Ionization energy of  Helium -like atom/ ions

       Due to the presence of second electron in the 1s orbit, it gets enlarged by the additional repulsive force between  the electrons. In its innermost orbit it is equal to e²/4Kr₁². Let Ze be the nuclear charge in the helium like ions . Following Bohr's theory of hydrogen atom Ze²/Kr₁² e²/4Kr₁² = (4Z-1) [e²/4Kr₁²] = mv₁²/r₁ where r_n = 4a_o/4Z-1. i.e., the first orbit in helium is (4/7) times of 1s orbit in hydrogen. The kinetic, potential and total energy of the 1s electron are e²/4Kr₁(4Z-1), e²/2Kr₁(-4Z+1) and e²/4Kr₁(-4Z+1) = - [e²/2Ka_o][(4Z-1)²/8]

     In many electron system, the orbital motion of  an electron is slightly perturbed due to the presence of other electrons which are very close to each other. An orbiting electron feels a resultant centripetal force where electron -electron interaction is superimposed over nucleus electron interaction. It is accounted by screening constant .The hydrogen-like ions, the ionization energy is directly proportional to Z²  i.e., I _{hydrogen-like} = Z² x 13.595 eV. The helium-like ions ,the (Z-1)^th ionization energy may have a similar formula. If we assume I_{helium like} = (Z-k)² x 13.595 eV, where k is a constant.                                   

      The value of k is determined from the known I ^st  ionization energy of helium like ions.

                                   (2-k)² 13.595 = 24.481  gives k = 0.6581                

The first ionization energy of helium is 24.481 eV, which gives s = 0.658. Using this value of s,      (Z-1)^th ionization energy of helium-like ions can be estimated.  The second ionization energy of lithium is I₂ = (3-0.658)² x 13.595 = 74.567 .   The Table . gives the  calculated value of (Z-1) ^th ionization energy along with experimental value

Table. Ionization energy of helium-like ions

                        I = (Z-0.6581)²   13.595 eV

................ ...........................................................

  Helium -like   Z      I_Theory                     I_practical

             ions                      .......in eV........

...........................................................................

         Li⁺     3          74.56               75.62

         Be⁺⁺     4         151.83           153.85

        B⁺⁺⁺      5         256.29           259.30

       C⁴⁺         6         387.94           391.98

       N⁵⁺         7         546.79           551.92

         O⁶⁺          8         732.82           739.11

.....................................................................

 

      The total energy required to strip out both the electrons from the helium atom  is the sum of its first and second ionization energy  I_total  = I₁ + I_{2      =} [(Z-k)²  + Z² ] 13.595 eV

Table.3.5: Sum of first and second ionization energies of helium like -ions

...........................................................................................................................

Z           I_total  = (2Z²  + k² - 2Zk) (13.595)  =   Z th     +   (Z-1) th = Total

..............................................................................................................................

2                                    78.878                        54.40            24.5    =  78.90

3                                    196.91                        75.62          122.42 =  198.04

4                                    369.36                       158.85        217.66  =    376.51

5                                   596.18                        259.30        340.13   =  599.43

6                                   877.19                         391.98        489.84   = 881.82

................................................................................................................................

     Due to the presence of second electron in the 1s orbit ,  the resultant centripetal force is reduced. . Let Ze be the nuclear charge of helium like ions. The resultant centripetal force is the sum of nuclear attractive force and the repulsive electron-electron interaction.

                                          mv²/r = Ze²/ K r²  - e² / 4Kr²  

                                                                   m² v²  = (m e² /4Kr) (4Z-1)

The condition on the allowed orbits restricts its radius as they  contain only an integral number of wave-length of waves associated with the orbital electron. It gives all the permitted orbits with radius r_n =(4 n² h² ε_o)/ π m e² (4Z -1) = 4 a_o / (4Z-1). The 1s orbit of helium is 4/7 times of 1s orbit in hydrogen .

The kinetic energy of the two electrons in the helium atom = mv²  =  (e² /4Kr)  (4z-1)

The potential energy of the system =  (e² /2Kr)  (- 4Z +1) which gives the total energy as (e² /4Kr)( -4Z +1)  where r = 4a_o/ (4Z-1).

The total energy of the helium-like ions = -  (e² /2Ka_o)[(4Z -1)²/8] eV Using this relation the total energy with any helium like atom/ion can be determined.

Helium Z = 2   Total energy = - (49/8) 13.595 = -83.269 eV

Lithium Z =3  - (121/8) 13.595 = - 205.624 eV

Beryllium Z = 4  - (225/8) 13.595 = -382.36  = -153.85 - 217.65 = -371.5 eV

Boron  Z=5   - (361/8) 13.595 = -613.7 ; -259.3 - 340.1 = -599.4 eV

Carbon Z = 6, -(529/8) 13.595 = -899.3 ; -391.98 - 489.84 = -881.82

Nitrogen Z = 7; -(729/8) 13.595 = -1239.3 ; -551.92 - 666.53 = -1218.75

      The first ionization energy of helium can be determined by finding the difference in the total energy of the normal helium atom with two electrons in its 1s state and helium ion with single electron in the same orbit.

Total energy of normal helium atom = - (7/4)²[e²/Ka_o] = - 83.27 eV

In the helium ion He⁺    the radius of the 1s orbits gets changed due to the absence of second electron . Its radius r = a_o/ 2 . The sum of its kinetic energy  e² /K r and potential energy - 2e²/Kr gives the total energy associated with the electron  and is equal to - 4 [e² /2K a_o] = 4 x 13.595 = 54.4 eV. The first ionization energy of helium = 83.27 - 54.4 =    28 .87eV

The kinetic energy of the helium-like ions   (e² /4Kr) [4z-1]

Potential energy of the system  -(e²/2Kr) [4Z - 3]

Total energy of the system -(e² /4Kr) [4z -5]

 when Z = 1, the total energy becomes positive wich means there is no binding  and consequently the second electron in H^- move away from the nucleus to keep its potential energy minimum .

With the concept of screening constant the ionization energy of helium atom  and helium like ions can be estimated. The nuclear charge as seen by the orbital electrons is less due to the presence of the other electrons . This is the consequence of electron-electron interaction within the system .Let the effective charge of the nucleus as seen by the orbital electron is Z^*  = (Z - s) , where s is the screening constant .

           (Z-s) e² /Kr²  - e² / 4Kr²  = mv² /r

           me² /4Kr [ 4Z - 1 -4k] = m² v²

The condition that the orbit can contain only an integral number of waves associated with the electrons gives r = 4a_o/ (4Z-4k-1)

The kinetic energy of both the electrons = (e² /4Kr)[ 4Z-4k -1] 

The potential energy of the system = -2(Z-s k)/Kr + e² /2Kr = (e² /2Kr)[ -4Z + 4k +1]                                                                                                                                                                                    Total energy associated with the electrons is - (e² /4Kr)[ 4Z-4s -1]                                                        Substituting the value for r in terms of a_o  it becomes  - (e² /2Ka_o)[ 4Z-4k -1]² / 8]                           In the case of helium Z= 2 , I₁ + I₂  = 78.884 eV which gives the mean screening constant s = 0.0467 Since the I₂  for helium is  Z²  x 13.595 eV , I₁  = I_total - I₂  = {Z²- [4Z-4k -1]² / 8]} (13.595)=  8.884 - 54.38 = 24.504 eV

Using the relation the (Z-1)^th ionization energy of helium like ions can be estimated. 

  Table.  (Z-1) ^th Ionization energy of helium like ions

  ..................................................................................................................

   element                                             ionization energy in eV

                                                               calculated             practical

 .....................................................................................................................

lithium-3,  (13.595)[(10.8132 )² - 8x9]/8= 76.345               75.619

Beryllium-4 (13.595)[(14.8132)² - 8x16]/8= 155.375       153.85

Boron-5 (13.595)[(18.8132)² - 8x25]/8= 261.596             259.298

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Semi-empirical formula for the ionization energy of helium like ions

     For all hydrogen like ions, the ionization energy is directly proportional to Z²

                                                              I_{H like ions}  = Z² x 13.595 eV

For all helium like ions, the (Z-1)^th ionization energy is supposed to be directly proportional to (Z-k)²  where k is a constant. The value of k is first determined from the known value of first ionization energy of helium atom and second ionization energy of lithium atom.

                            (2-k)² = 4 -4k + k² = 24.48/13.595 = 1.8

                            (3-k)² = 9 -6k + k² = 75.62/13.595 = 5.562

Solving for k we get k = 0.6192. Using this value the (Z-1)^th ionization energy is estimated for helium like ions.

                                                  Be²⁺         155.354         153.85

                                                  B³⁺               260.91          259.30

                                                  C⁴⁺              393.53          391.98

                                                  N⁵⁺          553.52          551.92

                                                  O⁶⁺          740.64         739.11

To derive a semi-empirical formula for the ionization energy of helium like ions, let us assume the nuclear charge be Ze . When a single electron is present in the inner most orbit ,the radius of the orbit r₁= a_o/Z and the velocity v₁ = Z e²/2hε_o and total energy of He⁺ like ions = - Z² [e²/2Ka_o]. When two electrons are present in the innermost orbit of helium like ions, the radius of the orbit r₁₁ = [4/(4Z-1)]a_{o ,}velocity v₁₁ = [(4Z--1)/4] e²/2hε_o  and total energy of He like ions - [(4Z-1)²/8] e²/2Ka_o. The difference in the total energy of the He like ions and He⁺ like ions gives the (Z-1)^th ionization energy of He like ions.           BE_Z - BE_Z-1= TE_Z-1 - TE_Z= [ -(4Z-1)²/8 + Z²] [[e²/2Ka_o] and  I_He-like^(z-1) = {[8Z(Z-1)+1]/8}{e²/2Ka_o}                                         

 It is noted that the ionization energy I_Z-1 of helium like ions  is little greater than the experimental value and the deviation is greater , greater the nuclear charge. It indicates that the difference must depend upon the nuclear charge Z.   The binding per electron is increased when half -filled orbital is transformed into completely filled orbital. In the case of helium the binding energy of a single 1s electron is 4[e²/2Ka_o]   whereas the binding energy per electron in a system with two 1s electrons is (49/16)[e²/2Ka_o]= 3.0625 [e²/2Ka_o], the increment per electron is 0.9375[e²/2Ka_o] . The completely filled orbits provides mo total energy

                     I_He-like^(z-1) =   {[-8Z(Z-1) +1]/8 + C Z}{e²/2Ka_o}                                                                    where C is a constant. The mean value of C is worked out as 2.54  . The ionization energy is calculated   with equation (1) and (2) and tabulated below for comparison with the experimental values.

                Table.      . Ionization energy of helium like ions                                                                                                                                                                              ........................................................................................................................

       Z                                                            I_z-1(eV)              

                                  _{[8Z(Z-1) +1]/8                           [8Z(Z-1)+1]/8 - kZ                    experimental value}

_{.....................................................................................................................................................................................}

      _{2                              28.89                                             23.81                                       24.48}

     _{3                              83.25                                             75.63                                        75.62}

     _{4                            164.80                                             154.64                                    153.85}

     _{5                            273.60                                             260.9                                      259.30}

    _{6                            409.55                                             394.3                                      392.00}

_{.......................................................................................................................................................................................}