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In hydrogen atom, energy of first excited state is -3.4eV. Then find out KE of same orbit of hydrogen atom

In hydrogen atom, energy of first excited state is -3.4eV. Then find out KE of same orbit of hydrogen atom

Grade:12th pass

1 Answers

Vikas TU
14149 Points
7 years ago
that means Potential Energy is -3.4 ev.
Now from closest distance approach,
PE/KE = -2
Therfore,
K.E = PE/-2 = > -3.4/-2 = 1.7 eV.

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