The total energy of an electron in the first excited state of the hydr – InterviewSolution

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1.	The total energy of an electron in the first excited state of the hydrogen atom is about -3.4 eV. What is the kinetic energy of the electron in this state ?
Answer» We know kinetic energy of electron `=(Kze^(2))/(2r)` and P.E of electron `=(-Kze^(2))/(r)` P.E.=-2 (kinetic energy) In this calculation electric potential and hence potential energy is zero at infinity. Total energy = PE + KE = -2KE + KE = -KE In the first excited state total energy= -3.4 eV. `therefore K.E=-(-3.4 eV)=+3.4 eV.`

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