Question

In a hydrogen atom, the energy of the first excited state is - 3.4 eV. Find out the K.E. of the same orbit of the hydrogen atom.

Answer 1

The ground state energy of hydrogen atom is -13.6 eV.
Ground state energy of hydrogen atom, E = -13.6 eV
This is the total energy of a hydrogen atom. Kinetic energy is equal to the negative of the total energy.
Kinetic energy $=-E=-(-13.6)=13.6eV$
Potential energy is equal to the negative of two times of kinetic energy.
Potential energy $=-2\times \left(13.6\right)=-27.2eV$.