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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. |
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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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