Non-relativistic model of helium atom In the helium atom the two electrons in the 1s orbit are stable with positive binding energy. They have both kinetic energy due to its motion in the circular orbit and potential energy due its position in the electrostatic field within the atom. The kinetic energy of an electron is (1/2) mv2 and for both the electrons which are identical in the system K.E is mv2. The electron is stable in the orbit by balancing out the resultant electrostatic force with the centrifugal force due to circular motion. Let us suppose that the two electrons are in the 1s orbit itself having radius r1.The centripetal force acting on an orbital electron due to nuclear attraction is 2e2 /Kr12 . The other electron in the same orbit is repelled by the electron already existing there and both of them are positioned at opposite end of a diameter of the orbit. If electron-electron interaction is not allowed in the same orbit, the relative position of these two electrons may vary from time to time. Then different helium atom with different placement of electrons in the orbit may exist .Since all helium atoms have identical spectral feature the electrons in an orbits take fixed relative position that is invariant with respect to time. The electrostatic force of repulsion between the two electrons which are diametrically opposite is e^2/4Kr1^2. The resultant force is balanced by centrifugal force i.e., 2e^2/Kr1^2 - e^2/ 4Kr1^2 = (7/4)e^2/Kr1^2 = mv1^2/r1 . The radius of the orbital electron in helium atom can be determined from the condition on its quantized angular momentum. It gives rn = n^2(4/7) ao. = 0.5714 n^2 ao. If we ignore the electron-electron interaction rn = ao/2 .The radius of the 1s orbit in He+ ion is increased by 0.0714 ao due to mutual electron-electron interaction in helium atom. Total kinetic energy of the electrons in the helium atom = mv1^2 = (7/4) e^2/Kr1 kinetic energy per electron = (1/2) mv1^2 = (7/8)e^2/Kr1 = (7/4) e^2/2Kr1 Potential energy of electron 1 = - 2 e^2/Kr1 Potential energy of electron 2 in the presence of electron 1 = - 2 e2 /Kr1 + e2 /2Kr1 Total potential energy of the electrons in the helium atom = - (7/2) e2/Kr1 Sum of K.E and P.E is - (7/4)e^2/Kr1 = - (49/8)[e^2/2Kao]= -6.125 x 13.595 = -83.269 eV which gives enough binding energy to the system. It gives the first ionization energy 83.269 - 54.368 = 28.901 eV. The experimental value of first ionization energy of helium is 24.481 eV which is 4.41 eV smaller than the theoretically predicted value.