In hydrogen atom, energy of first excited state is `- 3.4 eV`. Then, `KE` of the same orbit of hydrogen atom is.

In hydrogen atom, energy of first excited state is-3.4eV. The K.E. of the same orbit of H-atom is

In hydrogen atom, energy of second state is-3.4 eV. The, KE of electron in same orbit of hydrogen atom is

In hydrogen atom, energy of first excited state is -3.4 eV. Then find out the K.E. of the electron in the same orbit of hydrogen atom.

If the energy in the first excited state in hydrogen atom is 23.8 eV then the potential energy of a hydrogen atom in the ground state can be assumed to be

If the energy in the first excited state in hydrogen atom is 23.8 eV then the potential energy of a hydrogen atom in the ground state can be assumed to be

Show that the energy of first excited state of He^(+) atom is equal to the energy of electron in the first orbit of hydrogen atom.