Question

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 and potential energy of the electron in this state?

Answer 1

the first exited state of hydrogen atom is at energy -3.4 eV.

naturally this energy value is total energy.

E = T + V where T and V ar kinetic and potential energy of the system.

This energy is the eigen value of the Hamiltonian operator in ground state.

Therefore T = E -V ; as V = -2.T

will be E = -3.4+ 2.T therefore T = +(3.4 eV)

therefore the K.E. = 3.4 eV and potential energy will be - -6.8 eV

This may be you can't understand due to lack of knowloge in Hamiltonian operator And Schrödinger's equation

(electrons in orbitals closer to the nucleus hare higher kinetic energy.)