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The total energy of an electron in the first excited state of hydrogen atom is about -3.4 eV. Its kinetic energy in this state is
Options
(a) 3.4 eV
(b) 6.8 eV
(c) -3.4 eV
(d) -6.8 eV
Correct Answer:
3.4 eV
Explanation:
KE. = |(1/2) P.E.|
But P.E. is negavite
.·. Total energy = |(1/2) P.E.| – P.E. = – P.E. / 2 = – 3.4 eV
.·. K.E. = + 3.4 eV
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