1.

The total energy of an electron in second excited state of hydrogen atom is -1.51eV. Calculate (i) KE of electron (ii) PE of electron in this state.

Answer» As is known,
P.E. of electron`=-2KE` of electron.
Total energy `=PE+KE=-2KE+KE=-KE`
`-1.51=-KE :. Ke=1.51eV`
and `PE=-2KE=-2xx1.51eV=-3.02eV`


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