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 ?

Explanation

In Bohr.s model , `mvr = (nh)/(2pi) and (mv^2)/r = (Ze^2)/(4 pi epsilon_0 r^2)`
Which give
`T = 1/2 mv^2 = (Ze^2)/(8 pi epsilon_0 r) , r = (4 pi e_0 h^2)/(Z e^2 m) n^2`
These relations have nothing to do with the choice of the zero of potential energy. Now, choosing the zero of potential energy at infinity we have `V = – (Z e^2/4 pi epsilon_0 r)` which gives `V = –2T and E = T + V = – T`
(a) The quoted value of E = – 3.4 eV is based on the customary choice of zero of potential energy at infinity. Using E = – T, the kinetic energy of the electron in this state is + 3.4 eV .
(b) Using V = – 2T, potential energy of the electron is = – -6.8 eV
(c) If the zero of potential energy is chosen differently, kinetic energy does not change. Its value is + 3.4 eV independent of the choice of the zero of potential energy. The potential energy, and the total energy of the state, however, would alter if a different zero of the potential nergy is chosen.