Ionization Energy of Hydrogen and hydrogen-like ions                                                                        Hydrogen atom is a simple system having only nucleus-electron interaction where the question of electron-electron interaction and its interference with the system do not arise.The orbits of atomic electron cannot be arbitrary but specific due to the quantum condition the circumference of all allowed orbits contain an integral number of wave length of waves λ associated with the moving electron having momentum mv called de Broglie wave lenth λ = 68 h/mv 2π rn= n λ = nh/mv. The dynamic stability of the electron states that Ze^2 /4πεorn^ 2 = mvn^2/ rn  Solving for rn rn = n^2 h^2 εo/Z mπ e^2 = n^2 a0 /Z  where ao is the radius of the innermost orbit n = 1 called Bohr radius. The permitted electronic orbits in the hydrogen atom (Z = 1) have radii rn = n2 ao The orbital electron has kinetic energy by virtue of its circular motion and is equal to Ze^2/8πεornand potential energy by virtue of its position in the nuclear field and is equal to -Ze^,2/4 πεorn The total energy Enof the orbiting electron is the sum of its kinetic and potential energies and is equal to -Z e^2 / 8 πεorn. Substituting the value for rn En = -(Z^2 /n^2) [e^2/8πεoao] = -13.595 (Z^2 /n^2 ) eV ... (3.4) When n = 1 and Z= 1 (for hydrogen) the total energy of the orbital enectron is -13.595 eV. This is the energy required to pull out the electron from the hydrogen atom in its ground state and is called its ionization energy I H = 13.595 eV. When the hydrogen atom is excited and the orbital electron is in its n th orbit, then the required ionization energy is dropped IH* = 13.595/n^2 eV The equation (4) can be used to find out the Z th ionization energy of hydrogen like ions. For He+ , the second ionization energy is 2^2 x 13.595 = 54.28 eV, Like wise the third ionization of lithium is 3^2x 13.595 = 122.36 eV , and the fourth ionization energy of Beryllium is 4^2x 13.595 = 217.52 eV.On generalization it gives a formula for the Zth ionization energy of an element having atomic number Z is Z^2 x 13.595 eV.Using this relation one can determine the first ionization energy of hydrogen atom and Z th ionization energy of hydrogen-like ions. 

Table:  . Ionization energy of hydrogen atom and hydrogen-like ions                                                           -----------------------------------------------------------------------                                                                  Z Symbol Z2 (13.595) observed value .....................eV..........................                                                        -----------------------------------------------------------------------                                                                  1 H 13.59 13.59                                                                                                                                            2 He 54.38 54.40                                                                                                                                        3 Li 122.36 122.42                                                                                                                                     4 Be 217.52 217.66                                                                                                                                     5 B 339.87 340.13                                                                                                                                     6 C 489.42 489.84                                                                                                                                        7 N 666.16 666.83                                                                                                                                    8 O 870.10 871.12 69                                                                                                                                 9 F 1101.20 1103.12                                                                                                                                 10 Ne 1359.50 1362.20                                                                                                                            11 Na 1645.00 1648.70                                                                                                                             12 Mg 1957.68 1962.66                                                                                                                               13 Al 2297.56 2304.14                                                                                                                               ----------------------------------------------------------------------------------                                                                                           The deviation from I(Z-1) = Z2 [e^2 /2Kao] is well noticible as Z increases . This may be due to the change in the radius of the electronic orbit by the bulkyness of the nucleus. Besides the nuclear charge ,the size of the nucleus also has some influence in determining the orbits of the electrons.For example, the 1s orbit in hydrogen has radius ao , the Bohr radius, the 1s orbit in uranium has radius ao/92. When the number of nucleons increases, the size of the nucleus is enlarged.When the radius of the oribit of the electron increases due to bulkyness of the nucleus, its orbital velocity decreases , which results in the reduction of kinetic energy and addition of potential energy. Consequently the ionization energy is decreased. It gives an account why the first orbit of hydrogen unlike helium does not contain two electrons. The 1s orbit provides additional binding when it is completely filled with 2 electrons That is why the helium is more stable . But the hydrogen H- ions with two electrons in its 1s orbit is unstable. If one more electron is introduced in the first orbit of hydrogen , the total energy which is responsible for its binding with the nucleus becomes negative or equal to zero. Due to the presence of another electron in the close proximity , the electron -electron interaction is inevitable, The fact that two or more orbital electrons in any atomic orbits cannot be placed arbitrarily implies that there must be mutual interaction between the orbital electons. The two electrons in the 1s orbit must be diametrically opposite to each other.The electron -electron interaction reduces the nuclear force on the electron. Consequently the inermost orbit gets enlarged little , which reduces the velocity of the electron , The resultant force acting on the orbital electron is e^2/4πεo r^2 - e^2/4(4πεo)r^2 = (3/4) e^2/4πεo r^2 . As it is counterbalanced by the centrifugal force mv2 /r , the kinetic energy of both the electrons in the system becomes mv^2 = (3/4) e^2 /4πεo r. The potential energy of the first electron in the innermost orbit is -e2/4πεo r . The second electron is brought to the same orbit without doing any work or with negligible work. The work done when it is placed diametrically opposite to the first electron is e2 / [2(4πεo)r] so that the net potential energy of the electron becomes -e^2/[2(4πεo)r]. Since the total energy of the system is (1/4) e2/[(4πεo)r] which makes the binding energy to be posititive. The 1s orbit can accommodate a maximum of 2 electrons. But in hydrogen 1s orbit cannot have more than 1 electron.It can be filled with 2 electrons only when the nuclear charge is numerically equal to or greater than the sum of the electronic charges of the orbital electrons. Two electrons caanot occupy the 1s orbit of the hydrogen atom because of the Pauli's exclusion principle. According to this principle. 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 orbital (1s) and it can accomodate two electrons which musthave opposite spins (spin up and spin down).In H- ion , the kinetic energy associated with both the electrons K.E = 3e^2/4Kr1 , potential energy od the first electron = - e^2/Kr1. The second electron is brought in field free space and makes no contribution to potential energy. The energy due to electron-electron interaction is e^2/2Kr1. Total energy of the system is 3e^2/4Kr1 -e2/Kr1 + e^2/2Kr1= e^2/4Kr1Since the total energy is positive, it means there is no binding at all. Hence H-is theoretically possible only