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 .