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

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.`