ஆங்கில இலக்கணத்தில் active voice passive voice என்று இருப்பதைப்போல மனிதர்களில் செயல் வீரன் , செய்யா வீரன் என்று இரு வகைகள் உண்டு. முன்னவன் சுறுசுறுப்பாக ஆக்கப்பூர்வமான செயல்களில்  செயல்படு பவன் பின்னவன் மந்தமாகச் செயல்படுபவன். இலக்கணம் மற்றும் மனிதர்களைப் போல செலவினங்களிலும் நிறை பயன் செலவு (active expenses) என்றும் குறை பயன் செலவு (Passive expenses) என்றும் இரு வகை உண்டு.. முதல் வகைச் செலவுகள் தொடர்ச்சியான முன்னேற்றத்தை வழங்கும். இரண்டாம் வகை இதற்கு எதிராகத்  தொடர் பின்னேற்றத்தை கொடுக்கும்.   ஒரு செலவு செய்தால் அதிலிருந்து பயன் கிடைக்கவேண்டும். அந்தப்பயன் சமுதாயம் தழுவியதாக இருந்தால் பயன் அடுத்தடுத்து பயன்களைக்கொடுக்கும் . தனி மனிதர்களுக்கு இந்தப் பயன் என்பது மனத்திருப்தி மட்டுமில்லை ,மீண்டும் செலவு செய்யும் தகுதியைக் குன்றாமல் தருவதுமாகும் .படித்து  நல்ல திறமைகளை வளர்த்துக்கொள்வதற்குச் செய்யும் செலவு,தொழில் உற்பத்தியைப் பெருக்குவதற்கு திட்டமிட்டுச் செய்யும் செலவு நிறை பயன் செலவு.இன்றைக்கு மக்களும் சரி அரசாங்கமும் சரி குறை பயன் செலவுகளை அதிகரித்துக் கொண்டே வருகின்றார்கள் . இதனால் மக்களும் அரசாங்கமும் எது சரியில்லையோ அதைச் சரி என்று நினைக்கத் தொடங்கியிருக்கிறார்கள் . இந்த எண்ணத்தை மாற்றுவது அவ்வளவு எளிதில்லை தன் பாதிப்பு வரும்காலத்தில் பல சிக்கல்

களை உருவாக்கும்.

 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.

 

 இன்றைக்கு நம்மை ஆள்பவர்கள் தாங்கள் மட்டுமே ஆளவேண்டும் என்று ஆளவிரும்பும் மற்றவர்களை எதிரிகளாக ப் பார்க்கின்றார்கள் . நம்மை ஆண்டு மடிந்தவர்களைக் கூட  மதிக்க மறந்துவிடுகிறார்கள். இவர்கள் எப்படி மக்களுக்கு நல்லது பண்ணப்போகின்றார்கள் ?  அவர்களுக்கு வார்த்தைகளே செயல் . 

 மனிதர்களால் உண்டாக்கப்பட்ட கடவுள் . தொழுகை முறை , சடங்குகள் மதத்திற்கு மதம் வேறுபட்டிருந்தாலும் இயற்கை அவர்களுக்குள்ளும் உள்ளார்ந்த ஒற்றுமையை சுட்டிக்காட்டத்  தவறவில்லை .  உணவின் சுவை நிறம் திடம் மாறுபட்டிருந்தாலும் ஆற்றல் தேவைக்காக எடுத்துக்கொள்ளும் உணவை மென்று சாப்பிட்டு ஒரே மாதிரியாகத்தான் ஜீரணிக்கின்றார்கள் . தவித்தால் தாகத்திற்கு நீர்தான் குடிக்கிறார்கள் . இறந்த பிறகு எல்லோருடைய உடலும் எந்த வேற்றுமையையும் காட்டுவதில்லை .கூடி இருப்பவர்கள் தான் வேற்றுமை காட்டுகின்றார்கள் .இனப்பெருக்கத்திற்கு எல்லோரும் ஒரே வழிமுறையைத்தான் பின்பற்றுகிறார்கள் . எல்லோருக்கும் ரத்தம் அதே சிவப்பு. வியர்வை உப்பு கரிக்கின்றது. கழிவுகள் ஒரே மாதிரியாகத்தான் வெளியேறுகின்றன.   

 Application of Bohr's Theory to L2+ion

In Li 2+ a single 1s electron is present It is a two body problem nucleus with charge 3e+and an electron with charge e-.For the stability of the orbital electron in its nth allowed orbit, the electrostatic force acting on the electron must be equal to centrifugal force 3me^2 /Krn ^2 = m^2vn ^2/rn, or 3e^2/K = mvn^2 rn The circumference of the orbit must contain some integral multiples of wave length characteristic of the electron in the orbit. 2πrn= nλ = n[h/mnvn] or mnvn rn = nh/2π By solving these two relations mnrn = n^2 h^2εo/3 π e^2 rn = (n^2/3)[h^2εo/ moπ e^2][1 -vn ^2/c^2]^1/2 = (n^2/3)ao[1 -vn^  2/c^2]^1/2,    vn = [3e^2/4πεo][2π/nh] = 3e^2/2nhεo The relativistic increase of mass of the orbital electron is not taken into account . rn = n^2 ao/3. .i.e., the innermost orbit of the Li2+ion is one third of the innermost orbit of the hydrogen atom Under identical situations, the radius of the inner most and the higher orbits are (1/3) ao,(4/3) ao, (9/3)ao By summing up the kinetic and potential energies of the lithium ion total energy of the system can be computed. Kinetic energy of the electron (1/2) mvn^2 = (3/2) e^2/Krn and its potential energy  is -3e^2 / Krn which together gives its total energy En = -(3/2) e^2/ Krn .Substituting the value for rn , En = (1/n^2) [(3^2)(e^2/2Kao) ]= 9(1/n^2)(13.595) eV which is 9 times greater than the corresponding value for hydrogen atom.  

     When the electron is in the innermost orbit of lithium ion Li2+ , its energy is 9 x 13.595 = 122.355 eV , The third ionization energy of Li2+ ion is 122.42 eV which is in close agreemnt with the computed value. Its spectral feature can be studied by determing the energy of transition and the corresponding wavelength of radiation emitted. ΔE = Eexcited /initial - E ground / final =9[13.595] [(1/n1^2- 1/n2^2)] , n2 > n1 The frequency of the emitted radiation ν = ΔE / h = 9[13.595/h][(1/n1^2- 1/n2^2)] The wavelength of the emitted radiation λ = C/ν= hc/9(1/n1^2- 1/n2^2) 13.595 The Lyman series of Li2+ ion corresponds to electronic transition from n ≥ 2 to n =1. The wavelength of emitted radiation λ = [ 6.626 x 10^-34 x 2.998 x 10^8] / 9 [13.595 x 1.602 x 10^-19][1/ (1/n1^2- 1/n2^2) = 0.1013 x 10-7 x ][1/(1/n1^2- 1/n2^ 2). λn2→n1 ; 0.1013 x 10^-7x 4/3 = 13.5 nm . All the series of spectral lines of Li2+ ion is 9 times smaller than the corresponding lines in hydrogen spectrum. The same result can be derived from the proposed relativistic model applied to Li2+.. The energy corresponding to the relativistic change of mass of the electron in L2+, dmnc^2 ≃ (1/2) mo vn ^2 . Substituting the value for vn= 3 e^2/2nhεo , dmnc^2 ≃ (1/n^2)(9/8) mo e^4/h^2εo^2 .When the electron jumps from n th to first orbit, ΔEn→ 1 = dm1c^2 - dmnc^ 2 = (9/8) mo [e^4/h^2εo^ 2][(n^2 -1)/n^2] = 9 [e^2/2Kao] [(n^2-1)/n^2]. The first line of Lyman series of L2+ ion λ2→1 is 13.51 nm which is due to the difference in binding energy of the system ΔBE2→1 =91.766 eV.  

Aliter 

     Let us now estimate the total energy of excited helium atom with one electron in 1s orbit and another electron in the n th orbit.The radius of nth electronic orbit is n^2ao/2. Kinetic energy of the electron in 1s orbit 2e^2/Kr1 ^2- e^2/K(r1+rn)^2 = mv1^2 /r1 or (1/2) mv1^2 = e^2/Kr1- e^2/2Kr1(1 +rn/r1)^2 = e^2/Kr1[1 -1/2(1+rn/r1)^2].substituting rn = n^2r1 , KE of 1s electron = 4e^2/2Kao [ 1- 1/2(1+n^2)^2] = 4 x [(2n^4 +4n^2 +1)/2(1+n^2)^2] x e^2/2Kao Kinetic energy of the electron in n th orbit mvn ^2/rn = 2e^2/Krn ^2 e^2 /K(r1 +rn)^2 or (1/2) mvn^2 = e^2/Krn - e^2/2Krn(1+r1/rn)^2 = e^2/Krn[1 -1/2(1+r1/rn)^2].Substituting rn = n^2r1 , KE of the electron in the nth orbit is (1/n2)(4e^2/2Kao)[1- 1/2(1+1/n^2)^2]= (1/n^2)(4e^2/2Kao)x [n^4 +4n^2 +2]/[2(1+n^2)^2] Total kinetic energy of the system with one electron in 1s orbot and other electron in the nth orbit {2[e^2/2Kao]/[(1+n^2)^2]}{(2n^4 +4n^2 +1) +(1/n2)(n^4 +4n^2 +2)}.when n = 2; (2/25)[e^2/2Kao][49 + 17/2]= (23/5)[e2/2Kao].Negative potential energy of the 1s electron = -2e^2/Kr1 and the corresponding value for the electron in the nth orbit = - 2e^2/Krn . Total negative potential energy of both the electrons -[2e^2/Kr1] [1 + 1/n^2)]= -8 [(n^2 +1)/n^2][e^2/2Kao]. When n =2 , -10 [e^2/2Kao].Positive potential energy due to electron-electron interaction is e^2/K(r1 + rn) = e^2/Kr1((1 +n^2).When n = 2, (4/5)[e^2/2Kao].Adding together all the components of energy, we get the net energy associated with the system .E1n = [e^2/2Kao][2(2n^4 +4n^2 +1) +(1/n^2)(n^4 +4n^2 +2)]/[(1 +n^2)^2] -8[(n^2 +1)/n^2] +4/(1+n^2)] and E11= (49/8)[e2/2Kao].The energy radiated out during the electronic transition E2→ 1 = E11 - E12 = [e^2/2Kao][49/8 + 23/5- 10 + 4/5] = [e^2/2Kao][49/8 - 23/5] =(61/40)x 13.595 =1.525 x 13.595 = 20.732 eV and λn2→n1 = 59.81 nm.

  

 

 

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 .