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

Explanation

In Bohr.s model
`mvr=(nh)/(2pi)` or `mv=(nh)/(2pir)`
where m is the mass of electron
v is the velocity of electron in `n^(th)` energy level
r is the radius of `n^(th)` level orbit
Also, velocity of electron in `n^(th)` energy level is
`v=(Ze^2)/(2epsilon_0nh)`
So, kinetic energy is given by
`K=1/2mv^2=(mZ^2e^4)/(8epsilon_0 h^2 n^2)`
The above relations are independent of the choice of the zero of potential energy. Now , taking the zero of potential energy at infinity the potential energy of electron is given by
`U=-(Ze^2)/(4piepsilon_0r)` which gives
U=-2K
and total energy is given by E=K+U
(a) The given value of E = -3.4 eV is based on the choice of zero of potential energy at infinity. Using E=-K, the kinetic energy of the electron in the state is +3.4 eV.
(b)Using U = - 2 K, the potential energy of the electron is U =-6.8 eV.
(c) If the zero of potential energy is taken at some other location, the kinetic energy does not change. Its value remains +3.4 eV, being independent of the choice of the zero of potential energy. The potential energy and the total energy will change.