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.