The total energy of an electron in the first excited state of the hydrogen atom is about –3.4 eV.
(a) What is the kinetic energy of the electron in this state?
(b) What is the potential energy of the electron in this state?
(c) Which of the answers above would change if the choice of the zero of potential energy is changed?

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.