by Experts

The correct Answer is:
C
`K.E. = -E = + 3.4 eV, P.E. = -2xxK.E. =-6.8 eV`
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MARVEL PUBLICATION-ATOMS, MOLECULES AND NUCLEI -TEST YOUR GRASP
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