Application of Bohr's Theory to  Li⁺ ion 

      In Li ⁺  two electrons are present in the 1s orbit. The radius of the orbit is not equal to a_o/3 and different due to electron (1s) - electron (1s) interaction. The stability of the electron in its n^th allowed orbit   3e²/Kr_nn² - e²/4Kr_nn²=  (11/4) e²/Kr_nn² =m_n v_nn²/r_nn

     (11/4)e²/K =  m_n v_nn² r_nn  and   nh/2π = m_nv_nn r_nn

By solving as done before, m_nr_nn = (4/11)n²  h²ε_o/π e²

r_nn =(4/11)n²  [h²ε_o/m_oπ e²] (1- v_nn²/c²)^1/2 = (4/11)n² a_o (1- v_nn²/c²)^1/2  and v_nn = (11/4) [e²/2nhε_o]

In terms of r_n and v_n, r_nn = (4/11)n² a_o (1- v_nn²/c²)^1/2 =  (12/11) [n² a_o/3](1- v_nn²/c²)^1/2 =  (12/11)[(c²- v_nn²)/(c²- v_n²)]^1/2r_n.. In non-relativistic approach r_nn = (12/11) r_n

v_nn = (11/4) [e²/2nhε_o] =  (11/4) [3e²/2nhε_o](1/3) = (11/12) v_n

In the same situations, the radius of the inner most and the higher orbits are (4/11)a_o, (16/11) a_o , (36/11) a_o.  

     The Li+ ion is analogous to helium atom except the nuclear charge which is one unit higher than that. So the method of half-half mixing can be adopted to study its spectral feature. Let us suppose that the two electrons are in the n^th orbit having radius r_n . The centripetal force acting on an orbital electron due to nuclear attraction is 3e²/Kr_n² .   The electrostatic force of repulsion between the two electrons which are diametrically opposite is e²/4Kr_n² . The resultant force is balanced by centrifugal force i.e.,

  3e² /Kr_n² - e²/ 4Kr_n² = (11/4) e²/Kr_n² = mv_n² /r_n                                                                                                      

Total kinetic energy of the electrons in the Li+ ion = mv_n² = (11/4) e²/Kr_n                                                            

Kinetic energy per electron = (1/2) mv_n ² = (11/8) e²/Kr_n  = (11/4) e²/ 2Kr_n                                                                         

Potential energy of electron 1 = - 3 e² /Kr_n                                                                                                                                  

Potential energy of electron 2  in the presence of electron 1 = - 3 e²/Kr_n + e²/2Kr_n                                     otal potential energy of the electrons in the Li+ ion = - (11/2) e²/Kr_n                                                              Sum of K.E and P.E is  - (11/4) e²/Kr_n which provides the required binding energy to Li⁺ ion.

    (2π r_n)² = 4π² r_n²  =  n²  λ²  =  n²  h² / m² v_n² = (4/11)  n² h²  4 πε_o r_n / m e²

                                               r_n  =   (4/11) n²h² ε_o / mπ e²  = (4/11) n²  a_o

Substituting the value for r_n the total energy becomes - (11/4)²  e² / n² K a_o  =  - (121/8) (1/n² ) 13.595 eV. It gives the total energy associated with the Li+ ion as - 205.624 eV. The sum of the second and third ionization energies of Lithium atom is 75.62 + 122.42  = 198.04 eV. The estimated value of total energy of Li⁺ ion is 7.584 eV higher..   

           Hyper excitation and de-excitation may happen in Li⁺ ion when sufficient energy is available. In the hyper states both the electrons have equal contribution to its binding energy.

    Using the relativistic model,  BE₁₁ = dm₁₁ c² = 2 (dm₁) c² = 2[(1/2)m_ov₁₁²] = m_ov₁₁² and BE_nn = dm_nnc² = 2(dm_n) c² = m_o v_nn² . The difference in the binding energies BE₁₁ - BE_nn  of the two states is radiated out during transition. During  hyper de-excitation, both the electrons simultaneously jump into  to any inner or innermost orbit. BE₁₁ - BE_nn = m_o [v₁₁² - v_nn²]. Substituting the values for v₁₁ and v_{nn ,} BE₁₁ - BE_nn = m_o[121/8][e²/2Ka_o][(n²-1)/n²]. Since λ = hc/ΔBE, where ΔBE is in joules, the corresponding wavelength of radiation emitted in any transition is given by λn_2* → n_1* is 12.4 x 10^-7/205.624[(1/n₁²) - (1/n₂²)]m.= 6.03 {1/[(1/n₁²) - (1/n₂²)]}nm. Using this formula, the wavelength of radiation emitted can be predicted For n₂* → n₁* the wavelength is 8.03 nm.. Even though its probability is very little, it is not zero.

           The spectral lines of Li⁺ ion can be studied in the same way as we followed in helium atom. By estimating the total energy of Li⁺ ion in two different states E₁₁and E_1n, the wavelength  for any electronic transition from 1n to 11 level can be derived. There are many ways by which the Lyman series of Li⁺ ion may happen. The first case is transfer of electron from n^th orbit to the innermost orbit with single electron. The energy of a system with one electron in the innermost orbit and the other electron in the n^th orbit can be calculated by half-mixing of two states In half-half maxing  E₁₁ and E_nn are the contributors to E_1n As done in helium atom, E_1n is the sum of three components negative potential , kinetic and electron-electron interactional energies. Both the negative potential and kinetic energies are the mean negative potential and kinetic energies of the states E₁₁ and E_nn.

                   NPE_1n = - [3e²/K][1/r₁₁ + 1/r_nn] =  - [3e²/Kr₁₁][(n¹ +1)/n²]

                   KE_1n = (11/8)(e²/K)[1/r₁₁ + 1/r_nn]  = (11/8)(e²/Kr₁₁)[(n¹ +1)/n²]

                 (e-e)_1n = e²/Kd_1n = e²/K(r₁₁ + r_1n) = 2(e-e)₁₁/(1+n²)

E_1n = [e²/Kr₁₁][ -(13/8)(n¹ +1)/n² +(1/(1+n²)] = (11/2)[e²/2Ka_o][ - (13/8) (n¹ +1)/n² +1/(1+n²)]. When n →_∞, E_1∞ = (11/2)(13.595)(13/8) = 121.50 eV. The observed third ionization energy of lithium atom is 122 eV. Knowing E₁₁  and E_{1∞  ,} the second ionization energy of lithium atom can be predicted E₁₁  - E_1∞ = 84.125 eV. The observed second ionization energy of lithium is 75.62 eV, which is 8.51 eV smaller than the theoretical value.

          E_1n = (1/2) [KE₁₁ +N PE₁₁ + KE_nn + PE_nn] + (e-e)₁₁ [1/(1+n²)] + (e-e)_nn [n² /(1+n²)]. When an electron in the n th orbit jumps to the innermost orbit, the transition energy is given by ΔE = E_1n E₁₁ = {(1/2) [ .KE_nn +N PE_nn - KE₁₁ - PE₁₁ } + (e-e)₁₁ [(1/(1+n²) - 1] + (e-e)_nn [n² /1+n² ]} (e²/ 2Ka_o.) = { (1/2) [ KE_nn +N PE_nn - KE₁₁ - PE₁₁ } + [n² /1+n² ]  [(e-e)_nn - (e-e)₁₁]}(e²/ 2Ka_o.) = {(1/2) [(1/n²) -1]

(121/8) - (1/2)[(1/n²) -1]33 +  [n² /1+n² ]  [(1/n²) -1] (11/4) }(e²/ 2Ka_o.) = {[(1- n²) /2n²] [121/8 -33] + [n² /1+n²] (11/4) [1- n²]/n²]}(e²/ 2Ka_o.) = { - (143/8)[(1- n²)/2n²] + (1-n²)/(1+n²) (11/4)} (e²/ 2Ka_o.)

= {(n² -1)/2n² (143/8) - (n² -1)/(n² +1) (11/4)} (e²/ 2Ka_o.)

ΔE_{21- 11} =   [(3/8) (143/8) - (3 /5) (11/4)] (13.595) = 5.0531 x 13.595 = 68.6972 eV λn₂₁ → n₁₁ is 12.4 x 10^-7/ 68.6972  = 18.05 nm

        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 + PE₂₂] +(4/13) (e-e)₂₂ +(9/13)(e-e)₃₃]                                                                          

E₁₂ = (1/2)[KE₂₂ +NPE₂₂ +KE₁₁ + PE₁₁] +(1/5) (e-e)₁₁ +(4/5)(e-e)₂₂]                                                                    E₁₃  = (1/2)[KE₃₃ + NPE₃₃ +KE₁₁ + PE₁₁] +(1/10) (e-e)₁₁ +(9/10)(e-e)₃₃]

E₂₃ → E₁₂   = (1/2)[KE₃₃ + NPE₃₃ -KE₁₁ - PE₁₁] - (1/5) (e-e)₁₁ - (32/65) (e-e)₂₂ +(9/13) (e-e)₃₃

Substituting the values for each components, we have

E₂₃ → E₁₂   = {[121/144 - 33/18 - 121/16 + 33/2] -(11/20) - (32/65) (11/16) + (9/13) (11/36)} 13.595 =

7.2675 x 13,595 = 98.802eV and  λ₂₃ →₁₂  = 12.55 nm.

Likewise

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

           ={[121/64-33/8-121/16 +33/2] -11/40 + (4/13)(11/16) - (27/130)(11/36)]}13.595    .         = 89.414 eV and λ₂₃ →₁₃   =13.86  nm.

     The radius of Li⁺⁺ ion is a_o/3 =0.3333 a_o and is (4/11) a_o = 0.3636 a_o for Li⁺ ion. This change Δr = 0.0303 a_o (= a_o/33) is due to the additional electron-electron interaction among 1s electrons. The kinetic energy of the 1s electron in Li⁺⁺ ion is (3/2) e²/Kr = 9 (e²/2Ka_o) and the negative potential energy of Li⁺⁺ ion is - 3 e²/Kr = - 18 (e²/2Ka_o)  where as the kinetic energy of a single 1s electron in Li⁺ ion is (11/8) e²/Kr = (121/16) (e²/2ka_o) = 7.5625 (e²/2Ka_o) and the negative potential energy of Li⁺ ion is - 6 e²/Kr = -33 (e²/2Ka_o).  The changes  Δ (KE) = 1.4375 (e²/2Ka_o) and Δ(PE) = (3/2)(e² /2ka_o) are  due to electron-electron interaction. The Li⁺ ion has additionally positive potential energy due to electrons in the same orbit. It is equal to (11/4) (e² /2ka_o)

Relativistic model of Li⁺ ion

    According to this model {Ref.11], 6e²/Kr_nn - e²/2Kr_nn = (11/2)e²/Kr_nn =m_nnv_nn² + dm_nn c^2. .It gives

dm_nn c^2. = (11/4) e²/Kr_nn. When both the electrons are in the innermost orbit of Li⁺ ion,  dm₁₁c² = (11/4) e² /Kr₁₁ As derived before the radius of the n^th allowed orbit is (4/11) n² a_o (1 - v_nn²/c²)^1/2 and the velocity of the electron in that orbit is (11/4) e²/2nhε_o Hence dm₁₁c² = BE₁₁ = (121/8) [e²/2Ka_o] [1+ v₁₁²/2c²] = 205.624 [1+ (121/128)(e⁴/4h²ε_o²c²)]_.= 205.6343 eV. This can be substantiated by estimating the relativistic increase of mass of each electron in Li^+. dm = m₁ - m_o  ≃ (1/2)m_o v₁₁²  = (1/2)9.108 x 10^-31 x (121/16)(1.602 x 10^-19)³[4x (6.626 x 10^-34)² x (8.85 x 10^-12)²] = 102.94 eV and the two innermost electrons contribute a quantum of binding energy 205.88 eV.

              The spectral lines o Li⁺ ion can be studied by using the relativistic model. When the two electrons are in the same allowed orbit, 3e²/Kr_nn² - e²/4Kr_nn² = (11/4) e²/Kr_nn² = (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  =(4/11)n² a_o (1- v_nn²/c²)^1/2 and v_nn = (11/4)e²/ 2nhε_o

       When the electrons are in different orbits the electron in the orbit n₁ with radius (r_n1)_n1-n2  has mass (m_n1)_n1-n2 and _,velocity (v_n1)_n1-n2 whereas the electron in the orbit n₂ with radius (r_n2)_n1-n2  has mass (m_n2)_n1-n2 and _,velocity (v_n2)_n1-n2 .

  3e²/K(r_n1)² - e²/K(r_n1 + r_n2)^{2  =} m_n1v_n1²/r_n1; 3e²/K- e²/K[1/(1+r_n2/r_n1)]² = m_n1v_n1²r_n1                                                  [e²/K][3 - [1/(1+n₂²/n₁²]²  = [e²/K](2n₁⁴ +3n₂⁴ +6n₁²n₂²)/(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](2n₁⁴ +3n₂⁴ +6n₁²n₂²)/(n₁² + n₂²)²

                       r_n1 =  n₁² a_o [(n₁² + n₂²)²/(2n₁⁴ +3n₂⁴ +6n₁²n₂²)] (1- v_n1²/c²)^1/2

Consider the electron in orbit n₂

      3e²/K(r_n2)² - e²/K(r_n1 + r_n2)²  = m_n2 v_n2²/ r_n2 ; 3e²/K- e²/K[1/(1+r_n1/r_n2)]² = m_n2v_n2²r_n2                                                           [e²/K][3 -[1/(1+n₁²/n₂²]²  = [e²/K](3n₁⁴ +2n₂⁴ +6n₁²n₂²)/(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](3n₁⁴ +2n₂⁴ +6n₁²n₂²)/(n₁² + n₂²)²

                       r_n2 =  n₂² a_o [(n₁² + n₂²)²/(3n₁⁴ +2n₂⁴ +6n₁²n₂²)] (1- v_n2²/c²)^1/2 

Knowing the orbital velocity , the binding energy of the orbital electron  in _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][(2n₁⁴ +3n₂⁴ +6n₁²n₂²)²/n₁²+(3n₁⁴ +2n₂⁴ +6n₁²n₂²)²/n₂²]

    = [e²/2Ka_o][1/(n₁² + n₂²)⁴][(2n₁⁴ +3n₂⁴ +6n₁²n₂²)²/n₁²+(3n₁⁴ +2n₂⁴ +6n₁²n₂²)²/n₂²]

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

n₁= n₂= 1 ;[e²/2Ka_o] (1/16)[ 242] =(121/8)[e²/2Ka_o] =205.624  eV

n_{1 = 1;} n₂ = 2 ; [e²/2Ka_o] [1/(5)⁴] [(74)² +(59/2)²] = 133.215 eV

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

The binding energy difference and the first line of Lyman series of Li⁺ ion are

BE₁₁ - BE₁₂ = 205.624 - 133.215  = 72.41eV and  λ₁₂→₁₁ = 17.12 nm 

This can be confirmed by using the relativistic model of atom. BE₁₁ - BE_1n = dm₁₁c² - dm_1n c²= 2 (dm₁)₁₁ c²  - (dm₁)_1n c²  - (dm_n)_1n c²   . dm₁₁ c² = m_o v₁₁² =  m_o [(121/16) e⁴/4h²ε_o²] = (121/8)[e²/2Ka_o] .The energy equivalent of relativistic increase of mass of the electron in the inner most orbit when the other electron is in the nth orbit is (dm₁)_1n c² = (1/2) m_o (v₁)²_1n

                                           3 e²/Kr₁² - e²/K(r₁ + r_n)² = m₁(v₁)²_1n /r₁

          The optimization of Bohr's Theory of hydrogen atom to atoms of higher atomic number Z can be tested by computing the total energy content of helium like ions and the sum Z^th and (Z-1)^th ionization energies. Considering a helium like ion having atomic number Z. From the relations obtained by using the Bohr's postulates e²/K [(4Z-1)/4] = m_nv_n²r_n and nh/2π = m_nv_nr_n  , the radius of the n^th  orbit of the 1s electrons r_n = n² a_o [4/(4Z-1)] (1-v_n²/c²)^1/2  and its velocity v_n = [e²/2nhε_o][(4Z-1)/4]. The relativistic increase of mass of an electron in its innermost orbit dm₁c² and for both the electrons it is 2 dm₁ c² =2[m₁ -m_o] c² = m_o v₁². On simplification after substituting value for v_1, we get [(4Z-1)²/8] [e²/2Ka_o]. The binding energy of the helium like ions is derived and compared with the sum of (Z-1)^th  and Z^th ionization energies.  

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

.................................................................................................................................                                                                                                                           .                                                                                    ionization energy (eV)

He like ions  (4Z-1)/8   Binding energy(eV)    (Z-1)^th    Z^th     [(Z-1)^th + Z^th]       ^Diff

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

He                   49/8             83.269                  24.48        54.40         78.88          4.39

Li⁺                   121/8          205.624                 75.62       122.42        198.02        7.60

Be⁺⁺                      225/8           382.275                153.85       217.65       371.50         10.77

B⁺⁺⁺                 361/8          613.470                259.30        340.13      599.47        14.00

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

The above table  shows that the difference between the theoretical and experimental values of binding energies of helium like ions increases as Z increases. It indicates that besides electron-electron interaction there must be Z dependent interaction which is responsible for the observed discrepancies.