Aliter-1 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. 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 e^2/4Kr1^2 and have an initial potential + e^2/2Kr1. 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 e^2/K (x^2 +r1^2) and its component along the direction of displacement is 2e^2 cosφ /K (x^2 + r1^2 ) = 2 e^2 x /K (x^2 + r1^2 )^3/2 . The resultant component due to both the electrons is 4 e^2 x/K (x^2 + r1^2)^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 =[4e^2/K] ∞ ∫o x dx /(x^2 + r1^2)^3/2 = - [ 4e^2 /K (x^2 + r1^2)^1/2]∞o = -- 4 e^2/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 e^2/Kr1 + e^2/2Kr1 = - (7/2) e^2/Kr1. The parallel component of electrostatic force acting on electron-1 towards the center of the orbit is 2e^2r1/K(x^2 + r1^2)^3/2 .At the end of the displacement of the nucleus this force becomes maximum and is equal to 2e^2/Kr1^2. Taking into account the initial electron-electron repulsion the total centripetal force 2e^2/Kr1^2 - e^2/4Kr1^2 = (7/4) e^2/Kr1^2 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)e^2/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. (2πrn)^2 = 4π^2 rn^2 = n^2 λ^2 = n^2 h^2/m^2 vn^2 = (4/7) n^2 h^2 4 πεo rn /m e^2 rn = (4/7)n^2h^2εo/mπe^2 = (4/7) n^2 ao ...... (2.3) Substituting the value for rn the total energy becomes - (7/4)^2 e^2/n^2 Kao = - (49/8) (1/n^2) 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. There is a little difference of 4.385 eV which requires some correction in the above treatment. Whenever an electron leaves or enters the orbit, the radius of the orbit gets changed due to mutual electron-electron interaction among electrons in the same orbit. The change of orbital radius makes changes in all the components of energy of the system. It needs a small correction to both the kinetic and potential energies of the orbital electrons. Aliter-2 The same result can be arrived by computing the component of energy in assembling the helium atom with its nucleus. The kinetic energy of the 1s electrons is 2 x (1/2)mv1^2 = mv1^2 = (7/4)e^2/Kr1 = (48/8)e^2/2Kao The potential energy of the first electron in the 1s orbit = - 2e^2/Kr = - 4e^2/Kao = - 8(e^2/2Kao).The potential energy of the second electron in the orbit = - 2e^2/Kr1 = -14e^2/4Kao = - 7(e^2/2Kao). The potential energy induced on the first electron by the second electron = - 7(e^2/2Kao) +8(e^2/2Kao) = e^2/2Kao. The positive potential energy due to electron -electron interaction in the same orbit = e^2/2Kr1 = (7/4)e^2/2Kao. The sum of all the components of energy becomes (-14 +7/4) e^2/2Kao = (49/4) e^2/2Kao .Adding the kinetic energy, total energy becomes (-49/4 + 49/8) e^2/2Kao = -(49/8) e^2/2Kao Total energy of electrons in the helium atom assembled Let us suppose a helium atom is assembled with a nucleus having 2 units of positive charge and two separate electrons. There are 2 stages in the assembling. (1) e-1 is placed at r1 (= ao/2) to form a helium ion and (2) when e-2 is brought from infinity to the orbit having radius r2 (= 4 ao/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 spinning nucleus, it gains acceleration due to nuclear force of attraction and it starts making a circular motion around the nucleus with centrifugal force. 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 e^2/Kr1 = mv1^2 /r1 or mv1^2 = 2e^2/K r1 Kinetic energy = (1/2) m v1^2 = e^2/K r1 All permitted orbits in atomic systems must satisfy the condition-1 of Bohr's theory of hydrogen 4π^2 r1^2 = h^2/m^2 v1^2 = h^2 4πεo r1/2m e^2 r1 = h^2εo/2mπe^2 = ao/2 In terms of Bohr radius the kinetic energy becomes 4[e^2/2Kao] Potential energy of the electron is determined by evaluating the work done in taking the electron from infinity to the assigned orbit. - 2 e^2 /Kr1 = - 8 e^2/2Kao Total energy of the system is the sum of its kinetic and potential energies T.E = K.E + P.E = 4[e^2/2 Kao] - 8e^2/2K ao = - 4e^2/2Kao = - 4 x 13.595 = 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 Zth ionization energy of hydrogen-like ions of all elements. If Z is the atomic number, then Z e^2 /Kr1^2 = m v1^2 / r1 or m v1^2 = Z e^2 / K r1 Kinetic energy = (1/2) m v1^2 = Z e^2/2Kr1 4π^2 r1^2 = h^2/m^2 v1^2 = h^2(4πεo) r1/Z m e^2 r1 = h^2(εo)/Zmπe^2 = ao/Z Kinetic energy becomes Z^2 e^2/2K ao Potential energy = - Z e^2 /Kr1 = - Z^2 e^2/Kao Total energy of the system is the sum of its kinetic and potential energies T.E = K.E + P.E = Z^2e^2/2K ao - Z^2 e^2/Kao = -Z^2 e^2/2 K ao = - Z^2[e^2/2Kao] = - Z^2 x 13.595 eV In the ground state of helium atom, both the electrons are in orbit having radius r2.The attractive force experienced by e-2 due to the central nucleus = 2e^2 / Kr2^2.The repulsive force experienced by e-2 due to the presence of e-1 with an intermediate distance 2r2 = e^2 /4Kr2^2 .The resultant force of attraction = 2e^2/Kr2^2 - e^2/4Kr2^2 = (7/4)e^2/Kr2^2 = mv2^2/r2 Kinetic energy of both the electrons = m v2^2 = (7/4)e^2/Kr2 Kinetic energy per electron = (7/8) e^2/Kr2 The condition-1 of Bohr's Theory of hydrogen atom predicts 4π^2r2^2 = h^2/m^2 v2^2 = h^2 (4/7) 4πεo r2/ m e^2 r2 = (4/7) [h^2(εo)/mπe^2] = (4/7) ao 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 r2 .The potential energy of e-2 in the presence of nucleus is ∞∫r2 [2e^2/Kx^2] dx = [2e^2/Kx] ∞r2 = 2e^2 /Kr2 = - 7[e^2/2Kao] . 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) ao . 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)ao]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 - e^2 dx/K (x+ (15/28)ao)^2] . [e^2 /K (x+ (15/28)ao)]∞r2 = e^2 /K [r2+ (15/28)ao] = (e^2/K) {1/[(4/7) ao+ (15/28)ao]} = 28 e^2/31K ao = (56/31) [e^2 /2Kao] This induced potential is shared by both the electrons, each electron has an increase of potential by (56/62)[e^2 /2Kao] When e-2 reaches its assigned orbit of radius r2, 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 2e^2/Kr2^2 - e^2/4Kr2^2 = (7/4) e^2/Kr2^2 = mv2^2 /r2 The kinetic energy gained by e-2 is (1/2) m v2^2 = (7/8)e^2/Kr2 = (49/16) [e^2/2Kao]. Due to induction, the kinetic energy of e-1 is decreased. It is dropped from 4 [e^2/2Kao] to (49/16) [e^2/2Kao]. It is equal to - (15/16) [e^2/2Kao]. The decrease in kinetic energy of e-1 is (1/2) m v1^2 - (1/2)m v2^2 = e^2/ Kr1 - (7/8)e^2 /Kr2 = 4e^2 /2Kao - (49/16) [e^2/2Kao] =(15/16) [e^2/2Kao] 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 - [2e^2 / K x]r1r2 = - 2e^2/K[1/r2 - 1/r1] = - 2e^2/K[ (r1 - r2)/ r1 to r2]= - 2e^2/K( - 1/4ao) = [e^2/2Kao] When e-2 is r2 , the change in the potential energy of e-1 due to nucleus is [e^2/2Kao] and due to the presence of e-2, [e^2/K][1/(x+r2 )]r1-r2 = [e^2/K][1/2r2 - 1/(r1+ r2)] = [e^2/K] [ 7/8ao - 14/15ao] = [e^2/2Kao][-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 r2, 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 [e^2/2Kao] and when e-2 is at r2 is [e^2/2Kao] - [7/60][e^2/2Kao] which give a mean as [1 - 7/120][e^2/2Kao].= (113/120)[e^2/2Kao] Kinetic energy of e-1 when e-2 is present [e^2/2Kao][4 - 15/16] = (49/16) [e^2/2Kao] Potential energy of e-1 when e-2 is present [e^2/2Kao][ -8 +(56/62) +1 -7/240] = - 6.126 [e^2/2Kao] Kinetic energy of e-2 [e^2/2Kao](49/16) Potential energy of e-2 [e^2/2Kao][ -7+ (56/62) - 7/240 ] = - 6.126[e^2/2Kao] Total energy of the helium atom is the sum of all the four components and is equal to 2[3.0625 - 6.125][e^2/2Kao] = -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 Ze^2/Kr1^2 - e^2/4Kr1^2 = (Z-1/4) e^2/Kr1^2 [(4Z-1)/4]e^2/Kr1 = mv1^2 which gives r1 = [4/(4Z-1)] ao Kinetic energy of the electrons = mv1^2 = [(4Z-1)/4]e^2/ Kr1 = [(4Z-1)^2 /8]e^2/2Kao Potential energy of the electrons = -2Ze^2/Kr +e^2/2Kr = (e^2 /Kr)[ -2Z + 1/2] = - (e^2 /2Kr) [4Z-1] = - [(4Z-1)^2/4]e^2/2Kao Total energy of the system = - [(4Z-1)^2 /8]e^2/2Kao 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 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 ..............................................................................................................................................

Amazon kdp is no longer allowed to publish and sell my ebooks and paperback. if sold it would be illegal Dr. M. Meyyappan, author

நாட்டு மக்களில் இரு பிரிவினர் - சேவை செய்பவர்கள் என்று சொல்லிக்கொள்ளும் ஆள்பவர்கள் என்றாலும் அரசர்கள் அதிகாரமும், செல்வாக்குமிக்கவர்கள். மற்றொரு பிரிவினர் ஆள்பவர்களைத் தேர்ந்தெடுக்கும் உரிமை கொண்டுள்ள ஆளப்படுபவர்கள். அரசியவாதிகளும் ,அரசு அதிகாரிகளும் ஆட்சியாளர்களுக்கும் ,அரசு வருவாய்துறை சார்ந்த பணியாளர் களும் அரசு அதிகாரிகளுக்கும் காலப்போக்கில் தொண்டர்களாக மாறி சங்கிலித் தொடர் போன்ற ஒரு பெரிய கட்டமைப்பை ஏற்படுத்திக்கொண்டுவிடுகின்றார்கள். முதல் பிரிவினர் வசதி மற்றும் பொருள் மீது கொண்டுள்ள தீராத ஆசைக்கு மக்களின் உழைப்பைச் சுரண்டுகின்றார்கள் .அரசின் நீதியைத் தனதாக்கிக் கொள்கின்றார்கள் . செலவழித்து விட்டதாக கணக்குக் காட்டி நிதி ப் பற்றாக்குறையை மக்கள் மீது சுமத்திவிடுகின்றார்கள் .இரண்டாம் பிரிவினருள்மேல்தட்டு மக்கள் அரசின் அனுமதியாலும் அரசின் தயவாலும் அதிகம் சம்பாதிப்பவர்கள் .இவர்கள் எந்த நிலையிலும் அரசின் தவறுகளைச் சுட்டிக்காட்டுவதில்லை. மாறாக ஆள்பவர்களின் புகழ்பாடி தங்களுக்கு வேண்டிய காரியங்களைச் சாதித்துக் கொள்கின்றார்கள். இடைத்தட்டு மக்கள் எப்போதும் இருவேறு வகையினர் . எதாவது ஒரு காரணத்தை முன்னிறுத்தி எப்போதும் விவாதம் பண்ணிக்கொண்டே இருப்பார்கள் ஒருவர் அரசுக்கு ஆதரவாகப் பேசினால் மற்றொருவர் எதிர்த்துப் பேசுவார் .கீழ்த்தட்டு மக்கள் அரசு ஏதாவது இலவசம் தறாதா என்று ஏங்கிக் கொண்டே இருப்பார்கள் . என்றைக்காவது பயன் கிடைக்கும் என்று அரசை எப்போதும் புகழ் பாடிக்கொண்டே இருப்பார்கள். இடைத்தட்டு மக்களில் அரசுக்கு ஆதரவாளர்களும் ,கீழ்த்தட்டு மக்களில் அரசின் போலி வாக்குறுதிகளை நம்புகின்றவர்களும் ஆட்சியாளர்களுக்கு நடமாடும் விளம்பரங்களாகத் திகழ்கிறார்கள்.

அரசியல்வாதிகள் தங்களுடைய அரசியல் எதிரிகள் தொடர்பான எந்தப் பிரச்சனைகளிலும் அரசியல் ரீதியாக அணுகுவதில்லை. இதனால் அரசியல் பிரச்சனைகள் ஒரு முடிவுக்கு வராமல் புகைந்து புகைந்து ஒரு காலகட்டத்தில் நெருப்பைக் கக்கு கி ன்றது. அரசியலில் இருக்கும் ஒரே பிரச்னை யார் அரசின் நிதியை உரிமையோடு கொள்ளையடிப்பது. ஆட்சியா ள ர்களோடு தொடர்புடையவர்களாக இருந்தால் குற்றத்தை மறைத்து விட்டு தொடர்ந்து கொள்ளையடிக்க அனுமதிப்பார்கள். எதிரிகள் செய்தால் அதை தடுப்பார்கள். அதை ஊடகத்தில் பரப்பி அவர்கள் பெயரை கள ங்கப்படுத்துவார்கள்.ஆனால் சட்டத்தின் அடிப்படையில் தண்டிக்க மாட்டார்கள்

இருவர்ககிடையே கருத்து வேறுபாடு ஏற்பட்டு விட்டால் எதிர்த்து நின்று வெல்வதா அல்லது நட்புக்கரம் நீட்டி சமாதானம் செய்து கொள்வதா என்று சிந்தித்து செயல்படவேண்டும். எதிர்ப்பதாக இருந்தால் பலமுறை சிந்திக்கவேண்டும் சமாதானம் எப்பொழுது வேண்டுமானாலும் செய்துகொள்ளலாம் . ஆனால் எதிர்த்து விட்டு பின்னர் சமாதானம் செய்து கொள்வது அழகல்ல. அப்படிச் செய்யும் போது அந்தச் சமாதானம் மதிப்பிழ ந்ததாக இருக்கும்.. ஏனெனில் சமாதானம் சமாதான த்திற்காக மட்டுமே பின்பற்றப் படவேண்டும். தோற்றுப்போனதற்கு மாற்று நடவடிக்கையாக இருக்கக்கூடாது

Non-relativistic model of 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 r1.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. If electron-electron interaction is not allowed in the same orbit, the relative position of these two electrons may vary from time to time. Then different helium atom with different placement of electrons in the orbit may exist .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 e^2/4Kr1^2. The resultant force is balanced by centrifugal force i.e., 2e^2/Kr1^2 - e^2/ 4Kr1^2 = (7/4)e^2/Kr1^2 = mv1^2/r1 . The radius of the orbital electron in helium atom can be determined from the condition on its quantized angular momentum. It gives rn = n^2(4/7) ao. = 0.5714 n^2 ao. If we ignore the electron-electron interaction rn = ao/2 .The radius of the 1s orbit in He+ ion is increased by 0.0714 ao due to mutual electron-electron interaction in helium atom. Total kinetic energy of the electrons in the helium atom = mv1^2 = (7/4) e^2/Kr1 kinetic energy per electron = (1/2) mv1^2 = (7/8)e^2/Kr1 = (7/4) e^2/2Kr1 Potential energy of electron 1 = - 2 e^2/Kr1 Potential energy of electron 2 in the presence of electron 1 = - 2 e2 /Kr1 + e2 /2Kr1 Total potential energy of the electrons in the helium atom = - (7/2) e2/Kr1 Sum of K.E and P.E is - (7/4)e^2/Kr1 = - (49/8)[e^2/2Kao]= -6.125 x 13.595 = -83.269 eV which gives 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.

Application of Bohr's Theory to He+ The line spectrum of the He+ ion will resemble that of the hydrogen atom (H) because both are one-electron system. They behave similarly at the quantum level despite different nuclear charges. Due to strong attractive force, the orbital electron in helium comes more closure to the nucleus and has more binding energy. Although the energy levels and wavelengths for He+ will be different (shifted to shorter wavelengths) due to the stronger nuclear pull, the overall pattern of spectral lines from electron transitions remains analogous. The electrostatic force on the electron equals with centrifugal force 2e2/K rn2 = m vn2/rn, m vn^2 rn = e^2/2πεo = constant ...... (2.1) Bohr's first condition-1 concerned with angular momentum requires mvnrn = nh/2π . Solving for rn, one can show that rn = n2 h2 εo / 2 mπ e2 = n2ao/2 .i.e., the radius of the innermost orbit in the helium ion is half the corresponding value of hydrogen atom. It can be shown that the radius of the innermost orbit in hydrogen like ions with atomic number Z is ao/Z. The orbital electron has both kinetic energy due to its orbital motion and potential energy by virtue of its position in the electrostatic field of the nucleus. Kinetic energy of the electron in the innermost orbit = (1/2) m v12 = e2 /Kr1 and its potential energy is -2e2 / Kr1 which together gives its total energy T.E = - e^2/ Kr1 .Substituting the value for r1 = h^2 K/8π^2 m e^2 , E1→∞ = [2e^2 /Kao ]= 4 (13.595) eV. In the case of hydrogen, it is E1→∞ = e^2/2Kao = 13.595 eV, which is 4 times lower than that of the helium ion. When the electron is in the innermost orbit of helium ion He+, its energy is 4 x 13.595 = 54.38 eV , The second ionization energy of helium atom is 54.40 eV which is in close agreement with the computed value. Its spectral feature can be studied by determining the electron transition energy and the corresponding wavelength of radiation emitted. ΔE = E excited /initial - E ground / final =4[13.595][(1/n1^2 - 1/n2^2)] , n2 > n1 The frequency of the emitted radiation ν = ΔE/h = 4[13.595/h][(1/n1^2 - 1/n2^2)] The wavelength of the emitted radiation λ = C/ν = hc/4(1/n1^2 - 1/n2^2)13.595 The Lyman series of helium 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]/4[13.595 x 1.602 x 10^-19][1/(1/n1^2 - 1/n2^2) = 0.2280 x 10-7 x ][1/(1/n12 - 1/n22) n2→n1; 0.2280 x 10^-7 x 4/3 = 30.40 nm n3→n1; 0.2280 x 10^-7 x 9/8 = 25.65 nm n4→ n1; 0.2280 x 10^-7 x 16/15 = 24.32 nm The limit of the series n∞→ n1: 22.80 nm The Balmer series of helium ion He+ contain the following wavelengths n3→ n2 ; 0.2280 x 10^-7 x 36/5 = 164.16 nm n4→ n2; 0.2280 x 10^-7 x 16/3 = 121.60 nm n5→ n2; 0.2280 x 10^-7 x 100/21 = 108.57 nm The limit of the series n∞→ n2: 91.20 nm The Paschen series of helium ion He+ contain the following spectral lines n4→ n3 ; 0.2280 x 10^-7 x 144/7 = 469.02 nm n5→ n3; 0.2280 x 10^-7 x 225/16 = 320.62 nm n6→ n3; 0.2280 x 10^-7 x 12 = 273.60 nm The limit of the series n∞→ n3: 205.20 nm The wavelengths of the He+ ion spectrum are specific to the electron transitions between energy levels and are calculated by using the Rydberg formula, for example the 2p to 1s transition gives a wavelength around 30.3 nm. In the Balmer series, the second line (n=4 to n=2) has a wavelength of approximately 121.6 nm. The Paschen series of He+ consists of infrared wavelengths . The hydrogen, helium ion and helium atom all together contribute the ultraviolet (UV) spectrum of the sun usually it ranges from approximately 100 to 400 (nm) in wavelength. This spectrum is divided into three main bands: UVA (315–400 nm), UVB (280–315 nm), and UVC (100–280 nm). Most of the UVA and UVB radiation reaches the Earth's surface, while the ozone layer absorbs almost all UVC radiation. Relativistic Bohr Model of He+ Both the non-relativistic and relativistic approaches to helium ion Bohr's model gives same result. Applying the law of conservation of energy 2e^2/Krn = (1/2)mn vn^2 + dmnc^2 = (1/2) mo vn^2 + dmn c^2/(1-vn^2/c^2)^1/2) e^2/Krn = dmn c^2 = (mn - mo) c^2 ≃ (1/2) mo vn^2 , where rn = (n^2/2)ao [1-vn^2/c^2]^1/2 .By solving the relations for the nuclear attractive force and the orbital angular momentum vn = e2/nhεo .Substituting this value in dm c^2 dmnc^2 = mo e^4/2n^2h^2 εo^2 = (1/n^2)4 [e^2/2Kao] = [54.38/n2] eV = (1/n^2)[4e^2/2Kao]/[1 - vn^2/c^2]^1/2 Where vn = e^2/nhεo. When the electron is in the innermost orbit of helium ion, dm1c^2 will represent the atomic binding energy . dm1c^2 = [4 x 13.595][1 - e^4/h^2εo^2c^2]^1/2 = 54.38 x 0.99989 = 54.37 eV The spectral lines are the electromagnetic radiation due to the difference of binding energies of the final and initial positions of the electron in the system. [(dm)f - (dm)i]c^2 = (1/2)mo[vf^2 - vi^2] = (1/2) mo[e^4/h^2εo^2][1/nf^2 - 1/ni^2] = 4[e^2/2Kao][1/nf^2 - 1/ni^2] For a known spectral line in the spectrum of He- ion, the possible electronic transition between the energy levels can be predicted. For example λ = 320 nm, Using the above relation one can identify which transitions will give this wavelength . ΔE = hc/λ = 54.38[1/nf^2 - 1/ni^2] (ni^2 - nf^2)/ni^2 x nf^2 = 0.07125 For the positive values of both ni and nf , the best amicable value of nf =3. ni^2 - 9 = 0.07125 x 9xni^2 = 0.64125 ni^2 or 0.35875 ni^2 = 9 or ni = 5.0087 .It becomes 5 after adjusted to nearest whole number.If there are two or more possibilities for a any particular line, that would be more intense.