நம் அரசாங்கத்தின் செலவுக் கணக்கு எப்படி இருக்கின்றது   என்று என்னைக் கேட்டால் ஒரு கதை மூலம் செல்வேன் .ஒரு நாளைக்கு ச் சூரியனை கிழக்கில் உதிக்க வைத்து பொழுதை விடிய வைக்க  செலவு  4 லட்சம் . அதை நாள் முழுதும் ஊர்வலமாய் எடுத்துச் செல்ல ஒரு மணிக்கு 1 லட்சம் வீதம்  (6 - 6 ) 12  மணி நேரத்திற்கு 12 x 1 = 12  லட்சம். மலையில் சூரியனை அடக்கம் செய்ய செலவு 4 லட்சம் .மொத்தம் 1 நாளைக்கு செலவு 20 லட்சம் . ஒரு மாதத்திற்கு செலவு 30 x  20 = 6 கோடி . GST  18 % = 1 .008  கோடி . ஆக மொத்தம் ஒரு மாதச் செலவு 7 .008 கோடி . நிதிப்பற்றாக்குறை ஏற்படுவதால் சொத்து வரியை உயர்த்தவேண்டிய கட்டாயம் .மக்கள் அரசாங்கத்துக்கு ஒத்துழைப்பைத் தருமாறு கேட்டுக்கொள்கின்றோம்      

 செயற்கையை விரும்பும்போது இயற்கை காணாமற் போய்விடுகின்றது.  அதிக ஓய்வு உழைப்பை உறங்க வைத்துவிடுகின்றது . அளவுக்கு மீறிய அமிர்தம்  நஞ்சாக மாறிவிடுகின்றது . செயற்கை நுண்ணறிவின் வரவால் இயற்கை அறிவை இழந்துவிடப்போகின்றோம் .அது சமுதாய வளர்ச்சிக்காகப் பயன்படுத்தும் நல்லவர்களுக்கும் , சுயநலத்திற்காகப் பயன்படுத்தும் தீயவர்களுக்கு இடையிலான போட்டியாக இருக்கப்போகிறது . சைபர் கிரைம் குற்றங்கள் பெருகும் வாய்ப்பே அதிகம் .

   மனிதர்கள் இன்றைக்கு AI என்ற செயற்கை நுண்ணறிவை ப் பயன்படுத்த த் தொடங்கியிருக்கிறார்கள் . ரோபோக்கள் மனிதர்களுக்குப் பதிலாக நடமாடி வருகின்றன.  இயற்கைக்கு ஒவ்வொரு அணுவும் ,மூலக்கூறும்  ரோபோ தான் . இதைக்கொண்டு இந்த பிரபஞ்சத்தையே உருவாக் கியிருக்கிறது. நம்முடைய ரோபோக்கள் அச்சமூட்டும் இயற்கையின் ரோபோக்கள் அப்படியில்லை . ஏனெனில் மனிதர்கள் எதையும் அதற்காக மட்டுமே பயன்படுத்திக்கொள்வதில்லை 

 

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

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

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}

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

 

நாட்டை ஆளவேண்டும் என்று விரும்பும் அரசியல் வாதிகளுக்கு தன்னலத்தை விட மக்கள் நலமே முக்கியம் . வேறொருவன் தன்னைவிட மக்களை சிறப்பாக கவனித்துக் கொள்கின்றான் என்றால் , அவனுக்கு ஒத்துழைப்பு கொடுக்கவேண்டுமே   ஒழிய அவனையே ஒழித்துக்கட்டுவதில் ஆர்வம் காட்டக்கூடாது. ஆனால் அரசாங்கம் தரும் அளவில்லாத சுகங்களை தான் மட்டுமே  அள்ளிப் பருக ஆசைகொண்டு மதியிழந்து செயல்படுகிறார்கள் . ராமனும் ராவணனும் நல்லவர்கள் என்றால் என்னைப் பொறுத்த வரையில் ராமன் ஆண்டாளும் சரி ராவணன் ஆண்டாளும் சரி.

 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

 

இந்தியாவின் அரசியல் பிற நாடுகளிலிருந்து மாறுபட்டிருக்கிறது. இங்கே அரசியல் தலைவர்களை அண்டிப்பிழைப்பவர்கள் அவர்களை ஒரு வரம்பின்றி புகழ்ந்து தள்ளுவார்கள்.  அந்தத் தலைவரால் தனக்கு எதாவது பதவி மற்றும் சம்பாதிக்கும் வாய்ப்பு கிடைக்கும் வரை அளவின்றி புகழ்வதை வழக்கமாகக் கொண்டிருப்பார்கள் .இது அரசியல் தலைவர்களின் உண்மையான  முகத்தை , மறுபக்கத்தை மூடி மறைத்து விடுகின்றது. இந்தியாவில் அரசியல்தலைவர்கள்  அளவில்லாத சுதந்திரம் , அதிகாரத்தை எடுத்துக்கொள்கிறார்கள்..இதை யாரும் தடுப்பதில்லை என்பதால் ஓர் இலக்கண வரம்பின்றி அரசியல் தலைவர்கள் உருவாகிறார்கள் . நாளுக்கு நாள் அவர்களின் எண்ணிக்கை தொடர்ந்து அதிகரித்துக்கொண்டே வருகின்றது .இவர்கள் மக்கள் நலனில் அக்கறை கொள்வதை விட மறைவொழுக்க நடவடிக்கைகளில் விருப்பம் கொண்டு பொருள் சம்பாதிப்பதை மட்டுமே வாழ்நாள் குறிக்கோளாக க் கொண்டுள்ளார்கள். இது எல்லோருக்கும் தெரியும் என்றாலும் தவறான வளர்ச்சியை த் தடுக்க யாரும் சட்ட ரீதியிலான முயற்சி மேற்கொள்ள முன்வராததால் இந்திய அரசியல் மேலும் மேலும் கீழ்நோக்கியே சென்று கொண்டிருக்கின்றது .உழைத்து முன்னேறமுடியாது இனிமேல் மறைவொழுக்க நடவடிக்கைகளால் மட்டுமே வாழ முடியும் என்ற நம்பிக்கையை வளர்த்துக்கொண்டுள்ளார்கள்.

 ஒரு காலத்தில் மது அருந்தினால் குற்றவாளி என்ற நிலை இருந்தது.இன்றைக்கோ மது விற்பனை அரசாங்கத்தின் வருமானம் .அதனால் மது  அருந்துவதை அரசாங்கம் மறைமுகமாக  ஊக்குவிக்கின்றது. பூரண மதுவிலக்கை இனி யாராலும் சமுதாயத்தில் சேதாரமின்றி கொண்டு வரமுடியாது . அதுபோல செயற்கை நுண்ணறிவு இன்றைக்கு வளர்ந்து வருகின்றது. இது எதிர்காலத்தில் நம்பமுடியாத அளவிற்கு சமுதாயக் கேடுகளை த் தரலாம். வளர்த்து விட்டபிறகு  அதைவிரும்பாத நிலையில்  பயன்படுத்த க் கூடாது என்று கட்டுப்படுத்தவே முடியாது . அதைத் தவறான வழியில் பயன்படுத்தி பொருள் சம்பாதிக்கும் கூட்டம் இருக்கும் . கட்டுப்படுத்த வேண்டிய அரசாங்கம் வழி தெரியாமல் விழிக்கும் நிலையே அங்கும் தொடரும் 

 Application of Bohr's Theory of hydrogen to Beryllium

      The beryllium has two orbits 1s and 2s each with two electrons. For the stability of each electrons and nucleus, the two electrons are diametrically opposite in both the orbits, so that its diameters are perpendicular to each other.

                                                           Beryllium atom with 1s²2s²

      Considering 1s electrons   4e² /Kr_11s²  - e² /4Kr_11s²  = (15/4) e²/K = m v_11s²r_11s. and h/2π = m v_{11 s} r_11s The radius of the 1s orbit with two electrons becomes (4/15) a_o. and  the velocity of the electron v_11s = (15/4) e²/2hε_o. Since the electronic structures of both the orbits are same r_n = n² r₁ so that r_22s = (16/15)a_o. Since the inner orbital electrons are more tightly bound with the nucleus, its structure will remain unaltered.  Total energy of the system is sum of energies contributed by both the 1s and 2s electrons. The kinetic energy associated with the 1s electrons mv_11s² =  4e²/Kr_11s - e²/4Kr_11s = (15/4)e²/Kr_11s ^.The negative potential energy  is -8 e²/Kr_11s and the positive potential energy due to electron-electron interaction is e²/2Kr_11s.  Total energy associated with the 1s electrons is {15/4 - 8 +1/2] e²/Kr₁ = - (15/4) e²/Kr₁ = - (225/8)(e²/2Ka_o) = - 382.36  eV

     If there is no screening of nuclear charge, the radius of the 2s orbital becomes r_22s = 4 r_11s =

(16/15)a_o. Total energy associated with the 2s electrons is {15/4 - 8 +1/2] e²/Kr_22s = - (15/4) e²/Kr_{22 s} = - (225/32)[e²/2Ka_o] = - 7.03125 x 13.595 = -95.589 eV. The sum of first and second ionization energy of Beryllium atom is 9.32 +18.21 =27.53 eV.

     If we assume full screening of nuclear charge, the 2s electrons will realize only 2e⁺ . The stability of the 2s electron in its orbits requires 2e²/K - e²/4K = (7/4)e²/K = mv_22s²r_22s and h/π = m v_22s r_22s which together provide r_22s = (16/7) a_o

    The kinetic energy associated with the 2s electrons mv_22s²  =  (7/4) e²/Kr_22s .Negative potential energy  - 4 [e²/Kr_22s] _.  Positive potential energy of 2s electrons due to  the electron-electron interaction e²/2Kr_22s Total energy associated with the 2s electrons is -(7/4)[e²/Kr_22s] = -(49/32) x  [e²/2Ka_o]= -1.53125 x 13.595 = -20.8173 eV . It is little closer to the sum of the observed first and second ionization energy of the beryllium atom 9.32 + 18.21 = 27.53 eV

    The  first ionization energy of beryllium is determined from the knowledge of total energy associated with beryllium and beryllium ion Be^+.

Total energy possessed by beryllium atom = - 382.36 + - 20.82 = -403.18 eV , the observed experimental value is -398.03 eV.  In Be⁺ , the 2s electron has  kineetic energy  (1/2) m v_2s²  = e²/Kr_2s and negative potential energy - 2e² /Kr_{2s ,} Adding togethertotal energy  becomes - e²/Kr_2s

= - e²/2Ka_o = - 13.595 eV. It gives the first ionization energy of Beryllium as (20.82 - 13.60) 7.22 eV .       

 Lithium atom

       When a single electron revolves round the nucleus at high speed, the nucleus is attracted by the orbital electron equally in all directions and the nucleus is stable at the center as if unperturbed by the electron. When two electrons are present in the innermost orbit, both of them cannot be at the same point but separated apart with maximum distance of separation within the same orbit. For this requirement they are held diametrically opposite to each other.

     Lithium atom has three orbital electrons. There are two ways by which the lithium atom can be constructed. When third electron e-3 is present, one can suppose that it may be in the innermost orbit along with the existing two 1s electrons with electronic configuration 1s³. In another possibility, the third electron takes next higher orbit with electronic configuration 1s²,2s¹.  In all the atoms, the nucleus is stable at its center in the presence of all the orbital electrons. If all the three electrons are in the innermost orbit, they will be identical in all respect and hence each electron will realize the whole nuclear charge. The electron-electron interactions keep them with an angular separation of 120^o . In this arrangement, the resultant force experienced by the nucleus due to the orbital electrons will be zero at all time. When the third electron is in different orbit, the 1s electrons may be unperturbed so that they will be diametrically opposite to each other. This assumption is valid as the inner electrons are held strongly by the nucleus. If n is the number of electrons present in the innermost orbit, the angular separation between any two electrons will be same and is equal to 360^o/n.

Fig.22.The structure of innermost orbit with one or more electrons

   When a third electron is pulled to stay in the 1s orbit, the mutual interaction between the orbital electrons keep them away with maximum distance of separation so that they become symmetric with respect to the central nucleus .As a result, the resultant force acting on the central nucleus by all the surrounding electrons with angular separation of 120^o will be zero at all instant of its motion .Let us suppose the usual electronic configuration 1s² ,2s¹ for lithium atom. The electrostatic force acting on the nucleus by one of the 1s electrons is 3e²/Kr₁²  and by the 2s electron it is equal to 3 σ e²/Kr₂²   each along its radius vectors respectively. Where σ is fraction of nuclear charge as felt by the outer orbital electrons due to the presence of inner orbital electrons. If 2θ is the angle between the radii of 1s electrons, the resultant force acting on the nucleus by both the 1s electrons is 2F_1s cos θ, which must be equal and opposite to the force exerted on the nucleus by the 2s electron F_2s = 3σ e²/Kr₂².  For stability of the nucleus, the resultant force due to 1s electrons must be equal to the force due to the 2s electron 

                                               6 e²/ Kr₁² cos θ  =  3σ e²/Kr₂²   

When the 2s electron is taken away from the system, r₂ →_{∞ ,} cos θ = 0 or   θ = 90^oi.e., the two 1s electrons are diametrically opposite to each other in the orbit. When r₁ = r_{2 ,} all the three electrons are in the same orbit, σ = 1 and cos θ = 1/2 or   θ = 60^o. The stability of central nucleus requires symmetric arrangement of electrons about the nucleus with radii vectors making an angle 120^o in between them.

                                                       6 e²/ K r² cos θ  =  3e²/ Kr²                                                                

                    cos θ  = 1/2   or θ  = 60^o 

When e-3 is absent, r→ ∞ for 2s electron_,   cos θ = 0, θ = 90^o i.e., the introduction of third electron keeps the 1s electron in the same orbit but pushes away from it. In this shifting the 1s electrons are brought closer which increases its potential energy of the system. This may be the reason for its small first ionization energy in lithium. If all the three electrons are in the same innermost orbit, its potential energy is greater than the system with two electrons in the innermost orbit and one electron in the next higher orbit. i.e., the former arrangement is unstable with respect to the later configuration. If all the three electrons are in the innermost orbit, the electron-electron interaction by increasing the potential energy reduces the binding energy of the system.  

Total energy of the Li atom with three electrons in the innermost orbit

      Let us consider a case where all the three electrons are in the innermost orbit of the lithium atom. In the presumed case, each electron is acted upon by the central nucleus with charge 3e+.For want of stability, they are separated apart with equal angular displacement of 120^o.

     

Fig.23. Lithium atom with 1s³ configuration

        The nuclear attractive force acting on an electron is 3e²/Kr²  and the  electron-electron repulsive force     e²/Kd² = e²/3Kr²  .The resultant components of  electron-electron repulsive forces  acting on the electron is [2e²/3Kr²] cos 30^o  = e²/ (3)^1/2Kr^2. The resultant force is balanced with the centrifugal force.

                                                         3e²/Kr_n² - e²/ (3)^1/2Kr_n²  =    m_nv_n²/r_n

                                                (e²/K)[3 - 1/(3)^1/2]=    m_nv_n²r_n

                                                           nh/2π     =   m_nv_nr_n

By solving, the radius of the orbit as r_n = {[(n² (3)^1/2 a_o)/(3(3)^1/2 - 1)]}(1- v_n²/c²)^1/2  = [1.732/4.196] n²a_o (1- v_n²/c²)^1/2 = 0.41277a_o n² (1- v_n²/c²)^1/2  and v_n = [e²/2nhε_o] [3(3)^1/2 -1]/(3)^1/2. The radius of the innermost orbit r₁ = (3)^1/2a_o/[3(3)^1/2- 1].  The kinetic energy contributed by all the three identical electrons is (3/2) mv₁² = (3/2) (e²/Kr₁)[3 - 1/(3)^1/2] = [(3)^1/2] (e²/2Kr₁) [3 (3)^1/2  - 1].  

Negative potential energy is not equal to - 9e²/Kr₁ = -18 [e²/2kr₁]. We get the same result when the positive charge 3e+ is taken from infinity to the center of an inscriped equilateral triangle with side (3)^1/2 r₁ and electrons in all of its vertices.

  

Fig.24. constructing Lithium atomic model                                                                                                  with electronic configuration 1s³   

In an intermediate stage the positive charge 3e⁺ is at distance of x vertically above from the plane of the electronic orbits. The force between the positive charge and an electron in the plane is 3e²/K(x²+ r₁²). Its component along the direction of motion is 3e² x /K(x²+r₁²)^3/2. For all the three electrons in the plane it becomes 9e²x/K(x²+r₁²)^3/2 The work done in taking the positive charge from infinity to the center of the plane 9e²/K   _∞∫⁰ [x /(x²+r₁²)^3/2] dx. By substituting y =(x²+ r₁²) NPE= (9/2)(e²/K) _∞∫⁰ dy/y^3/2= - (9/2)(e²/K){[2/(x² +r₁²)]^1/2} _∞⁰= - 9e²/Kr₁

The positive potential energy between e-1 and e-2 electrons is e²/(3)^1/2 Kr₁ = [2/(3)^1/2](e²/2Kr₁) 

The third electron also contributes positive potential energy. When e-3 is brought from infinity to the innermost orbit, the existing electrons are drifted in the opposite side without changing its orbit. When e-3 is at infinity the angle between the radii vectors of 1s electrons is 180^{o ,} when it reaches the innermost orbit it becomes 120^o. Hence the potential energy of e-3 due to electron-electron interaction is calculated by assuming mean bending angle of 1s electrons as 150^o.   

         When e-3 is at an intermediate distance x from the innermost orbit, the distance between e-3 and any one of the electrons in the innermost orbit is d² = (x+ r + r cos 75^o)² + (r sin 75^o)² = x² + 2r(1+cos 75^o) (x+r). The electrostatic force between the electrons e-3 and e-1 or e-2 is  e²/K[x² + 2r(1+cos 75^o)(x+r)] ^. The component of force F(e-e)₁₃ along the direction of displacement = 2e² cosφ /K[x² + 2r(1+cos 75^o)(x+r)]. Substituting the value of cosφ = x+r(1+cos 75^o)/[x² + 2r(1+cos 75^o)(x+r)]^1/2 

                          F(e-e)₁₃   = 2e²[x +r(1+cos 75^o]/K[x² + 2r(1+cos 75^o )(x+r)]^{3 / 2}

The total workdone in bringing the electron e-3 to the innermost orbit is _∞∫⁰ 2e² [x +r(1+cos 75^o] dx/ K[x² + 2r(1+cos 75^o ) (x+r)]^3/2 = [2e²/K]{1/ x² + 2r(1+cos 75^o ) (x+r)}^1/2 = [2e²/K]{1/[r² + 4r²(1.2588)}^1/2 = [2e²/Kr][1/6.0352]^1/2 = [e²/2Kr][1.6282] eV

    The sum of all the components of energy  [(3)^1/2(3x3^1/2 - 1) - 18 + 2/(3)^1/2 + 1.6282] [e²/2Kr]    = -7.949 x [e²/2Kr] = -[7.949 x 4.19615/1.732] [e²/2Ka_o] eV= -19.2578 x 13.595 = - 261.81 ^. It gives the first ionization energy of lithium atom as 261.81 - 205.62 = 56.2 eV, which is not in agreement with the experimental value of 5.39 eV. It implies that the third electron in the lithium atom cannot be housed in the innermost orbit along with the two 1s electrons. It predicts that there must be a cause to provide additional positive potential energy to reduce the binding energy of the system. In quantum physics it is accounted by Pauli's exclusion principle which states that no two identical fermions (particles with half-integer spins, like electrons, protons, and neutrons) can simultaneously occupy the same quantum state within a quantum system.

      Let us now study the pragmatic analysis in details to prove that the lithium atom with all the three electrons in the innermost orbit is less stable than the lithium with electronic configuration 1s², 2s¹.  The radii of the electronic orbits can be derived from the condition imposed on the permitted orbits.

Total energy of the Li atom with two electrons in 1s orbit and one electron in 2s orbit

        When more than one electron is present in an orbit, due to intra-electronic interaction, they are separated apart with a maximum distance possible within the given orbit itself.. It makes the electronic configuration to be symmetric with respect to the central nucleus. The radii of the electronic orbits in an atom are predetermined by the effective nuclear charge acting on the electron. It is self-modified only by the intra-electronic interaction. In helium atom the radius of the  1s orbit with single electron is a_o/2,when one more electron is added in the orbit, the radius gets modified to (4/7) a_o.  Since the electrons in different orbits revolve with different angular velocity, the relative positions of  them will vary continuously and cyclically as well. This periodic variation of inter-electronic interaction provides a wave-like motion in the allowed orbits without changing its radius. As the resolved component of inter-electronic interaction is same in all direction, it is dropped to zero in any particular direction and it can be supposed that   the electrons in different orbits  are non-interactive. Since electrons moving in the inner orbits of an atom are very close to the nucleus, the radii of its orbits are not altered by electrons moving in the outer orbits. Furthermore, since the strength of the interaction between electrons orbiting in inner and outer shells at different velocities is significantly weaker than the interaction between stable electrons with a fixed distance of separation, Therefore the radii of the inner and outer electronic orbits do not undergo any change. With these assumptions, the optimization of Bohr's theory of hydrogen atom is extended to atoms of higher atomic number. 

        Let the 1s and 2s electrons be in the orbit with radius r₁ and r₂ respectively. This is the simple case where the electron (1s) - electron (2s) interaction with electrons in different orbits occurs. Total energy of the normal lithium atom is calculated in two steps-energy associated with Li⁺ ion and energy contributed by the 2s electron. Usually the orbital electrons in atom have three different components-kinetic and negative potential energies of the orbital electrons and electron-electron interactional energy. The three components of energy of Li⁺ ion are available from previous section. The kinetic energy of the 1s electrons = (121/8) (e²/2Ka_o), negative potential energy =  - 33 (e²/2Ka_o) and the positive potential energy due to electron(1s)- electron (1s) interaction = (11/4) (e²/2Ka_o).Its summation gives - 205.624 eV and by adding the energy associated with the 2s electron, the energy of normal lithium atom can be computed.

         There are many presumptions in the atomic models of the lithium atom out of which one must be very close to the real picture.  The presence of the third electron in the 2s orbit may induce different structural changes which has its own impact in the determination of binding energy of the 2s electron. At first, there are two different   suppositions - the electron (2s) electron (1s) interaction is (1) forbidden and (2) allowed .It is forbidden because the different orbital electrons move with different angular velocity. It is allowed because the potential energy of the 2s electron in the presence and in the absence of intermediate 1s electron must be different.   The computation of total energy associated with the 2s electron will be different in accordance with the effective nuclear charge as seen by the 2s electron. Considering the interactional status and the effective nuclear charge as seen by the 2s electron, the pragmatic analysis is undertaken to find the real picture of the lithium atom and each case has three different situations and they are (i) there is no screening of nuclear charge by the intermediate electrons i.e., all the electrons in all the atomic orbits realize the same nuclear charge (ii) there is complete screening of nuclear charge. The effective nuclear charge  as seen by the 2s electron may be the sum of nuclear charge 3e⁺  and the electronic charge of the inner sphere 2e^- i.e., 3e⁺- 2e^- = e⁺ after full screening of nuclear charge with the intermediate two 1s electrons and (iii) there is partial screening of nuclear charge by the intermediate orbital electrons. It is partial within certain limit inside the atomic space as the (e-e) interaction is differential and beyond that limit the interaction is due to the resultant charge of the inner sphere. By using Slater's rule the effective nuclear charge experienced by the outer electron can be estimated.

1 . (e-e) inter-interaction is forbidden

        When the (e-e) interaction is forbidden, the radii of both the inner and outer orbits remain   unchanged due to inter electronic interaction and the 1s electrons are still being diametrically opposite in the inner orbit. This is because the inner orbital electrons are more tightly bound with the nucleus and become more rigid due to strong nuclear attraction than the outer orbital electron. It prevents any positional variation among 1s electrons due to (e-e) interaction. The radii of the innermost and its successive electronic orbits in an atom with atomic number Z are given by n²a_o/Z.   

1(i) : Z_eff = 3e⁺, 

Fig.  25. Lithium atom with no screening of nuclear charge                                                                               .             and no inter electronic interaction

        If we assume that there is no shielding of nuclear charge and no inter electronic interaction  all the electrons in all the successive orbits may perceive the entire and same nuclear charge .i.e., in lithium atom both 1s electrons at r_1s and 2s electrons at r_2s sense the same nuclear charge 3e⁺_.  Both the 1s and 2s electrons remain in their respective allowed orbits where the mutual interaction is inhibited as they are moving fastly with different speeds. The change of orbit is allowed only under the condition of integral multiplicity of angular momentum of the orbital electron. The radii of the 1s and its higher orbit with single electron when the nuclear charge Z = 3e⁺ can be worked out as follows.

  3e²/Kr_n²  = mv_n²/r_n  or 3e²/K = mv_n² r_n                                                                                                                                                                                                                        nh/2π = mv_n r_n                                                                                                                                                                                                                                                       v_n  =[3e²/4πε_o][2π/nh] = 3e²/2nhε_o ; v₁= 3e²/2hε_o                                                                                                                                                       m r_n = [n²h²/4π²][4πε_o/ 3e²] = n²h²ε_o/3πe² ; r_n =  n² a_o/3                                                                                                      v_n/r_n  = [3e²/2nhε_o]/[ n² a_o/3] = [9/n³] [e²/2hε_oa_o]                            r₁(Li²⁺) = a_o/3 . The radius of its next higher orbit(2s orbit with single electron) is 4 r₁ = 4 a_o/3

The radius of the 1s orbit with two electrons when the nuclear charge Z = 3e⁺  is                                                    3e²/Kr₁₁²  - e² /4Kr₁₁² =   (11/4) e²/K r₁₁² = mv₁₁²/r₁₁                                                                                                                                             (11/4) e²/K = mv₁₁² r₁₁                                                                                                                                                 h/2π = mv₁₁ r_{11    ;}   v₁₁  =[11/4][e²/4πε_o][2π/h] = (11/4) e²/2hε_o                                                                             mr₁₁ = [h²/4π²][4/11][4πε_o/e²] = (4/11)[h²ε_o/πe²]  ; r₁₁ = (4/11)a_o                                                                                           v₁₁/r₁₁ =  (11/4) [e²/2hε_o][11/4a_o] = [121/16]  [e²/2ha_o ε_o]

Total energy of the system Li⁺ ion = KE + NPE + PPE = mv₁₁² - 6 e²/Kr₁₁ + e²/2Kr₁₁ = (11/4)e²/Kr₁₁- (11/2) e²/Kr₁₁ = - (11/4) e²/Kr_{11 =} - (121/16) e²/Ka_o = - (121/8) [e²/2Ka_o] =-15.125 x 13.595 = - 205.624   eV.

     The stability of 2s electron in its orbit requires 3e²/K = mv_2s² r_2s and h/π = mv_2sr_2s  which together give r_2s = 4a_o/3. The sum of kinetic and negative potential energy of the 2s electron is (3/2) e²/Kr_2s - 3 e²/Kr_2s = - (3/2)e²/Kr_2s = -(9/4) e²/2Ka_o = - 30.588 eV.  It gives the total energy of the lithium atom -205.624 - 30.588 = - 236.212 eV and first ionization energy 30.588 eV. But the practical value of first ionization energy of lithium atom is 5.39 eV only.  The inter-electronic interaction and the reduction of nuclear charge due to screening by intermediate orbital electrons have some influence to make up this discrepancy. 

 1.(ii) : Z_eff  = e⁺,

      In this case there is maximum shielding of nuclear charge and no inter electronic interaction. Here the presence of 1s electrons in between the nucleus and the 2s orbital is considered to be equivalent as 2 electrons stay within the nucleus. The radius of the 1s orbit with single electron in H is a_o. which predicts  the radius of the 2s orbit with single electron with effective nuclear charge e⁺ would be 4a_o. The stability of 2s electron in its orbit gives the same result. e²/K = mv_2s² r_2s  and h/π =  mv_2sr_2s  which together give r_2s = 4a_o,v₂ =   (e²/4πε_o)(π/h) = (1/4) (e²/hε_o) and v₂/r₂ = (1/16)(e²/hε_o a_o)._. The K.E and NPE of the electron 2s in lithium atom are, KE = (1/2)m v_2s² = (1/2)e²/Kr_2s  and NPE = - e²/Kr_2s  which give total energy  TE = - (1/2)[e²/Kr_2s]= (1/4)[e²/2Ka_o]=  - 0.25 (13.595) = - 3.39875 eV . It is somewhat closer to the practical value, however it is still lower than by an amount -1.99125 eV.

Screening by inner orbits

       The effective nuclear charge as seen by the outer orbital electrons is reduced by screening caused by the presence of the inner orbital electrons .This is well hinted by the neutral atom not    responsive to any external electromagnetic fields. Since the atom as a whole is electrically neutral, any charged particles existing outside the atom will not realize the nuclear charge and get accelerated. The screening is maximum when the orbit is completely filled with electrons, and partial if not . Due to screening of nuclear charge, the interaction between the nucleus and the outer orbital electrons is reduced, it makes changes in its orbital velocity , radius of the orbit , potential energy and mutual electron-electron interaction.

Shielding constants -Slater's rule

          The shielding constant or screening constant σ  is a value used to estimate the effect of inner electrons on the attraction between the nucleus and outer electrons in an atom. It's used in the calculation of the effective nuclear charge, which is the net positive charge experienced by an electron. A higher shielding constant means the outer electrons are more effectively shielded from the nucleus's attraction, resulting in a lower effective nuclear charge

         Slater's Rules  are commonly used to estimate the shielding constant and successfully employed in the determination of diamagnetic susceptibility of atoms  The formula, based on Slater's rules, is σ = ∑(n_i x s_i), where  ni is the number of electrons in the i^th shell, s_i is the shielding contribution of electrons in the i^th shell. The shielding contribution (s_i) depends on the electron's shell and its position relative to the electron of interest (the one whose shielding is being calculated). For an electron in the nth shell: Electrons in the same shell (n) as the electron of interest: Each electron contributes 0.35, except for the 1s orbital where the contribution is 0.30. If the electron of interest is in an s or p orbital, all other electrons in the s and p orbitals of the n-1 shell contribute 0.85. If the electron of interest is in a, d or f orbital, all other electrons in the n-1 shell contribute 1.00. Let us try to understand the estimation of shielding constant with a specific example by calculating the shielding constant for a 3s electron in Magnesium (Z=12). The electronic configuration of Mg is 1s²2s²2p⁶3s². Identify electrons and its contribution to shielding constant are exemplified below.

The electron of interest is the 3s electron.                                                                                                                  

There is 1 electron in the 3s 3 rd shell.                                                                                                           There are 8 electrons in the 2rd shell (2s², 2p⁶)                                                                                              There 2 electrons in the inner shells (1s²).

The shielding constant for a 3s electron in magnesium is  σ =   (0.35 x 1) + (0.85 x 10)   =                                0.35 +  8.5 = 8.85. The shielding constant for a 3s electron in Mg is 8.85.    Due to shielding effect, the nuclear charge as seen by the 2s electron in lithium atom is 32(0.30) = 2.4  e.

1.(iii): Z_eff = 2.4 e⁺ , (e-e) interaction is forbidden

   This is a case with partial shielding of nuclear charge as per Slater's rule and the absence of inter electronic interaction. The radius of the orbit with single 1s electron r_1s with effective nuclear charge 2.4 e⁺ = a_o/2.4  The radius of the orbit with single 1s electron r_1s with effective nuclear charge 3 e⁺ =  a_o/3.The radius of the orbit with single 2s electron with effective nuclear charge 2.4 e⁺, r₂ = 4a_o/2.4=

1.667ao                                                                                                                                                           The radius of the orbit with two 1s electrons r_11s with effective nuclear charge 3 e⁺ = (4/11) a_o = 0.3636

K.E = (1/2)m v_2s² =  (2.4/2)e²/Kr_2s                                                                                                               NPE = -2.4e²/Kr_2s                                                                                                                                        

Total energy = - (1.2) [e²/Kr_2s] = - 1.2 x[(2.4)e²/4Ka_o] = -1.44 x 13.59 5 = -19.58 eV . The vast indifference indicates that besides the screening of nuclear charge, there must be inter-electronic interaction, which changes the radius of the electronic orbit and its total energy.

2.(e-e) interaction is allowed

      If we assume that there is no shielding of nuclear charge, all the electrons in all the successive orbits may sense the entire and same nuclear charge .i.e., in lithium atom both 1s electrons at r_1s and 2s electrons at r_2s are attracted by the same nuclear charge 3e⁺_. The 1s electrons are more tightly bound with the nucleus and have less tendency to make changes in its relative position. i.e., they are unaffected by the outer electrons where the changes caused by the mutual electron-electron interaction are resulted with the displacement of outer orbital electrons. If the 1s electrons remain in the same orbit  as they are more tightly bound with the nucleus it may refuse to undergo any change in its position, ie., the inner orbital electrons are unaffected by the outer electrons where the changes caused by the mutual electron-electron interaction are resulted with the displacement of outer electrons as it is loosely bound with the nucleus. The 2s electron gets drifted away from the nucleus which makes its radius little increased. It is resulted with a reduction in its both kinetic and negative potential energy.

      The expansion of 2s orbit may be inhibited for want of satisfying the condition on the circumference of the orbit must hold integral number of wavelength characterizing the orbital electron.  If the electron-electron interaction between electrons in different orbits is allowed, due to mutual interaction, the 1s electrons may drift away from the 2s electron. As a consequence of which the 1s electrons in the same orbit become little closer and provide little positive potential energy to the 2s electron, which reduces its binding energy with the nucleus. This drifting makes the resultant force acting on the central nucleus by the both 1s and 2s electrons to be equal to zero.

2.(ia) Z_eff= 3e⁺,  2s orbit expands.

 

Fig.26. no screening of nuclear charge                                                                                                         

and 2s orbit expands due to (e-e) interaction

      The radii of the 1s and its higher orbit with single electron when the nuclear charge Z =3e⁺ can be shown as r_n = n²a_o/3  i.e., r₁(Li²⁺) = a_o/3 .and the radius of its next higher orbit (2s orbit with single electron) is 4 r₁ = 4 a_o/3 .The radius of the 1s orbit with two electrons when the nuclear charge Z = 3e⁺ is  r_11s= (4/11)a_o (1- v_1s²/c²)^1/2 and it is approximately equal to (4/11) a_o. The intermediate orbital electrons has two effects on the outer electrons - screening the nuclear charge and the (e-e) interaction If the interaction between different orbital electrons is allowed , the radius of the 2s orbit gets modified where as that of 1s orbit remains same , as they are more tightly bound with the nucleus. Let r_2s  be the radius of 2s orbit with forbidden interaction with 1s electrons, then 3e²/K = mv_2s² r_2s and h/π =  mv_2sr_2s which together gives r_2s= (4/3)a_o. The radius of 2s electron is modified due to the electron (1s) - electron (2s) interaction as it is free to move outward . Let r_2s' be its modified radius. According to Slater, the effect of 1s orbital electrons on the 2s electron is equal to 0.6 electronic charge at the center.  The resultant electrostatic force of attraction acting on the 2s electron  3e²/Kr_2s'² - 0.6e²/Kr_2s'² = (2.4)e²/Kr_2s'²  =m v_2s'²/r_2s'  and  h/π =  mv_2s'r_2s' ,which together give r_2s' = [4/(2.4)]a_o. By comparing the stability conditions of 2s electron with (e-e) interaction forbidden and allowed, we have v_2s²r_2s /v_2s'²r_2s' = v_2s/v_2s' = 3/2.4 since v_2sr_2s

=h/π = v_2s'r_2s' i.e.,  v_2s/v_2s' = r_2s'/r_2s = 3/2.4 or r_2s' = (3/2.4)r_2s = (4/2.4) a_o

The kinetic energy associated with the 2s electron revolving round the lithium nucleus in an orbit with radius r_2s' is    (1/2) mv_2s'² = (1.2) e²/Kr_{2 s'}

Negative potential energy = - 3e²/Kr_{2 s'}

1s electrons in its orbital is equal to 0.6 e^- at the center which gives positive potential energy = 0.6 e2 /Kr2 s'

Total energy of the 2s electron = - [e²/Kr_2s'](1.2 - 3 + 0.6) = - (1.2) [e²/Kr_2s']           

Substituting the modified radius of the 2s orbit r_2s'= [4/2.4]a_o

    T.E  =-(1.2)² [e²/2K a_o]   = -1.44 x 13.595 = -19.5768 eV

  Again it is not in agreement with the experimentally observed value of first ionization potential of lithium atom.   

2.(ib) Z_eff= 3 e⁺ ,  1s electrons with angular displacement

       If the radii of the electronic orbits are not allowed to change by the (e-e) interaction and are fixed by the nuclear charge at the center, the repercussions  of the mutual electron-electron interaction is carried away by angular displacement of the 1s electrons without making any change in the radius of its orbit. The stability of the nucleus demands that the resultant force experienced by the nucleus in any direction at all time      must be equal to zero. i.e., the relative position of the orbital electrons surrounding the nucleus must be such that the resultant force acting on the central nucleus must be zero. If F_1s and F_2s denote  the electrostatic forces of attraction acting on the nucleus by the electron in 1s and 2s orbits respectively, the stability of the nucleus  requires  2 F_1s cos  θ = F_2s  where  60^o ≤ θ ≤ 90^o  , where  2θ is the angle between the radius vectors of 1s electrons.  θ = 90^o, when 2s electron is far away from the nucleus, and θ = 60^o , when the 2s electron exists in the same 1s orbit. When the electrons exist in different orbits θ lies in between 60^o and 90^o. If the orbits of both 1s and 2s electrons are not changed and only the 1s electrons get angular displacement in the same orbit  

          

 

  Fig.27. Lithium atom with 1s² 2s¹ configuration                                                                                                       with angular displacement of 1s electrons 

      The stability of the nucleus with F_1s and F_2s   demands 6 [e² /Kr_11s²] cos θ = 3 [e² /Kr_2s²] or cosθ =(1/2) (r_11s/r_2s)²  The radius of the innermost orbit of  lithium atom is r_11s = (4/11) a_o =0.3636 a_o. and the radius of  2s orbit with single electron is 4a_o/3.    By substituting these values in the condition for the stability of the nucleus we arrive at  cos θ =(1/2) (r_11s/r_2s)² = (9/242) = 0.0372 or θ = 87^o 48' .  The bending angle is (90 - 87^o 48')  = 2^o 12'. By virtue of this bending, the 1s electrons become closer from 2r_11s to 2r_11s sin θ without changing its radius of the allowed orbit.  The increase of its positive potential energy is [e²/2Kr_11s]{1/[sinθ)] - 1} = [e²/2Kr_11s][1/(cos2^o 12') - 1]  = [e²/2Kr_11s][0.0007]  = (11/4) [e²/2Ka_o] = 2.75 x 13.595 x 0.0007 = 0.0262 eV which reduces  the resultant binding energy of the 2s electron in lithium atom 

       The kinetic energy of the 2s electron = (1/2)m v_2s² = (3/2) e²/Kr_2s and its negative potential energy = - 3 e²/Kr_2s .Total energy associated with the 2s electron = - (3/2) e²/Kr_2s = - (9/4) e²/2Ka_o=-2.25 x 13.595 = - 30.588 eV . The resultant binding energy of the 2s electron becomes 30.588 - 0.026 = 30.562 eV.  The ionization energy of 2s electron in the lithium atom is 5.39 eV only which requires some modifications.   

2(iia) Z_eff= e⁺ , 2s orbit expands with maximum screening of nuclear charge

       The effective nuclear charge is equal to net charge of the nucleus and the electronic charge of the inner sphere. The 2s electron revolving round the nucleus in an orbit with radius r_2s is  supposed to be acted upon by  an effective nuclear charge +e .In the absence of (e-e) interaction  e²/K = mv_2s² r_2s and  h/π =  mv_2sr_2s  which give r_2s =4a_o If the interaction between different orbital electrons is allowed , the radius of the 2s orbit is changed to r_2s',when (e-e) interaction is  allowed ,the interactional effect of 1s electron on the 2s electron is equivalent to an interaction between the 2s electron and the total electronic charge of the inner sphere at the center. As per  Slater's rule it is equal to 0.6 e^-. It gives e²/K- 0.6 e²/K = 0.4 e²/K =  mv_2s'²r_2s'. and h/π =  mv_2s'r_{2 s'} Combining these two relations, we get r_2s' = 10 a_o.

         The sum of kinetic [(1/2) m v_2s'² = 0.2 e²/Kr_2s'], negative potential energy [-e²/Kr_2s']  and the positive potential energy  due to (e-e) interaction [0.6e²/Kr_2s'] of the 2s electron is -(0.2) [e²/Kr_2s'] .Substituting the value for r_2s' T.E = - 0.04 [e²/2Ka_o] =- 0.5438  eV

2(ii b) Z_eff= e⁺ , 1s orbit with angular displacement

      The stability of the nucleus with F_1s and F_2s   demands 6 [e² /Kr_11s²] cos θ =  [e² /Kr_2s²] or cosθ =(1/6) (r_11s/r_2s)²  The radius of the innermost orbit of  lithium atom is r_11s = (4/11) a_o =0.3636 a_o.  and the radius of  2s orbit with single electron sensing an effective nuclear charge e⁺ is 4a_o.    By substituting these values in the condition for the stability of the nucleus we arrive at  cos θ =(1/6) (r_11s/r_2s)² = (1/6)(1/121) = 0.00826 or θ = 89^o 30' . The  bending angle  is (90 - 89^o 30')  = 0^o 30'. By virtue of this bending, the 1s electrons become closer from 2r_11s to 2r_11s sin θ without changing its radius of the allowed orbit.  The increase of its positive potential energy is [e²/2Kr_11s]{1/[sinθ)] -1} = [e²/2Kr_11s][1/(cos 0^o30') - 1]  = [e²/2Kr_11s][0.000]   = 2.75 x 13.595 x 0.000 = 0.0 eV the resultant binding energy of the 2s electron in lithium atom remains same as 13.595 eV 

       The kinetic energy of the 2s electron = (1/2)m v_2s² = (3/2) e²/Kr_2s and its negative potential energy = - 3 e²/Kr_2s .Total energy associated with the 2s electron = - (3/2) e²/Kr_2s = - (9/4) e²/2Ka_o=-2.25 x 13.595 = - 30.588 eV . The resultant binding energy of the 2s electron becomes -30.588 + 0.026 = -30.562 eV.  The ionization energy of 2s electron in the lithium atom is 5.39 eV .It implies that the outer electron  is not attracted in the same manner as that of the inner electrons.

2(iiia) Z_eff = 2.4 e⁺,  2s orbit expands

2.4 e²/K =  mv_2s²r_2s and   h/π =  mv_2sr_2s give r_2s = (4/2.4) a_o

If the interaction between different orbital electrons is allowed, the radius of the 2s orbit is changed from r_2s to r_2s'. The resultant electrostatic force of attraction experienced by the 2s electron is 2.4 e²/Kr_2s'² -  0.6 e² /Kr_2s'² = 1.8 e²/Kr_2s'² = mv_2s'²/r_2s' or 1.8 e²/K = mv_2s'²r_{2s' .}From which the modified radius of the 2s orbit can be predicted, r_2s' =(4/1.8)a_o

The total energy associated with the 2s electron is sum of its kinetic energy and negative potential energy. K.E = (1/2)m v_2s'² = (0.9)e²/Kr_2s' . NPE = -2.4e²/Kr_2s' and PPE = 0.6 e²/Kr_{2s' .}Total energy

=  -[e²/Kr_2s'] [ 0.9] = -(0.9)² [e²/2Ka_o]=-11.01  eV

2 (iii.b) Z_eff = 2.4 e⁺,1s orbit angular displacement

    The stability of the nucleus with F_1s and F_2s   demands 6 [e² /Kr₁₁²] cos θ = 2.4 [e² /Kr_2s²] or cosθ = (2.4/6) (r₁₁/r_2s)²  The radius of the innermost orbit of  lithium atom is r_11s = (4/11) a_o.  and the radius of  2s orbit with single electron sensing an effective nuclear charge 2.4 e⁺ is 4a_o/2.4.    By substituting these values in the condition for the stability of the nucleus we arrive at  cos θ =(2.4/6) (r_11s/r_2s)² = (2.4/6)(5.76/121) = 0.01904 or θ = 88^o54' .  The bending angle is (90 - 88^o 54') = 1^o 6'. By virtue of this bending, the 1s electrons become closer from 2r₁₁ to 2r₁₁ sin θ without changing its radius of the allowed orbit.  The increase of its positive potential energy is [e²/2Kr_11s]{1/[sinθ)] - 1} = [e²/2Kr_11s][1/(cos 1^o 6') - 1]  = [e²/2Kr_11s][0.0002]   = 2.75 x 13.595 x 0.0002 = 0.0075 eV .The resultant binding energy of the 2s electron in lithium atom  is reduced by 

0.0075 eV.  

    The kinetic energy of the 2s electron = (1/2)m v_2s² = (2.4/2) e²/Kr_2s and its negative potential energy = - 2.4 e²/Kr_2s .Total energy associated with the 2s electron = - (2.4/2) e²/Kr_2s = - (2.4/2)

2.4e²/4Ka_o=-1.44 x 13.595 = - 19.58 eV . The resultant binding energy of the 2s electron becomes

19.58 - 0.01 = 19.57eV.  The ionization energy of 2s electron in the lithium atom is 5.39 eV only which requires some correction in the computation.

      The pragmatic analysis of the lithium atom shows that the case of maximum shielding of nuclear charge with no inter electronic interaction gives the first ionization very close to the observed value.      

Spectral lines of Lithium atom

     With the conclusion arrived from the pragmatic analysis of lithium atom, the first few spectral lines of the Balmer series are derived. The radius of the 2s electron in the next higher orbit is r₃ = 9a_o and its total energy TE = - (1/2)[e²/Kr₃]= - (1/2)x[e²/K 9a_o] =   - (1/9) (13.595) = 1.51055  eV . When electron jumps from orbit with radius r₃ to orbit with radius r_{2 ,} the transition energy ΔE = 1.8882 eVand its corresponding wavelength λ₃→₂ = 657.1 nm.. It represents the bright, dominant lines caused by transitions from higher p orbitals to the lowest s orbital. This includes the most famous lithium resonance line at 670.8 nm, which gives off the bright, crimson-red color seen in flame.

    In general the radius of the 2s electron of the lithium atom  at its n^th orbit r_n = n²a_o  and its total energy - (1/n²) 13.595 eV

                                   n = 4 to n =2, ΔE₄→₂ = 0.1875 x 13.595 = 2.549 eV ; λ₄→₂    486.5nm

                                              n=4 to n=3,  ΔE₄→₃ = 0.0486 x 13.595 = 0.661eV ; λ₄→₃    1875.94 nm

Lithium produces characteristic, brightly colored emission lines in the visible spectrum. The most prominent atomic transition corresponds to a bright crimson-red line at exactly λ = 670.78 nm. Another commonly observed spectral line lies at λ = 610.36 nm (orange-red)