The first (1s) orbit can accommodate two electrons , but in the hydrogen atom its first orbit cann't be filled with two electrons . It is possible when one more proton is added within the nucleus (Helium atom).
Two electrons cannot occupy the first orbit (or the 1s orbital) of the hydrogen atom because of the Pauli Exclusion Principle. This principle states that no two electrons in the same atom can have the same set of all four quantum numbers. In the first orbit, there is only one possible orbital (1s), and it can only accommodate two electrons, which must have opposite spins (spin up and spin down)
If one more electron is allowed in the first orbit of hydrogen , its total energy becomes positive or zero and hence it moves away from the nucleus.
The Kinetic energy of the electron = (1/2) m v^2 and for both the electrons which are identical K.E = m v^2
Electrostatic force (attraction) due to the nucleus = e^2 / Kr^2 where K = 4πεo . Since both the electrons are in the same orbit ,there must be mutual interaction between them, that is why they attain stability by staying exactly diametrically opposite. Electrostatic force (repulsion) due to the second electron e^2 / 4Kr^2. The resultant force e^2 / Kr^2 - e^2 / 4Kr^2 = (3/4) e^2 / Kr^2 .It is counterbalanced by centrifugal force mv^2/r .K.E of both electrons = (3/4) e^2 / Kr .
The potential energy of the first electron due to the presence of the nucleus only = - e^2 / Kr and the potential energy due to the presence of the second electron in the same orbit + e^2 /2Kr (electron-electron system) .This is shared equally by both the electrons. The total energy associated with the system = (3/4) e^2 / Kr - e^2 / Kr + e^2 /2Kr = ( e^2 /4Kr) . The second electron feels the effective charge to be zero Since there is no binding energy , the system gets transformed into normal hydrogen atom by expelling out the second electron.
In the helium atom the two electrons in the 1s orbit are stable with positive binding energy. The Kinetic energy of an electron = (1/2) m v^2 and for both the electrons which are identical in the system K.E = m v2
K.E = 2 e^2 /Kr^2 - e^2/ 4Kr^2 = (7/4) e^2/Kr^2 = mv^2 /r
Total kinetic energy of the electrons = mv^2 = (7/4) e^2/Kr
Potential energy of elecron 1 = - 2 e^2 /Kr + e^2 / 2Kr ( shared by both the electrons)
potential energy of electron 2 = - 2 e^2 /Kr
Total potential energy of the electron-electron system = - (7/2) e^2 /Kr
Sum of K.E and P.E = - (7/4) e^2/Kr
which gives enough binding energy to the system.
