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.