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 .