Energy of an electron in ground state of hydrogen atom is `-13.6eV`. Calculate its K.E. and P.E. in the same state.

Energy of the electron in the ground state of hydrogen atom is (-13.6eV) . Calculate the energy in 3^(rd) and 4^(th) orbit.

The energy of an electron in an excited state of hydrogen atom is -1.51 eV . Calculate the angular momentum of the electron in this state. h = 6.63 xx 10^-34 Js

The energy of the electron in the ground state of hydrogen atom is -13.6 eV . Find the Rydberg's constant and the wavelength of the first line of Paschen series

The energy in the ground state of hydrogen atom is -13.6 eV , then energy of 1st excited state is

The total energy of an electron in the first excited state of the hydrogen atom is about -3.4eV . What is the potential energy of the electron in this state?

Energy in the ground state of normal H-atom is -13.6 eV .Calculate its kinetic energy and potential energy.

The total energy of an electron in the first excited state of the hydrogen atom is about -3.4eV . What is the kinetic energy of the electron in this state?

The total energy of an electron in the first excited state of the hydrogen atom is about -3.4eV . Which of the answers above would change if the choice of the zero of potential energy is changed?

Find the velocity of the electron in the ground state of hydrogen atom.Hence find the velocities when the electron is in the second and fourth orbit.(Radius of the first Bohr orbit = 0.53xx10^-10m , mass of electron= 9.1xx10^-31 kg )

The ground state energy of hydrogen atom is -X eV . What will be the kinetic energy of the electron in this state?