Normal excited states and de-excitation 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 Enn be the energy of the helium atom in its ground state and in its nth 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.. 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 e^2/Kr11 + e^2/2Kr11 + (7/4) e^2/Kr11 = -(7/4) e^2/Kr11= (7/4)^2 e^2/Kao = 83.269 eV. When both the electrons are in the same orbit labelled by n, its total energy is Enn = NPEnn + (e-e ) nn + KEnn = - (7/4) e^2/Krnn = - (7/4)^2 (1/n^2)e^2/Kao. When one of the electrons is in the innermost orbit and other electron is in the nth orbit, its total energy E1n = NPE1n + (e-e)1n + KE1n. NPE1n and KE1n are half of the sum of total negative potential energies and kinetic energies respectively with electrons in the innermost and n th orbit NPE1n = -[2e^2/K][1/r11 + 1/rnn] = - [2e^2 /Kr11][1 +1/n^2] - [2e^2/Kr11][(n^2 +1)/n^2] KE1n = [(7/8)e^2 /K][1/r11 + 1/rnn] = [(7/8)e^2/K][(n^2 +1)/n^2] (e-e)1n = e^2/Kd1n where d1n is the distance between the two electrons when they are in two different orbits. But d1n = r11 + rnn and e^2/K = 2(e-e)11 r11 = 2[(e-e)11x r11]/(r11 + rnn) = 2(e-e)11/ (1+n^2) =[e^2 /Kr11][1/(n^2+1)] E1n = [e^2/Kr11]{[(n^2 +1)/n^2][(7/8)-2] + [1/(n^2 + 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 . E1∞ = [e^2/Kr11][ 1 x (-9/8)] = -[9x7/16] e^2/2Kao = 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 nth orbit , then the state E1n gets 1/2 share of kinetic and negative potential energies from each contributing states (E11 and Enn), 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 [m^2/(n^2+m^2)]. 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 [n^2/(n^2+m^2)]. This energy contributed by E11 is utilized by E1∞ to have more potential and kinetic energy with higher binding energy. E1∞. = (1/2) [KE11+ NPE11] + (e-e)11 [l/(1+∞)] =. (1/2) [ KE11+ NPE11 ] .In fact E1∞ 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 E1∞ is the first ionization energy i.e., (-83.269) + (53.53) = 29.739 eV. .Enn is formed by mixing of two identical states Enn each state contributes 1/2 of its energy and keep the energy same. When a hyper state Enn is formed by half-half mixing of two hyper-states Enn and Enn , then its energy by this method becomes 2 (1/2)[ KEnn + NPEnn] + 2(e-e)nn n^2/2n^2] = Enn = [KEnn + N PEnn + (e-e ) nn] . The above expression for Enm 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 Emn gets 1/2 share of kinetic and negative potential energies from each contributing states (Emm and Enn) , 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 [m^2/(n^2+m^2)]. 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 [n^2/(n^2+m^2)]. In helium atom (e-e)mm = e^2/2krm, (e-e)mm = e^2/2krn and (e-e)mn = e^2/k(rn + rm). Let us suppose that the fraction of contribution by the hyper excited is x. Then x[e^2/2krm] + (1-x) [e^2/2krn] = e^2/k(rm+ rn) .By solving, we can determine x, x [1/rm - 1/rn] = 2/(rm+ rn) - 1/rn ; x = rm /(rm + rn) and (1-x) = rn / (rm + rn) Since r is proportional to square of its orbital quantum number x = m^2/(m^2 + n^2) and (1-x) = n^2/(m^2 + n^2). Emn =(1/2)[KEmm + NPEmm + KEnn + NPEnn] + (e-e)mm [m^2/(m^2 + n^2)] + (e-e)nn[n^2/(m^2 + n^2)] Since (e-e)mm = (e-e)11/ m^2 and (e-e)nn = (e-e)11/n^2 , (e-e)mm = (m2/n2)(e-e)nn. Substituting this value in the above relation, we get, Emn =(1/2)[KEmm + NPEmm + KEnn + NPEnn] +2(e-e)mm [m^2/(m^2 + n^2)] Lyman series for Helium atom 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 E22. The radius of the inner most electronic orbit is (4/7)ao and the radius of the next higher orbit is (16/7)ao The three components of energy pertaining to the ground state of helium atom is kinetic energy KE11= (7/4) e^2/Kr11 ; negative potential energy NPE11 = - 4e^2 /Kr11 and the positive electron-electron interaction energy (e-e)11 = e^2/2Kr11. The corresponding components for next higher energy state are KE22 = (7/4) e^2 /Kr22, PE22 = - 4e2 /Kr22 and (e-e)22 = e^2/2Kr22 . 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 [e^2/2Kr11][ -(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 nth orbit and other electron in the innermost orbit. The energy associated with the system E1n can be computed as before from its contributors E11 and Enn .The energy contributed by E11 to E1n is (1/2)[ KE11 +NPE11], the energy contributed by Enn to E1n is (1/2)[ KEnn + NPEnn], and the electron-electron interactional energy in the resultant assembly is (e-e)1n = 2(e-e)11[1/(n^2 +1)] = 2(e-e)nn [n^2/(n^2+1)] ,. The energy associated with E1n is the sum of these three contributions and is equal to (1/2)[(KE11 + KEnn) + (PE11 + PEnn)] + [2(e-e)11 [(1/(n^2+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 E1n - E11 = (1/2)[(KE11 +NPE11) + (NPEnn + KEnn)] + 2(e-e)11 [(1/(n^2+1)] - [KE11 +NPE11 + (e-e)11] = (1/2) [KEnn - KE11 + NPEnn - NPE11] + 2(e-e)11 [(1/(n^2+1)]- (e-e)11 = (1/2)[ KEnn - KE11 + NPEnn - NPE11] + [(1-n^2)/(1 + n^2)][(e-e)11] Substituting the values of various components of energy we get (E1n - E11) ={(1/2)[49/8n^2 - 49/8 - 14/n^2 + 14]+ [(1-n^2)/(1 + n^2)](7/4)} (e^2/2Kao). On simplification, ΔE(1n→ 12) = [(1-n^2)/2n^2]{-63/8 +(7/2)[n^2/(1+n^2)]} (e2/2Kao). 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 E1n → E11, the transition energy is given by [(1-n^2)/2n^2]{-63/8 +(7/2)[n^2/(1 +n^2)]}(e^2/2Kao). Using this formula the Lyman series for helium atom can be predicted. Table 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.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) e^2/Kr22 - 4 e^2/Kr22 - (7/4) e^2/Kr11 + 4 e^2/Kr11] + (1/4)e^2/2Kr11 - (2/5) e^2/2Kr11 = (1/2)[e^2/Kr11][(7/16)-1- 7/4 + 4] + [e^2/2Kr11][(1/4) - (2/5)]= (7/4) [e^2/2Kao] [(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 Enn to E1n is Enn - E1n = (1/2)[KEnn + NPEnn - KE11 - NPE11] + [(e-e)nn - 2(e-e)11 /[1/(n^2+1)]N = [(35+63n^2)/16(n^2+1)][(n^2 -1)/n^2]. 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. 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 Emn and E1n or E1m . 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-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} (e^2/2Kao) = {(1/2) [ 5.0312] - 1.4807 }(e^2/2Kao) = 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. 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]}(e^2 /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]} (e^2 /2Kao) = {(1/2[0.6805 - 1.5555 - 1.5312+ 3.5] + (63/52)(-5/36)} (e^2 /2Kao) = 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, E1n = (1/2) [KE11 + NPE11 + KEnn + NPEnn] + [1/(n^2 +1)](e-e)11 + [n^2/(n^2 + 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) =E1n → E12 = (1/2)[KEnn + NPEnn - KE22 - NPE22] + [1/(n^2 +1)](e-e)11 - [1/(5)](e-e)11 (4/5)](e-e)22 + [n^2/(n^2 + 1)](e-e)nn =(1/2)[KEnn + NPEnn -KE22 -NPE22]+[(1/n^2+1) -1/5](e-e)11 -(4/5) (e-e)22 +[n^2/(n^2 + 1)](e-e)nn = {(1/2)[49/8n^2 - 49/32 -14/n^2 +14/4]- (7/4)[(4-n^2)/5(n^2+1)] -7/4 (1/5) +[n^2/(n^2 + 1)](7/4n^2)}(e^2 /2Kao) = (49/16)[(4-n^2)/4n^2] - 7 [(4-n^2)/4n^2] + (7/10) [(4-n^2)/(n^2 +1)] = (7/4) (4-n^2)[7/16 n^2 - 1/n^2 + (2/5) [1/(n^2+1)] = (7/2) (n^2 - 4) [13 n^2 + 45]/[16 n^2 (n^2 +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 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. E1n = (1/2) [KE11 + NPE11 + KEnn + NPEnn] + [1/(n^2 +1)](e-e)11 + [n^2/(n^2 + 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 - n^2)/10(n^2 +1)](e-e)11 - [9/10](e-e)33 + [n^2/(n^2 + 1)](e-e)nn = (1/2) [49/8n^2 - 14/n^2 - 49/72 + 14/9](e^2 /2Kao) +(7/4)[(9-n^2)/10(n^2+1)] -9/10 (7/4) (1/9) + 7/4 [1/(n^2+1)][e^2/2Kao] = [(9-n^2)/9n^2](-63/16) + (7/20)[(9-n^2)/5(n^2+1)][e^2/2Kao] = (7/4) (n^2 - 9) [(n^2 +5)/20n^2 (n^2 +1)][e^2/2Kao] 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 .

Hyper excited states and de-excitation Three states of 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. Hyper excitation and Hyper de-excitation Total energy of helium in its nth hyper excited state is given by - (1/n^2)(49/8) (e^2/2Kao). In two different hyper excited states with n = n1 and n2,the total energies are - (1/n1^2)(49/8)(e^2/2Kao)and - (1/n2^2)(49/8) (e^2/2Kao) 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) (e^2/2Kao)[(1/n1^2) - (1/n2^2)] = 83.269 [(1/n1^2) - (1/n2^2)] eV. Since λ = hc/ΔE, where ΔE is in joules, the corresponding wavelength of radiation emitted in any transition is given by λn2* → n1* is 12.4 x 10^-7/(49/8) (e^2/2Kao) [(1/n1^2) (1/n2^2)]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 Table. 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

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