Journey through an array of dots with one excluded is an interesting puzzle.

It is generalized that one can touch all the dots and paths once in one line drawn from beginning to end.It is possible if the array is in the form 2n x 2n or 2n x (2n+1) but not  (2n +1) x (2n+1). In the odd x odd array a portion will be left untouched.

 

இது போன்ற அமைப்புக்களில் ஒரு புள்ளியிலிருந்து  தொடங்கி கையை எடுக்காமல் ஒரே கோட்டில் அனைத்து தடங்களையும் கடந்து அனைத்துப் புள்ளிகளையும் இணைக்கவேண்டும்

இதில் உள்ள ஒவ்வொரு புள்ளியை அடைவதற்கும்  வெளியேறுவதற்குமாக பல  வழிகள் உள்ளன .ஒரு வழி மூலம் உள்ளே நுழைந்தால் மற்றொரு வழி மூலம் வெளியேற  முடிகின்றது. முதல் அமைப்பில் எல்லா ப் புள்ளிகளும் இரு வழிகொண்ட வை . தடத்தில் அல்லது புள்ளியில் எங்கிருந்து தொடங்கினாலும்  அனைத்துப் புள்ளிகளையும்  தடங்களையும் ஒரே கோட்டில் ஒருமுறை மட்டும் கடந்து நிறைவு செய்யமுடியும் . இரண்டாவது அமைப்பில் ஒரு புள்ளி ஒரு வழிப் பாதையையும் மற்றொரு புள்ளி மூன்று வழிப் பாதைகளையும் கொண்டுள்ளன . 1 அல்லது 3 லிருந்து தொடங்கினால் முயற்சி வெற்றிபெறும் . 2 லிருந்து தொடங்கினால் வெற்றிபெறுவதில்லை . மூன்றாவது அமைப்பில் 3 லிருந்து தொடங்க முயற்சியில் வெற்றி பெறமுடிகிறது .நான்கு மற்றும் ஐந்தாவது அமைப்புக்களில் ஒற்றைப்படையில் பாதை கொண்ட புள்ளிகள் இல்லை. எங்கிருந்து தொடங்கினாலும் வெற்றிபெறமுடிகின்றது . ஆறாவது அமைப்பு இதை உறுதி செய்கின்றது . 2 லிருந்து தொடங்கினால் ஒரு பகுதி விடுபட்டுப் போகின்றது . 5  லிருந்து தொடங்க அனைத்துப் புள்ளிகளையும் ,தடங்களை ஒரே கோட்டில் இணைக்க முடிகின்றது .

 

 Research article-2

Utilization of Bohr’s Theory of Hydrogen to   Helium atom

Dr.M.Meyyappan*

[Professor of Physics (Retd), Alagappa Government Arts College, Affiliated to Alagappa University, Karaikudi- 630003, Tamilnadu, India,  *email id: meydhanam@gmail.com Orcid No:0000 0002 5194 2992]

Abstract

[An attempt is made to optimize Bohr's atom model—originally developed for the hydrogen — to the helium atom, a simplest many electron system as a first step. Correction due to electron-electron interactions are taken into account.  The possibility of hyper excitation and hyper de-excitation is proposed where two electrons in the same orbit jump simultaneously into another orbit. Normal excited states and its de-excitation are also studied by proposing half-half mixing of two allowed states. With the help of this ingenious approach, various energy levels of helium atom are calculated and its spectral feature is studied by calculating the energy radiated out in certain selected allowed transitions. The result is compared with the observed data.  The branched de-excitation of excited states is possible when two or more orbital electrons are present in the atom. The branched de-excitation in helium atom is studied with and without an electron in the innermost orbit. The correction due to the relativistic variation of mass was applied to helium ion and   helium atom. It accounts for the non-existence of helium negative ion and explains how far the radius and orbital velocity of the orbital electron change in helium atom when the orbital electrons exist in different orbits.]

Introduction

      Bohr's theory introduced the concept of quantized energy levels in atoms and successfully explained the hydrogen spectrum. The model was successfully applied to hydrogen like ions. Due to some limitations [1) the non-relativistic Bohr’s theory of hydrogen atom failed to explain the spectral feature of atoms with two or more electrons.  However, it was later replaced by more advanced quantum mechanical models that provided a more complete understanding of atomic structure and behavior. When Bohr's atomic theory was reviewed with an idea of extending Bohr's atomic theory to atoms with higher atomic number, the first thing that struck the mind was the relativistic variation of mass of the orbital electrons moving rapidly in circular orbits. The Bohr’s theory of hydrogen atom is redone [2] with and without considering the relativistic variation of mass of orbital electron. When a free proton and an electron are allowed to form a hydrogen atom, half of the loss of its potential energy is converted into its kinetic energy required to move in a stable orbit. surrounding the proton and the remaining half is deposited as relativistic increase of mass. The mass-energy conversion is accomplished involuntarily within the space of nuclear field. The orbital electron binds with the proton where the atomic binding energy comes from the energy equivalent of the added mass of the system. This is the energy that is radiated out during the electronic transitions i.e., ΔE = Δm c² = (m_n1 - m_n2)c² = E_n2 - E_n1

           In this article, optimization of Bohr’s theory of hydrogen atom is described with helium atom .The results obtained with non-relativistic Bohr model of helium atom are compared with that of relativistic model. The hyper excitation and de-excitation is proposed where two electrons in one orbit jump simultaneously into another orbit. To estimate the total energy associated with helium atom in any one of its electronic states, a new method of computation called half-half mixing of two different states is proposed.  The normal excited states and its de-excitation is studied elaborately. The energy radiated out in allowed electronic transition and the corresponding wavelength of electromagnetic radiation emitted out are predicted. Yet another possibility of electronic transition called branched de-excitation is pointed out, where both the electrons which are initially in different orbits other than the innermost orbit and the electronic transition takes place from the outer orbit to the innermost orbit. The spectral feature of helium atom is compared with Relativistic Bohr model.         

Non-relativistic model of helium atom

      

         Fig.1.Helium atom

      In the helium atom the two electrons in the 1s orbit are stable with positive binding energy. They have both kinetic energy due to its motion in the circular orbit and potential energy due its position in the electrostatic field within the atom. The kinetic energy of an electron is (1/2) mv² and for both the electrons which are identical in the system K.E is mv². The electron is stable in the orbit by balancing out the resultant electrostatic force with the centrifugal force due to circular motion. Let us suppose that the two electrons are in the 1s orbit itself having radius r_1.The centripetal force acting on an orbital electron due to nuclear attraction is 2e² /Kr₁² . The other electron in the same orbit is repelled by the electron already existing there and both of them are positioned at opposite end of a diameter of the orbit. Since all helium atoms have identical spectral feature the electrons in an orbits take fixed relative position that is invariant with respect to time.    

      The electrostatic force of repulsion between the two electrons which are diametrically opposite is e²/4Kr₁².The resultant force is balanced by centrifugal force i.e., 2e²/Kr₁² - e²/4Kr₁² = (7/4)e²/Kr₁² = mv₁²/r₁ . The radius of the orbital electron in helium atom can be determined from the condition insisting quantized angular momentum. It gives r_n = n²(4/7) a_o.  = 0.5714 n² a_o. If we ignore the electron-electron interaction r_n = a_o/2 .The radius of the 1s orbit in He⁺ ion is increased by 0.0714 a_o due to mutual electron-electron interaction in helium atom.

      Total kinetic energy of the electrons in the helium atom = mv₁² = (7/4) e²/Kr₁.Potential energy of electron 1 = - 2 e² /Kr₁ and the potential energy of electron 2  in the presence of electron 1 = - 2 e² /Kr₁ + e²/2Kr₁ . It gives its total energy as  - (7/4) e²/Kr₁ = - (49/8) [e²/2Ka_o]=  -83.269 eV which provides enough binding energy to the system. It gives the first ionization energy 83.269 - 54.368 = 28.901 eV. The experimental value of first ionization energy of helium is 24.481 eV which is 4.41 eV smaller than the theoretically predicted value.       

      Alternatively very same result can be derived instead of moving the electrons towards the nucleus to place them in the prescribed electronic orbits, the nucleus with two units of positive charge is taken from infinity towards the center of the orbit with two electrons whose circular motion is induced as the nucleus moves closer towards the coupled electrons or the coupled two electrons move together  vertically  towards  the nucleus  until all of them come to a plane with nucleus at the center of the coupled electrons.

(a)                                                 (b)

 (a).Bring proton from infinity to the center of a smallest                                                                                  orbit  with two electrons diametrically opposite

(b).Bring an orbit with two electrons diametrically opposite                                                                      from infinity towards a proton until all of them are in a plane

          Fig.2. Constructing Helium atom

       As the two electrons are kept with a distance of separation 2r₁ initially the given system of coupled electrons repel each other with a force e²/4Kr₁² and have an initial potential + e²/2Kr₁. To find out the potential energy of the system, the nucleus with charge 2e⁺  is taken vertically  from infinity to the center of the coupled electrons and the total work done is determined . When the nucleus is at a distance x  from the center, the force experienced by  it  due to  electron-1 is 2 e²/K (x ²+r₁² ) and its component along the direction of displacement  is  2e² cosφ /K(x ² + r₁² )  =  2 e² x /K(x ² + r₁² )^3/2 . The resultant component due to both the electrons is 4 e² x/K (x² + r₁²)^3/2. The parallel component of electron-1 is nullified by the equivalent component of electron-2. Since F = - dU/dx , the change in potential energy  U =  -  _∞∫^o Edx = _∞∫^o Fdx   =[4e²/K] _∞ ∫^o x dx /(x² + r₁²)^3/2  = -[4e² /K (x² + r₁²)^1/2]_∞^o  = - 4 e²/Kr₁, where E is the electric intensity at x. Adding the initial potential energy associated with the coupled electrons the total potential energy of the system then becomes - 4 e²/Kr₁ + e²/2Kr₁  =   - (7/2) e²/Kr₁. 

    The parallel component of electrostatic force acting on electron-1 towards the center of the orbit is 2e²r₁/K(x² + r₁²)^3/2 .At the end of the displacement of the nucleus this force becomes maximum and is equal to 2e²/Kr₁^2. Taking into account the initial electron-electron repulsion the total centripetal force 2e²/Kr₁² - e²/4Kr₁ = (7/4)e²/Kr₁² which induces the circular motion with centrifugal force mv₁² /r₁ . It gives the kinetic energy of the system as (7/4)e²/Kr₁. By adding the kinetic energy of the electro ns with its potential energy, the total energy becomes - (7/4)e²/Kr₁.  Very same result can be obtained by keeping the helium nucleus at a point and the coupled electrons separated by a distance 2r₁ is moved from infinity towards the nucleus vertically.

     When the condition -1 of Bohr's theory is applied the circumference of the n th orbit must be equal to n times the wavelength of matter-waves associated with the orbiting electron.

         (2πr_n)² = 4π² r_n² =  n² λ² =  n² h²/m² v_n² =   (4/7) n² h² 4 πε_o r_n /m e²

                              r_n =  (4/7)n²h²εo/mπe² = (4/7) n² a_o                                    ...... (1)

Substituting the value for r_n the total energy becomes - (7/4)²  e²/n² K a_o  =  - (49/8) (1/n²) 13.595 eV.   It gives the total energy associated with the helium atom as - 83.269 eV. The sum of first and second ionization energies of helium is 24.481 + 54.403 = 78.884 eV.

Total energy of electrons in the helium atom assembled

     The same result can be arrived by computing the component of energy in assembling the helium atom. Let us suppose a helium atom is assembled with a nucleus having 2 units of positive charge and two separate electrons.

 

Fig.3.Helium ion + electron → Helium atom

     There are two stages in the assembling. (1) e-1 is placed at r₁ (= a_o/2) to form a helium ion and (2) when e-2 is brought from infinity to the orbit having radius r₂ (= 4 a_o/7) the e-1 at r₁ is shifted to take up a new orbit having the same radius r₂ 

     When electron (e-1) is brought closer to the nucleus, it gains acceleration due to nuclear force of attraction and it starts making a circular motion around the nucleus. In the first stage helium ion is formed.  Let r₁ be the radius of its orbit. The ionized helium atom with single electron has both kinetic and potential energies. Its kinetic energy can be determined by equating the electrostatic force and centrifugal force 2 e²/Kr₁  =  mv₁² /r₁    or mv₁² = 2e²/K r_1..Kinetic energy = (1/2) m v₁²  = e²/K r₁  where  r₁ =  h²ε_o/2mπe² = a_o/2 .In terms of Bohr radius the kinetic energy becomes 4[e²/2Ka_o]. Potential energy of the electron is determined by evaluating the work done in taking the electron from infinity to the assigned orbit. It is worked out as - 2 e² /Kr₁ = - 8 e²/2Ka_o  .Total energy of the system is  4[e²/2 Ka_o] - 8e²/2K a_o =  - 4e²/2Ka_o = 54.38 eV This is the second ionization energy of helium. This is in good agreement with the practical value 54.403 eV

     It gives a formula for the Z^th ionization energy of hydrogen-like ions of all elements. If Z is the atomic number, then Z e² /Kr₁²   = m v₁² / r₁ or m v₁² = Z e² / K r₁.Kinetic energy = (1/2) m v₁² = Z e²/2K r₁ = Z²  e²/2K a_o   where  r_{1 =}   h²ε_o /Zmπe²  = a_o /Z .Potential energy  = - Z e² /Kr₁ = - Z² e²/Ka_o  and the total energy of the system is      -Z² e²/2 K a_o  = - Z²[e²/2Ka_o] = - Z² x 13.595  eV .

Stage.1.  Helium ion + electron = helium atom 

    Now the second electron is placed in an orbit of radius r₂ . In the final assembly both the electrons are in the same orbit having radius r₂ and both of them have same kinetic and potential energies as they are identical in all respect. When e-2 is brought from infinity to r₂ , e-1 is shifted from r₁ to r_{2 .}The potential energy of e-2 in the presence of nucleus is  _∞∫^r₂  [2e²/Kx²] dx  =  [2e²/Kx] _∞^r2   = 2e² /Kr₂  = - 7[e²/2K a_o] . The increase in the potential energy of e-2 due to e-1 occurs when e-1 is displacing from r₁ to r₂ . In calculating the increase in potential energy of e-2 due to e-1 the electron e-1 is supposed to be at its mean position (1/2) (r₁ + r₂) = (15/28) a_o . When e-2 is at an intermediate distance x away from  the central nucleus , the  repulsive  force F experienced by it due to e-1 is   -e² /K [x+ (15/28)a_o]² .  For an infinitesimal small displacement dx towards the nucleus, the workdone dw = F dx. Total work done is stored as its potential energy.

Potential energy of the e-2 = _∞∫^r2  - e² /K (x+ (15/28)a_o)²] dx.

   [e² /K (x+ (15/28)a_o)]_∞^r2 = e² /K [r₂+ (15/28)a_o] =  (e² /K) {1/[(4/7) a_o+ (15/28)a_o]} = 28  e² / 31K a_o      =  (56/31) [e² /2Ka_o]

This induced potential is shared by both the electrons, each electron has an increase of potential by (56/62)[e² /2Ka_o]

Stage.2.shifting of e-1 from r₁to r₂

       When e-2 reaches its assigned orbit of radius r_2, the electron e-1 is shifted back from r₁ to r₂ .In the final position,   the condition of electrostatic force is equal to centrifugal force   requires

       2e²/Kr₂² - e²/4Kr₂² = (7/4) e²/Kr₂² = mv₂² /r₂

The kinetic energy gained by e-2 is (1/2) m v₂² = (7/8)e²/Kr₂  = (49/16) [e²/2Ka_o]. Due to induction, the kinetic energy of e-1 is decreased. It is dropped from 4 [e²/2Ka_o] to (49/16)

[e²/2Ka_o]. It is equal to - (15/16) [e²/2Ka_o]. The decrease in kinetic energy of e-1 is  (1/2) m v₁² -

(1/2)m v₂² =  e²/ Kr₁ - (7/8)e² /Kr₂ =  4e² /2Ka_o - (49/16)  [e²/2Ka_o] =(15/16)  [e²/2Ka_o] 

      When e-2 is at infinity, the potential energy of e-1 is increased due to its displacement from r₁ to r₂. 

A change in potential energy of e-1 occurs in the presence of nucleus only  _- [2e² / K x]_r1^r2

= - 2e²/K[1/r₂ - 1/r₁] = - 2e²/K[ (r₁ - r₂)/ r₁ r₂ ]=   ^- 2e²/K( - 1/4a_o) =  [e²/2Ka_o]

When e-2 is r₂ , the change in the potential energy of e-1  due to nucleus is  [e²/2Ka_o] and due to the presence of e-2,  [e²/K][1/(x+r₂ )]_r1^r2 = [e²/K][ 1/2r₂ - 1/(r₁+ r_{2 )}] = [e²/K] [ 7/8a_o - 14/15 a_o] = [e²/2Ka_o][-7/60] .  The actual change of potential energy of e-1 is taken as the mean of change of potential energy of e-1 when e-2 is at infinity and e-2 is at r_2, where the change of potential due to electron-electron interaction is equally shared by the participants.    The   change of potential energy of e-1 when e-2 is at infinity is [e²/2Ka_o] and when e-2 is at r₂ is [e²/2Ka_o] - [7/60][e²/2Ka_o] which give a mean as [1 - 7/120][e²/2Ka_o].= (113/120)[e²/2Ka_o]  

Kinetic energy of e-1 when e-2 is present [e²/2Ka_o][4 - 15/16] = (49/16) [e²/2Ka_o]                                                                 

Potential energy of e-1 when e-2 is present [e²/2Ka_o][ -8 +(56/62) +1 -7/240] = - 6.126 [e²/2Ka_o]                                          

Kinetic energy of e-2 [e²/2Ka_o](49/16)                                                                                                             

Potential energy of e-2 [e²/2Ka_o][ -7+ (56/62) - 7/240 ] = - 6.126[e²/2Ka_o]                                                          

Total energy of the helium atom is the sum of all the four components and is equal to 2[3.0625 - 6.125][e²/2Ka_o]  = -6.125 x 13.595 = -83.269 eV

     The relativistic variation of mass where energy can be exchanged with the mass of the moving electron, spin-spin interaction between electrons and with nucleus  may be responsible for this small difference between practical and theoretical values of total energy of the system. By studying the total energy of helium like ions, one can find the cause of deviation. Let Ze be the nuclear charge with two 1s electrons in the helium like ions. The radius of the innermost orbit is given by

     Ze²/Kr₁² - e²/4Kr₁² = (Z-1/4) e²/Kr₁²

      [(4Z-1)/4]e²/Kr₁ = mv₁²   which gives r₁ = [4/(4Z-1)] a_o

Kinetic energy of the electrons = mv₁² = [(4Z-1)/4]e²/ Kr₁ = [(4Z-1)² /8]e²/2Ka_o

Potential energy of the electrons = -2Ze²/Kr +e²/2Kr =(e² /Kr)[-2Z + 1/2]=- (e²/2Kr) [4Z-1]             

                                                           = - [(4Z-1)² /4]  e²/2Ka_o

Total energy of the system = -   [(4Z-1)² /8]e²/2Ka_o

Using this relation the theoretical value of total energy is worked out and compared with its experimental value [the sum of z ^th and (z-1) ^th ionization energy]

Table.1. z ^th and (z-1) ^th ionization energies of first few helium like ions

..............................................................................................................................................               Z     Symbol       x 13.595 eV      Total energy    Ionization     experimental    Difference                                   

     Z     (Z-1)          value          D        D/Z

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

2             He    -(49/8) = 6.125        - 83.269       54.403    24.481     -78.884         4.385     2.19

3             Li⁺     - (121/8)= 15.125  -205.624     122.419    75.619   -198.038         7.586     2.52

4             Be²⁺   -(225/8) = 28.125 -382.359       217.657   153.85   - 371.507       10.852    2.71

5             B³⁺  - (361/8) =  45.125   -663.474    340.127   259.298  -599.425       64.049    12.81

6             C⁴⁺   -(529/8) = 66.125   -848.969        489.84      391.986  -881.826      32.857   5.476

7             N⁵⁺ -(729/8) = 91.125  -1238.844       666.83     551.925  -1218.755     20.089   -2.87     

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

Hyper excited states and de-excitation 

Fig.4. Helium atom - normal, hyper excited and  normal excited states

     When sufficient energy is available both the electrons in the ground state of helium may be excited. Such excitation is called hyper excited.  When a helium atom is hyper excited, the process of de-excitation takes place in two ways. In the first way one of the electrons in the hyper excited state jumps to the inner orbits either directly or step by step with intermediate energy levels. Then the second electron follows its own de-excitation. In the second way called hyper de-excitation both the electrons jumps to the lower energy level. The hyper excitation of helium is possible in dense stars enriched with helium at high temperature which provide more probability for hyper excitation to happen. Since the instability of the atomic system is increased many-fold, the hyper de-excitation will be faster than the normal de-excitation. The hyper excitation and de-excitation are not possible in hydrogen atom, a single electron system.

Fig.5. Hyper excitation and Hyper de-excitation

         Total energy of helium in its n^th hyper excited state is given by - (1/n²)(49/8) (e²/2Ka_o). In two different hyper excited states with n = n₁ and n_2, the total energies are - (1/n₁²)(49/8) (e²/2K a_o) and  - (1/n₂²)(49/8) (e²/2Ka_o) respectively. During hyper de-excitation, both the electrons simultaneously jump into to any inner or innermost orbit.  In hyper de-excitation between any two states, the transition energy ΔE is given by   (49/8) (e²/2Ka_o) [(1/n₁²) - (1/n₂²)] = 83.269 [(1/n₁²) - (1/n₂²)] eV.  Since λ = hc/ΔE, where ΔE is in joules, the corresponding wavelength of radiation emitted in any transition is given by λn_2* → n_1* is 12.4 x 10^-7/(49/8) (e²/2K a_o) [(1/n₁²) (1/n₂²)]m. The star (*) is used to represent the hyper excited state. Using this formula, the transition energy and the corresponding wavelength of radiation emitted can be predicted. The energy of transition and the wavelength of radiation emitted in various hyper de-excitation are given in Table.2

                           Table.2. Hyper de-excitations and wavelengths in helium

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

                              transition       energy              wavelength                                                                      .           eV                        nm

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

          n₂* → n₁*         62.452                  19.85                                                     n₃*→ n₁*          74.017                  16.75                                                     n₄* → n₁*         78.065                  15.88                                                   n₃* → n₂*          11.565                117.22                                                    n₄* → n₂*          15.613                  79.42                                                     n₄* → n₃*         4.048                     306.32                                                 n₅* → n₃*         5.921                    209.42                                                    n₆* → n₃*           6.939                  178.70                                                  n₅* → n₄*          1.874                    661.69                                               n₆* → n₄*           2.891                   428.91

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

       The ultraviolet wavelength in the helium spectrum is most notably characterized by the strong atomic line at 58.4 nm. The strongest UV lines for astrophysical observation are at 30.38 nm and 58.43 nm. The spectroscopical study in UV region of helium atom shows that hyper excitation and hyper de-excitation have very little probability to happen under normal situations. In hot stars enriched with helium, hyper de-excitation may be one of the causes for the emission of UV radiation. Helium has several spectral lines in the ultraviolet (UV) region, which are defined as having wavelengths shorter than approximately 400 nm. Key UV lines for neutral helium include lines around 396.5 nm and 388.9 nm. It shows that the cause of transition in helium atom liberating energy in the UV radiation is not hyper de-excitation

Normal excited states and de-excitation

Fig.6. Normal de-excitation in helium atom

Method-I : Half - half mixing of two states

    Hydrogen has only one electron, so its emission spectrum is relatively simple with few lines. Helium, on the other hand, has two electrons, which means there are more possible transitions and therefore more emission lines.  When the electron jumps from the   n (n>1) to the first  orbit of neutral helium atom, the transition energy and the corresponding wavelength can be worked out by half-half mixing of two states involved in the transition. At first let us try to understand this technique. The spectral feature of helium can be studied if we are able to calculate the energy associated with an excited/ionized state of helium, where the two electrons are in two different orbits. In general the most probable transition is with the de-excitation of helium atom with one of the electrons in the innermost orbit and the other electron is in the nth orbit (n ≥ 2).  When electrons are in different orbits and making jumping  between the permitted energy levels, its relative position varies  which makes changes in its velocity which in turn alters  the radius of the electronic orbits as they have to obey the condition-1 of Bohr's theory of hydrogen.  Such perturbation in the system makes the problem of estimating the total energy of the system cumbersome. However one can solve by a technique called half-half mixing of two hyper-states.    Let E₁₁ and E_nn be the energy of the helium atom in its ground state and in its n^th  hyper-excited  state where  both the electrons are in the first and nth orbit respectively.

         In any state the system has three different components of energy- 1.kinetic energy of the electrons due to its orbital motion (KE), 2.negative potential energy of the orbital electrons by virtue of its position in the nuclear field (NPE) and 3.positive potential energy due to electron-electron interaction. (e-e).This energy is equally shared by the interacting participants.

Fig.7. Half-half mixing of two allowed states

     When both the electrons are in the innermost orbit the total energy of the system E₁₁ = NPE₁₁

+ (e-e)₁₁ + KE₁₁ = - 4 e²/Kr₁₁ + e²/2Kr₁₁ + (7/4) e²/Kr₁₁ = -(7/4) e²/Kr₁₁= (7/4)² e²/Ka_o = 83.269 eV. When both the electrons are in the same orbit labelled by n, its total energy is E_nn = NPE_nn +  (e-e ) nn + KE_nn = - (7/4) e²/Kr_nn = - (7/4)² (1/n²)e²/Ka_o. When one of the electrons is in the innermost orbit and other electron is in the n^th orbit, its total energy E_1n = NPE_1n + (e-e)_1n + KE_1n.  NPE_1n and KE_1n are half of the sum of total negative potential energies and kinetic energies respectively with electrons in the innermost and n th orbit

              NPE_1n = -[2e²/K][1/r₁₁ + 1/r_nn] = - [2e² /Kr₁₁][1 +1/n²] - [2e² /Kr₁₁][(n² +1)/n²]

                KE_1n = [(7/8)e² /K][ 1/r₁₁ + 1/r_nn] = [(7/8)e² /K][(n² +1)/n²]

             (e-e)_1n = e²/Kd_1n where d_1n is the distance between the two electrons when they are in two different orbits. But d_1n = r₁₁ + r_nn and e²/K = 2(e-e)11 r11

                        = 2[(e-e)₁₁x r₁₁]/(r₁₁ + r_nn) = 2(e-e)₁₁/ (1+n²) =[e² /Kr₁₁][1/(n²+1)]

E_1n = [e²/Kr₁₁]{[(n² +1)/n²][(7/8)-2] + [1/(n² + 1)]}

This expression can be verified by computing the total energy associated with the system with one electron in the innermost orbit and other electron is removed.

E_1∞ = [e²/Kr₁₁][ 1 x (-9/8)] = -[9x7/16] e²/2Ka_o = 53.53 eV. The observed second ionization energy of the helium atom is 54.4 eV.

     When an excited helium atom exists  with one electron in the innermost orbit and the other electron in the n^th orbit , then the state E_1n gets 1/2 share of  kinetic and negative potential  energies  from each contributing states (E₁₁ and E_nn), which  are due to the interaction with stable central nucleus. The sharing of electron-electron interactional energy is different. When an electron in lower energy state (m) donates its energy to higher energy state (n) it carries an amount of energy (e-e)_mm [m²/(n²+m²)]. When an electron from higher energy state (n) donates its energy to lower energy state (m) it carries an amount of energy + (e-e)_nn [n²/(n²+m²)].

         This energy contributed by E₁₁ is utilized by E_1∞ to have more potential and kinetic energy with higher binding energy. E_1∞. = (1/2) [KE₁₁+ NPE₁₁] + (e-e)₁₁ [l/(1+∞)] =_. (1/2) [ KE₁₁+ NPE₁₁ ]  .In fact E_1∞ refers the second ionization of helium atom and is equal to (1/2) [E₁₁ - (e- e)₁₁] = (1/2)[(-83.269) - (7/4)13.595]  = -53.53 eV   and the energy contributed by  E₁₁ to E_1∞ is the first ionization energy i.e., (-83.269) + (53.53) = 29.739 eV.  .E_nn is formed by mixing of two identical states E_nn each state contributes 1/2 of its energy and keep the energy same. When a hyper state E_nn is formed by half-half mixing of two hyper-states E_nn and E_{nn  ,} then its energy by this method  becomes  2 (1/2)[ KE_nn + NPE_nn] + 2(e-e)_nn   n²/2n²]  = E_nn = [KE_nn + N PE_nn + (e-e ) nn] . 

     The above expression for E_nm can be derived by yet another way. When an excited helium atom exists  with one electron in the inner orbit labelled m  and the other electron in the outer orbit   n , then the state E_mn gets 1/2 share of  kinetic and negative potential  energies  from each contributing states (E_mm  and E_nn) , which  are due to the interaction with stable central nucleus. The sharing of electron-electron interactional energy is different. When an electron in lower energy state (m) donates its energy to higher energy state (n) it carries an amount of energy (e-e) mm [m²/(n²+m²)].  When an electron from higher energy state (n) donates its energy to lower energy state (m) it carries an amount of energy + (e-e)_nn [n²/(n²+m²)].

         In helium atom (e-e)_mm = e²/2kr_m, (e-e)_mm = e²/2kr_n and  (e-e)_mn = e²/k(r_n + r_m). Let us suppose that the fraction of contribution by the hyper excited is x. Then x[e²/2kr_m] + (1-x) [e²/2kr_n] = e²/k(r_m+ r_n) .By solving, we can determine x,                                                                    x [1/r_m - 1/r_n] = 2/(r_m+ r_n) - 1/r_n   ;  x = r_m /(r_m + r_n)  and (1-x) = r_n / (r_m + r_n)

Since r is proportional to square of its orbital quantum number x = m²/(m² + n²) and (1-x) = n²/(m² + n²).

E_mn =(1/2)[KE_mm + NPE_mm + KE_nn + NPE_nn] + (e-e)_mm [m²/(m² + n²)] + (e-e)_nn[n²/(m² + n²)]

Since (e-e)_mm = (e-e)₁₁/ m² and (e-e)_nn  = (e-e)₁₁/n² , (e-e)_mm = (m²/n²)(e-e)_nn. Substituting this value in the above relation, we get, 

E_mn =(1/2)[KE_mm + NPE_mm + KE_nn + NPE_nn] +2(e-e)_mm [m²/(m² + n²)]

Lyman series for Helium atom 

                                   Fig.8.De-excitation from n=2 to n=1 (Type-I)

       On the basis of this description let us make an attempt to find the total energy associated with an ionized helium where one electron is in its innermost orbit and the other in next higher orbit. The energy associated with the system E₁₂ is derived from the calculated energies of the systems E₁₁ and E_{22. T}he radius of the inner most electronic orbit is (4/7)a_o and the radius of the next higher orbit is (16/7)a_o The three components of energy pertaining to the ground state of helium atom is kinetic energy  KE₁₁= (7/4) e²/Kr₁₁  ; negative potential energy NPE₁₁ = - 4e² /Kr₁₁  and the positive electron-electron interaction energy (e-e)₁₁ = e²/2Kr_11. The corresponding components for next higher energy state are KE₂₂ = (7/4) e² /Kr_22, PE₂₂ = - 4e² /Kr₂₂ and (e-e)₂₂ =  e²/2Kr₂₂ . The energy of excited helium in its state E₁₂ is {(1/2) [KE₁₁ + KE₂₂ +NPE₁₁ + NPE₂₂] + 2(e-e)₁₁ (1/5)  eV. Substituting the values of each component we get [e²/2Kr₁₁][ -(9/4) (5/4)+(2/5)] 

= 57.396 eV. When electron jumps from n=2 to n=1, the energy liberated ΔE_(12→11)  is 83.269 57.396 = 25.873  eV . The wavelength of this radiation corresponds to 12.4 x 10^-7 /25.873 = 47.926 nm

     Knowing this technique, one can derive a formula suitable for any electronic transition in helium atom. Consider an excited state of helium atom where one electron is in the n^th orbit and other electron in the innermost orbit.  The energy associated with the system E_1n can be computed as before from its contributors  E₁₁ and E_nn .The energy contributed by E₁₁  to  E_1n  is (1/2)[ KE₁₁ +NPE₁₁], the energy contributed by E_nn to E_1n is (1/2)[ KE_nn + NPE_nn], and the electron-electron interactional energy in the resultant assembly is (e-e)_1n  = 2(e-e)₁₁[1/(n² +1)] = 2(e-e)_nn [n²/(n²+1)] _,.  The energy associated with E_1n is the sum of these three contributions and is equal to (1/2)[(KE₁₁ + KE_nn) + (PE₁₁ + PE_nn)] +  [2(e-e)₁₁ [(1/(n²+1)] . When the electron jumps from n ^th orbit to the innermost orbit  _, the transition energy is the energy difference between these two states. It is  E_1n -  E₁₁ = (1/2)[(KE₁₁ +NPE₁₁) + (NPE_nn + KE_nn)] +  2(e-e)₁₁ [(1/(n²+1)] - [KE₁₁ +NPE₁₁ + (e-

e)₁₁] =  (1/2) [KE_nn - KE₁₁  + NPE_nn  - NPE₁₁] + 2(e-e)₁₁ [(1/(n²+1)]- (e-e)₁₁   = (1/2)[ KE_nn - KE₁₁  + NPE_nn  - NPE₁₁] + [(1-n²)/(1 + n²)][(e-e)₁₁]  Substituting the values of various components of energy we get (E_1n -  E₁₁) ={(1/2)[49/8n² - 49/8 - 14/n² + 14]+ [(1-n²)/(1 + n²)](7/4)} (e²/2Ka_o). On simplification, ΔE_{(1n→ 12)} = [(1-n²)/2n²]{-63/8 +(7/2)[n²/(1+n²)]}  (e²/2Ka_o). When an electron jumps from the orbit n=2 to the innermost orbit, the transition energy (E₁₂ -  E₁₁) is derived by using the above formula    ΔE_{(2→ 1)}  =(13.595) (-3/8)[ -(63/8)+(14/5)]  =  25.873 eV and  λ₂ →₁ = 12.4 x 10^-7 /25.873= 47.93  nm  

      For any transition E_1n → E_11, the transition energy is given by [(1-n²)/2n²]{-63/8 +(7/2)[n²/(1 +n²)]}  (e²/2Ka_o). Using this formula the Lyman series for helium atom can be predicted. Table.2 given below gives transition energy and the corresponding wavelength for various possible transition from orbits with n ≥ 2 to the innermost orbit with n =1 .

                            Table.3.Spectral lines in Lyman series of normal helium atom

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

                                   transition    energy    Wavelength

                                                       in eV           in nm

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

n₂→n₁   25.87          47.93                                         n₃→n₁    28.55         43.43                                       n₄→n₁  29.19       42.48                               ..........................................................                         

The radiation components in this region are the resonance transitions from atomic helium originating from the upper n  (n = 2, 3, 4,---) P state to the lower n= 1 , ground state S . The observation of atomic resonance emission shows 58.43 nm and 30.38 nm.(Lyman series in Helium atom)   The helium spectrum in the ultraviolet range includes several prominent lines, with the strongest being at 58.43 nm and 30.38 nm. Additionally, other lines can be found between 60-110 nm

      There is yet another way by which the Lyman series may take place, where both the electrons are in the same orbit with n greater than 1 and one of the electrons jump from the orbit to the innermost orbit.  For example, in the initial state the helium atom is in its first hyper-excited state, where both the electrons are in the second permitted orbits. During transition, one of the electrons jumps to the innermost orbit. Usually this transition will be followed by another successive transition where the remaining electron in the second orbit will jump to the innermost orbit with half-filled. When the hyper de-excitation is hindered by some reasons, the processes of de-excitation takes place in steps.

The energy of the atom in its initial stat E₂₂ = KE₂₂ + NPE₂₂ + (e-e)₂₂

The energy of the atom in its final state E₁₂ = (1/2)[KE₂₂ +NPE₂₂  + KE₁₁ + NPE₁₁] + 2(e-e)₁₁ (1/5) 

The transition energy is given by ΔE_{(22→ 12)  =} E₂₂ -E₁₂ = (1/2)[KE₂₂ + NPE₂₂  - KE₁₁ - NPE₁₁] + [(e-

e)₂₂ -(2/5) (e-e)₁₁].Substituting the values of the components of energy  we get the transition energy as (1/2)[(7/4) e²/Kr₂₂ - 4 e²/Kr₂₂ - (7/4) e²/Kr₁₁  + 4 e²/Kr₁₁] + (1/4)e²/2Kr₁₁ - (2/5) e²/2Kr₁₁  

= (1/2)[e²/Kr₁₁] [(7/16) -1 - 7/4 + 4] + [e²/2Kr₁₁][(1/4) - (2/5)]= (7/4) [e²/2Ka_o] [(27/16) - (3/20)]= 2.69 x 13.595 = 36.57 eV  and  wavelength of radiation emitted is  λ₂₂^* →₁₂ =  33.9  nm

The transition energy in jumping of an electron fron the energy level E_nn  to E_1n is    E_nn - E_1n = (1/2)[KE_nn + NPE_nn - KE₁₁ - NPE₁₁] + [(e-e)_nn - 2(e-e)_{11 /}[1/(n²+1)]N = [(35+63n²)/16(n²+1)] [(n² -1)/n²]. Using this formula the other possible transition can be studied. For example E₃₃ → E₁₃ gives 27.27 nm.

     The another possibility of this kind of transition is both the electrons are in different orbits other than the innermost orbit and the electronic transition take place from the outer orbit to the  innermost orbit.

Fig. 9.Branched De-excitation of excited states under Lyman series

The branched de-excitation may happen among the excited states of helium atom with and without an electron in the innermost orbit.

Without an electron in the innermost orbit transition may happen between E_mn and E_1n or E_{1m .} For example E₂₃ can undergo transition through either E₁₂ or E₁₃.

E₂₃ = (1/2)[KE₃₃ +NPE₃₃  +KE₂₂ + NPE₂₂] +(4/13) (e-e)₂₂ +(9/13)(e-e)₃₃]                                                                          

E₁₂ = (1/2)[KE₂₂ + PE₂₂ +KE₁₁ + PE₁₁] +(1/5) (e-e)₁₁ +(4/5)(e-e)₂₂]                                                                   

E₁₃ =   (1/2)[KE₃₃ + PE₃₃ +KE₁₁ + PE₁₁] +(1/10) (e-e)₁₁ +(9/10)(e-e)₃₃]

E₂₃ → E₁₂   = (1/2)[KE₃₃ + PE₃₃ -KE₁₁ - PE₁₁] - (1/5) (e-e)₁₁ - (32/65) (e-e)₂₂ +(9/13) (e-e)₃₃                       

        = [49/144 - 7/9 - 49/16 +7 - 14/65 + 7/52 - 7/20](13.595)= 41.725 eV                                  and  λ₂₃ →₁₂₁  = 30 nm.

Similarly

E₂₃ →E₁₃ = (1/2)[KE₂₂ + PE₂₂ -KE₁₁ - PE₁₁] -(1/10) (e-e)₁₁ +(4/13)(e-e)₂₂-(9/10)(e-e)₃₃ + (9/13)(e-e)₃₃ 

          = (1/2)[ KE₂₂ + PE₂₂ -KE₁₁ - PE₁₁]  -  (1/10) (e-e)₁₁ +(4/13) (e-e)₂₂ -(27/130)(e-e)₃₃

        = [49/64 -7/4 - 49/16 +7 + 7/52 -7/40 -21/520]13.595 = 39.05 eV                         λ23 →13   =  31.75 nm.

     Even though the possibility is very little there is yet another way for the transition under Lyman series to happen. Two electrons in two different orbits with n > 1 jump simultaneously to the inner most orbit. For example E₂₃ may undergo to E₁₁

E₂₃ =   (1/2)[KE₃₃ +NPE₃₃  +KE₂₂ + NPE₂₂] +(4/13) (e-e)₂₂ +(9/13)(e-)₃₃]                                                               

E11 = KE11 + NPE11 + (e-e)11                                                                                                                                                                                                   

E₂₃ →E₁₁ = (1/2)[KE₃₃ +NPE₃₃ + KE₂₂ + NPE₂₂ -KE₁₁ - NPE₁₁] -(e-e)₁₁ +(4/13)(e-e)₂₂+(9/13)(e-e)₃₃   = {(1/2)[49/72 -14/9 +49/32 -14/4 - 49/8 + 14] -(7/4) +(4/13)(7/16) + (9/13)7/36} 

           

(e²/2Ka_o)         = {(1/2) [ 5.0312] - 1.4807 }(e²/2Ka_o) = 14.069 eV                                                                                         λ23 →11   =  88.13 nm.

The whole Lyman series of helium atom falls in UV region. The wavelength of the emitted radiation in this series is around 30-60 nm

Balmer series of Helium atom

       There are few ways by which the Balmer series  in helium atom may arise. The first one corresponds to electronic transition in helium atom where one of the electrons is in the inner most orbit, while the other electron jumps from the orbit with n  ≥ 3 to n =2 The second one corresponds to electronic transition in helium atom where one of the electrons is in second orbit and the other electron jumps from the orbit with  n  ≥ 3 to n =2 .The former case is more probable than the other due to its different transient nature.  

   

      

Fig.10. Normal de-excitation from  n=3 to n =2

        Let us calculate the energy associated with systems denoted by E₁₃   and E₁₂ where one of the electrons is in the innermost orbit and other electron is in the third and second orbit respectively. By using the half-half mixing the energy content of the systems can be evaluated. For the system E₁₃, the contributors are E₁₁ and E₃₃ and for the system E₁₂ they  are E₁₁ and E₂₂ 

E₁₃ = (1/2) [KE₁₁ + NPE₁₁ + KE₃₃ + NPE₃₃] + (e-e)₁₁ (1/10) + (e-e)₃₃(9/10)

E₁₂ = (1/2) [KE₁₁ +NPE₁₁ +KE₂₂ + NPE₂₂] + (e-e)₁₁(1/5)+ (e-e)₂₂(4 / 5)

ΔE(₁₃→ ₁₂)  =E₁₃ - E₁₂ = (1/2)[KE₃₃ + NPE₃₃ - KE₂₂ - NPE₂₂ - (e-e)₁₁(1/10) - (e-e)₂₂(4/5) +(ee)₃₃(9/10)

Substituting the values for all the components of energy, we get

ΔE (₁₃→ ₁₂) = {(1/2) [49/72 - 14/9  - 49/32 + 7/2 ] +[- 7/40 + 7/40 - 7/20]}(e2 /2Kao) =  [0.1969]

13.595 = 2.6768 eV and λ₁₃ →₁₂ = 463.24 nm

In the Type II transition,  it is E₂₃ → E₂₂  , where

E₂₃ = (1/2) [KE₃₃ + NPE₃₃ + KE₂₂ + NPE₂₂]  +  (4/13) (e-e)₂₂ + (9/13) (e-e)₃₃ 

E₂₂ = KE₂₂ + NPE₂₂ + (e-e)₂₂

ΔE(₂₃→ ₂₂)  =E₂₃ - E₂₂ = (1/2)[KE₃₃ + NPE₃₃ - KE₂₂ - NPE₂₂] + (9/13)[(e-e)₃₃ - (e-e)₂₂]

               = {(1/2)[ 49/72 -14/9 - 49/32 + 7/2] + (9/13) [7/36 - 7/16]} (e² /2Ka_o)

                =  {(1/2[0.6805 - 1.5555 - 1.5312+ 3.5] + (63/52)(-5/36)} (e² /2Ka_o)

               = 0.5469 - 0.1682 = 0.3787  X 13.595 = 5.1484 eV

λ₂₃ →₂₂ =   240.85 nm

By driving a formula, one can determine the wavelengths of various spectral lines of Balmer series of helium atom.

For Type-I transition,

E_1n = (1/2) [KE₁₁ + NPE₁₁ + KE_nn + NPE_nn] + [1/(n² +1)](e-e)₁₁ + [n²/(n² + 1)](e-e)_nn

E₁₂  = (1/2) [KE₁₁ + NPE₁₁ + KE₂₂ + PE₂₂] + N[1/(5)](e-e)₁₁ + (4/5)](e-e)₂₂

ΔE(_1n→ ₁₂)  =E_1n  → E₁₂  = (1/2)[KE_nn + NPE_nn - KE₂₂ - NPE₂₂] + [1/(n² +1)](e-e)₁₁ - [1/(5)](e-e)₁₁ (4/5)](e-e)₂₂ + [n²/(n² + 1)](e-e)_nn

   =(1/2)[KE_nn + NPE_nn -KE₂₂ -NPE₂₂]+[(1/n²+1) -1/5](e-e)₁₁ -(4/5) (e-e)₂₂ +[n²/(n² + 1)](e-e)_nn

  = {(1/2)[49/8n² - 49/32 -14/n² +14/4]- (7/4)[(4-n²)/5(n²+1)] -7/4 (1/5) +[n²/(n² + 1)](7/4n²)}( e² /

2Ka_o) = (49/16)[(4-n²)/4n²] - 7 [(4-n²)/4n²] + (7/10) [(4-n²)/(n² +1)] = (7/4) (4-n²) [ 7/16 n² - 1/n2 +

(2/5) [1/(n²+1)] = (7/2) (n² - 4) [13 n² + 45]/[16 n² (n² +1)]

n = 3 ;  ΔE_{(13→ 12)} = [0.1987] (13.595)    = 2.6765 eV and λ₁₃ →₁₂ =  463.3 nm

n=4 ; ΔE_{(14→ 12)} = [0.2442] (13.595) = 3.320 eV and λ₁₄ →₁₂ = 373.5 nm   

  Fig.11.Helium emission spectrum

As the nuclear charge is twice that of hydrogen, all the electronic orbits are little closer to the nucleus and as a consequence of which the electronic transition between n ≥ 3 to n=2 emits more energy which fall in UV region

Paschen series are due to the transition between n ≥ 4 to n = 3. In the most probable transition one of the electrons is bound in the innermost orbit and the other electron make transitions from orbits with n ≥ 4 to n = 3. As the process of de-excitation is not completed usually it is followed by another successive transition.

 E_1n = (1/2) [KE₁₁ + NPE₁₁ + KE_nn + NPE_nn] + [1/(n² +1)](e-e)₁₁ + [n²/(n² + 1)](e-e)_nn

E₁₃ = (1/2) [KE₁₁ + NPE₁₁ + KE₃₃ + NPE₃₃] + (1/10)](e-e)₁₁ + [9/10](e-e)₃₃

ΔE(1n→ 13) =E1n - E13 =(1/2)[KEnn + NPEnn - KE33 - NPE33]+[(9 - n2)/10(n2 +1)](e-e)11 -  [9/10](e-e)33 + [n²/(n² + 1)](e-e)_nn

                          = (1/2) [49/8n² - 14/n² - 49/72 + 14/9](e² /2Ka_o) +(7/4)[(9-n ²)/10(n²+1)]

                                                                         -9/10 (7/4) (1/9) + 7/4 [1/(n²+1)][e² /2Ka_o]

                              = [(9-n²)/9n² ] (-63/16)  + (7/20)[(9-n²)/5(n²+1)][e² /2Ka_o]

                             =    (7/4) (n² - 9) [(n² +5)/20n² (n² +1)][e² /2Ka_o] 

when n = 4, ΔE_{(14→ 13)} =E₁₄ - E₁₃ = (49/4)[21/20x16x17)](13.595)= 0.0473 x 13.595 =0.643 eV and  λ₁₄ →₁₃ =   1928.5 nm

n = 5, ΔE_{(15→ 13)} =E₁₅ - E₁₃ = (7/4)[(16x 30)/(20x25 x26)](13.595)= 0.0646 x 13.595 =0.8784 eV and  λ₁₅ →₁₃ =   1411.6 nm

n=6  , ΔE_{(16→ 13)} =E₁₆ - E₁₃ = (7/4)[(27 x 41)/(20x36 x 37)](13.595)= 0.0727 x 13.595 =0.9884 eV and  λ₁₆ →₁₃ =   1254 nm

      Normal helium spectral lines in the visible range include wavelengths around 587.6 nm (yellow), 667.8 nm (red), and 706.5 nm (red). Other visible lines also exist, such as 447.1 nm (blue-green), 492.2 nm (blue-green), 501.6 nm (green), and 667.8 nm (red). The visible part of the helium spectrum falls roughly between 388.8 nm and 781.3 nm, while the invisible parts include ultraviolet (UV) and infrared (IR) radiation. Specifically, the visible helium spectrum contains lines at 388.8 nm, 447.1 nm, 471.3 nm, 492.1 nm, 501.5 nm, 504.7 nm, 587.5 nm, 667.8 nm, 686.7 nm, 706.5 nm, 728.1 nm and 781.3 nm, . UV radiation has wavelengths shorter than 380 nm, and IR radiation has wavelengths longer than 780 nm

   The successive secondary transition followed after a transition can be identified with the energy balance relation. .If a transition is split into two successive transitions,

hν₁   +  hν₂      =     hν₃                   1/λ₁ + 1/λ₂  = 1/λ₃                        or        λ₃ =  λ₁λ₂ / (λ₁  +λ₂)

For example 728.1 nm and 781 .3 nm are two visible radiations in helium spectrum. When it happens as a single transition its wavelength will be    (781.3 x 728.1)/ (781.3 + 728.1) = 376.88 nm .

Results and Discussion

         The radius and orbital velocity of orbital electron change in helium atom when the electrons are in different orbits. When they are in the same orbit 2e²/Kr_nn² - e²/4Kr_nn² = (7/4) e²/K = (m_e)_nn v_nn² r_nn and nh/2π = (m_e)_nn v_nn r_nn .By solving these two relations we get r_nn  =(4/7)n² a_o (1- v_nn²/c²)^1/2 and v_nn = (7/4)e²/ 2nhε_o

Let us suppose that one of the electrons is in orbit n₁ and the other electron is in orbit n₂ . The  electron in orbit n₁  has radius (r_n1)_n1-n2  mass (m_n1)_n1-n2 and velocity (v_n1)_n1-n2

                2e²/K(r_n1)² - e²/K(r_n1 + r_n2)² =2e²/K- e²/K[1/(1+r_n2/r_n1)]² = m_n1v_n1²r_n1                                          

                   [e²/K][2 - [1/(1+n₂²/n₁²]²  = [e²/K](n₁⁴ +2n₂⁴ +4n₁²n₂²)/(n₁² + n₂²)²  = m_n1 v_n1²r_n1                                                                      .                                                 n₁ h/2π = m_n1 v_n1 r_n1

By solving these two relations, v_n1 = [e²/2n₁hε_o](n₁⁴ +2n₂⁴ +4n₁²n₂²)/(n₁² + n₂²)²

                                                            r_n1 =  n₁² a_o [(n₁² + n₂²)²/(n₁⁴ +2n₂⁴ +4n₁²n₂²)] (1- v_n1²/c²)^1/2       

 The electron in orbit n₂  has  radius (r_n2)_{n1-n2 ,}  mass (m_n2)_n1-n2 and velocity (v_n2)_n1-n2

      2e²/K(r_n2)² - e²/K(r_n1 + r_n2)² =2e²/K- e²/K[1/(1+r_n1/r_n2)]² = m_n2v_n2²r_n2                             [e²/K][2 -[1/(1+n₁²/n₂²]²=[e²/K](2n₁⁴+n₂⁴+4n₁²n₂²)/(n₁²+n₂²)²= m_n2 v_n2²r_n2                                                                                         n₂ h/2π = m_n2 v_n2 r_n2

By solving these two relations, v_n2 = [e²/2n₁hε_o](2n₁⁴ +n₂⁴ +4n₁²n₂²)/(n₁² + n₂²)²

                                                         r_n2 =  n₂² a_o [(n₁² + n₂²)²/(2n₁⁴ +n₂⁴ +4n₁²n₂²)] (1- v_n2²/c²)^1/2 

Knowing the orbital velocity, the binding energy of the orbital electron in orbits n1 and n2 can be estimated from (1/2)m_o v_n1² _,  (1/2)m_o v_n2² respectively. Total binding energy                                     BE_n1-n2 =  (1/2)m_o[(e²/2hε_o)²/(n₁² + n₂²)⁴][(n₁⁴ +2n₂⁴ +4n₁²n₂²)²/n₁²+(2n₁⁴ +n₂⁴ +4n₁²n₂²)²/n₂²]

           = [e²/2Ka_o][1/(n₁² + n₂²)⁴][(n₁⁴ +2n₂⁴ +4n₁²n₂²)²/n₁²+(2n₁⁴ +n₂⁴ +4n₁²n₂²)²/n₂²]

Using this general expression, one can find out the binding energy of helium atom in any of its state.

n₁= n₂= 1 ;[e²/2Ka_o] (1/16)[ 98] =(49/8)[e²/2Ka_o] = 83.269 eV

n_{1 = 2;} n₂ = 3 ; [e²/2Ka_o] [1/(13)⁴] [(322/2)² +(257/3)²] = 15.8316 eV

n₁= n₂ = 2; [e²/2Ka_o] [1/(8)⁴] [(322/2)² +(257/3)²] =  20.8173 eV

BE₁₂ - BE₂₃ = 20.8173 - 15.8316 = 4.9857 eV and  λ₂₃→₁₂ = 248.72 nm It represents  the first line of Balmer series of helium atom .

Relativistic Bohr Model of Helium Atom

    Rejuvenation of Bohr's Theory of  Hydrogen atom with the relativistic change of mass of the orbital electron predicts that the energy corresponding to any electronic transition from orbit with quantum number n₂ to n₁ is equal to the energy equivalent of  the relativistic variation of mass of the electron i.e.,  ΔE  =   Δmc² = (m_n1- m_n2)c²_. The transition energy of the electron in the hydrogen atom is derived in terms of wavelength of matter waves of the electron in the concerned orbits. It is shown that the atomic binding energy is exactly equal to the energy equivalent of relativistic increase of mass of the orbital electron.

    Many literature is available for the application of Bohr's Theory to Helium atom [3-6].The proposed relativistic Bohr model is applied to helium to verify the observed experimental data and to study its spectral lines. According to conservation of energy, the loss of total potential energy of the system is equal to sum of its total kinetic energy and energy equivalent of relativistic increase of masses of the electrons in the same orbit labelled n

                      4e²/Kr_nn - e²/2Kr_nn = (7/2) e² /Kr_nn =  m_nn v_nn²   +  dm_nn c²

here dm_nn represents the relativistic increase of mass of both the electrons in the orbit n. Since m_nn v_nn²  = 2e²/Kr_nn - e²/4Kr_nn = (7/4) e²/Kr_{nn ,} dm_nn c² = (7/4) e²/Kr_nn, where m_nn, v_nn  are the mass and velocity of the electron in the orbit  n,when both the electrons are in the same orbit having radius r_nn. The loss of potential energy of both the electrons is equally shared by its kinetic energy and energy required for its relativistic increase of mass.

From the relations m_nn v_nn² r_nn = (7/4) e²/K and m_nn v_nn r_nn = nh/2π , the radius of the orbit of the electrons  and its orbital velocity in helium atom can be derived as r_nn = (4/7) n² a_o [1-v_nn²/c²]^1/2   ≃ (4/7) n² a_o [1- v_nn²/2c²] and v_nn = (7/4) [e²/2ε_onh] .  In the ground state of helium atom, substituting the values for r₁₁ and v₁₁  in dm₁₁ c^2, that measures its binding energy

       dm₁₁ c² = (7/4)² e²/ [Ka_o (1- v₁₁²/c²)^1/2]=  (49/8) x (e²/2Ka_o) x  [1/(1- v₁₁²/c²)^1/2]                               

                = 6.125 x 13.595 [1/(1- v₁₁²/c²)^1/2]

v₁₁²/c²  = (49/64)  e⁴/ε_o²h²c² = 0.000163 and (1- v₁₁²/c²)^1/2 = 0.9999184

         dm₁₁c²   =83.269 x [1/ 0.9999184] =  83.275 eV

This can be verified by computing dm₁₁c²  by calculating the relativistic increase of mass of each electron. For an electron dm = m₁ - m_o where m₁ = (4/7) h²ε_o/πe² r₁ and m_o = h²ε_o/πe² a_o. Substituting the value for r₁ we get m₁ - m_o = h²ε_o/πe² [4/7r₁ - 1/a_o] = h²ε_o/a_oπe² [(1/[(-v₁²/c²)^1/2  - 1] =  41.651 eV. For both the electrons it is 2 x 41.651 =83.302 eV.

          It is dmc² that measures the binding energy of the system. When both the electrons in helium atom are in the same orbit labelled n, dm_nnc² = (7/4) e²/Kr_nn. i.e., each electron contributes equally to the binding energy of the system. When one of the electrons is taken away from the system, the binding energy of the remaining electron will not be exactly half of the binding energy of the system. This can be verified with the practical values of first (24.48 eV) and second (54.40 eV) ionization energies of helium atom. The binding energy of helium atom is sum of its first and second ionization energies 78.88 eV.  It shows that when one of the electrons in the ground state of helium is taken away from the system, the binding energy is not halved. The outgoing or departing electron gives some fraction of its binding energy to the other electron that remains in the system. This is due to the absence of electron-electron interaction.  That is why (x+ 1)^th ionization energy is greater  than x^th.

      By using the relation BE_nn = dm_nnc²  =(439/8) [e²/2Ka_o](1/n²)[1 + v_nn²/2c² ], one can  study the process of hyper de-excitation and its spectral feature. BE₁₁ - BE_nn = (49/8) [e²/2Ka_o][(1 -1/n²) + (1/2c²) (v₁₁² - v_nn²/n²) ]. Substituting the values for v_nn = (49/64) [e⁴/h² ε_o² n²], BE₁₁ - BE_nn = (49/8) [e²/2Ka_o][(n² -1)/n²] [ 1 + (49/16) (1/m_oc²) ( e²/2Ka_o) = (83.269) [(n² -1)/n²] [1 .000081]. When n= 2 ΔE_{(22→ 11)} =BE₁₁ - BE₂₂ = 62.457 eV and the wavelength of the emitted radiation is 19.85 nm , when n =3 ,ΔE_{(33→ 11)} =BE₁₁ - BE₃₃ = 74.0232 eV and λ₃₃ →₁₁ = 16.75 nm. (Ref.Table.2)

      When the energy level of a state with the two electrons in the helium atom are in two different orbits labelled with m and n respectively the binding energy BE_mn  can be assessed by half-half mixing of two states with both the electrons are in m and n respectively. The binding energy of the mixed state with one electron in n = 1 and another electron in n = n is given by

           E₁₁ = [KE₁₁ +NPE₁₁ +(e-e)₁₁] = - dm₁₁c²

               Enn = [KEnn +NPEnn + (e-e)nn]  = -dmnnc2

          E_1n =  (1/2)[NPE₁₁ + KE₁₁ + NPE_nn + KE_nn] + (e-e)_1n  = - dm_1n c²

(e-e)_1n =  e²/K(r₁₁ + r_nn)  = 2(e-e)₁₁ r₁₁/(r₁₁ + r_nn) = 2(e-e)₁₁ /(1+n²)  = (e-e)_nn[n²/(1+ n²)]

         E_1n = (1/2)[NPE₁₁ + KE₁₁ + NPE_nn + KE_nn] +  2(e-e)₁₁ /(1+n²)  = - dm_1n c²

BE_1n = (1/2)[dm₁₁c² + (e-e)₁₁ + dm_nn c² + (e-e)_nn ] -2(e-e)₁₁/(1+n²)

BE₁₁ - BE_1n  = (1/2)[dm₁₁c² - (e-e)₁₁- (e-e)_nn - dm_nnc²] + 2(e-e)₁₁/(1+n²) 

           =  (1/2)[dm₁₁c² - dm_nnc²] + (e-e)₁₁ {-(1/2) _- (1/2n²) + [2/(1+n²)]}

Substituting the values for dm₁₁ c² _, dm_nn c²,  and (e-e)_{11 ,}

BE₁₁ - BE_1n =   (1/2) (49/8)[e²/2Ka_o] [(n² - 1)/n²]  + (7/8)[e²/2Ka_o][(-n⁴ +2n²-1)/n²(1+n²)]

                      = (7/8)[e²/2Ka_o][ 5n⁴ +4n² - 9]/ 2n²(n²+1)] 

Giving different values (n= 2,3,4,5......) the Lyman series of the helium atom can be determined.(Ref.Table 3)

  n                                  BE₁₁ - BE_1n                                            λ_{1n→ 11}                                    

n=2;            0.875 x13.595 x 2.175 =  25.873 eV            47.93 nm

n =3;          0.875 x13.595 x 2.4  =   28.550 eV              43.43 nm

n=4;          0.875 x 13.595 x 2.454 =29.192 eV              42.48 nm

 This can be verified by calculating the relativistic variation of mass of the orbital electrons in normal and excited helium atom.

The Lyman series of the helium atom can be studies by deriving a relation for its binding energy in the energy levels when the two electrons are in the states 11 and 1n

BE₁₁  = dm₁₁c²  ≃  2 [(1/2) m_o  v₁₁²] where  v₁₁  = m₁₁ v₁₁² r₁₁ /m₁₁ v₁₁ r₁₁ = (7/8) [e²/hε_o]. Substituting this value in the above relation BE₁₁  =[m_o (49/64)e⁴/h²ε_o²] = 2 x 41.681 = 83.374 eV

BE_1n  is the sum of binding energies of the electrons in two different orbits labelled with 1 and n. BE_1n = (dm₁)_1n c² + (dm_n)_1n c² 

  

Fig.12. Excited helium atom labelled 1n

First let us consider the electron in the innermost orbit. The resultant force acting on the electron is counter balanced by the centrifugal force  2e²/K - e²/K(1 +r_n/r₁)² = [e²/K][(2n⁴+4n² + 1)/(1 +n²)²]  = m₁ v₁² r_{1 .} Solving for (v₁)_1n with the condition for orbital angular momentum m₁ v₁ r₁ = nh/2π, we get (v₁)_1n = [e²/2hε_o] [(2n⁴ +4n²+1)/(1+n²)]. The   binding energy contributed by the electron in the inner most orbit is dm₁c² ≃ (1/2) m_o (v₁)_1n²   = (1/2) m_o [e³/h²ε_o²] [(2n⁴ +4n²+1)/2(1 +n²)]²  in terms of eV. When the other electron is in the orbit labelled with n the resultant force and the centrifugal force together keep it in stable orbit 2e²/K - e²/K(1+r₁/r_n)² = [e²/K][(n⁴+4n² +2)/(n⁴ +2n² +1)]  = m_n v_n² r_{n .} Solving for (v_n)_1n  as before, we get (v_n)_1n =  [e²/2nhε_o][ n⁴+ 4n² +  2/(1 +n²)²]. The binding energy contributed by the excited electron is dm_nc² = (1/2) m_o  [e³/h²ε_o²] [ (n⁴+ 4n² + 2)/2n (1+n²)²]². The binding energy of the system dm_1n c² is the sum of its contribution by both the electrons and is equal to  (1/2) m_o  [e³/h²ε_o²]{[(2n⁴ +4n²+1)/2(1+n²)]² + [(n⁴+ 4n² +  2)/2n  (1 +n²)]² = (54.45){[(2n⁴ +4n²+1)/2(1+n²)]² + [(n⁴+ 4n² + 2)/2n (1+n²)]² When the excited electron jumps to the innermost orbit, the change in the binding energy dm₁₁c² - dm_1n c². dm₁₁c² represents the energy equivalent of the relativistic change of mass of both the innermost electrons. dm₁₁ c² = m_o v₁₁² , substituting the value of v₁₁  = (7/8)[e²/hε_o], dm₁₁c² = (1/2)m_o [e⁴/h²ε_o²] (49/32) = 83.4378 eV.  dm₁₁c² - dm₁₂ c² = (1/2)m_o [e⁴/h²ε_o²]{(49/32) -[(2n⁴ +4n²+1)/2(1+n²)]² - [(n⁴+ 4n² +  2)/2n  (1 +n²)]²}. = (1/2)m_o [e⁴h²ε_o²]{(49/32) - [34/100]² + [49/50)]². When excited electron in 2^nd orbit jumps to innermost orbit, the binding energy difference is 83.4378 - 58.5882 = 24.786 eV which corresponds to the electromagnetic radiation 50.03 nm. If the excited electron is in the 3^rd orbit, the change in binding energy and the  wavelength of emitted radiation become   dm₁₁c² - dm₁₃ c² = (1/2)m_o [e⁴h²ε_o²]{(49/32) - [119/600]² + [199/200)]² = 83.4378 - 56.029 = 27.345 eV and λ_{13→ 11} =  45.34 nm.

Acknowledgement

The Author is grateful to RM.Alagappa Chettiar, the philanthropist and fonder of all  Alagappa educational institutions from where the author is educated.

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