Research article Revisiting Bohr's Theory of Hydrogen Atom M.Meyyappan* [Professor of Physics(Retd), Alagappa Government Arts College, Affiliated to Alagappa University,Karaikudi- 630003, Tamilnadu, India, *email id: meydhanam@gmail.com Orchid No:0000 0002 5194 2992] Abstract [Since the velocity of orbital electrons in atoms is of the order of 106 m/s and comparable with the velocity of electromagnetic wave in free space, its relativistic variation of mass cannot be ignored. The Bohr's theory ofhydrogen atom is studied again with and without considering the relativistic variation of mass of orbitalelectron. When a free proton and an electron combine to form a hydrogen atom, half of the loss of its potential energy is converted into its kinetic energy required to move in a stable orbit surrounding the protonand the remaining half is deposited as relativistic increase of its mass. The orbital electron binds with theproton where the atomic binding energy comes from the energy equivalent of the added mass of the system.This energy is radiated out as electromagnetic waves. It is shown that the energy released during the electronic transition from orbit with quantum number n2 to n1 is equal to the energy equivalent of relativisticvariation of mass of the electron. i.e., ΔE = Δm c2 = (mn1- mn2)c2. It gives same value by the difference in total energies of the orbital electron in n1and n2, ΔE = En2- En1]. Key words: Atomic structure- Bohr's theory of Hydrogen atom- spectral feature--Rydberg formula relativistic correction- modified Bohr model-atomic binding energy. Revisiting Bohr's Theory of Hydrogen Atom Introduction Bohr's atomic theory of hydrogen atom [1,2,3], while revolutionary, is primarily limited to mono electron systems such as Hydrogen (1H1) and hydrogen-like ions (2He1- and 3Li2-). This theory revolutionized atomic physics by introducing quantized energy levels, explaining the stability of atoms, and accurately predicting the hydrogen spectral lines and ionization energy. It resolved contradictions in classical physics by proposing stationary orbits where electrons do not radiate energy unless jumping between shells. Unfortunately Bohr's theory of hydrogen atom fails to explain atoms with higher atomic number, where the additional electron electron interactions introduce some corrections in the theory. These interactions depend on the speed of electrons moving in circular paths, its spin, distance between them, and its periodic changes during its continuous orbital motion. Since the radius of the orbit and orbital velocity of the electron are determined by the interaction within the system, any additional interactions disturb the configuration by changing them obviously. The non-relativistic Bohr's theory of hydrogen atom has some limitations [4]. It does not explain the fine structure and the splitting of spectral lines in an external magnetic field (Zeeman effect) or electric field (Stark effect). There is no explanation why some spectral lines in emission spectra are more intense than others. It contradicts the Heisenberg Uncertainty Principle as Bohr assumed electrons move in well-defined circular orbits with a known radius and momentum simultaneously. This model treats electrons solely as particles and fails to account for its wave-particle duality. It fails to explain why atoms form chemical bonds or how they share/transfer electrons to form molecules. All the micro particles have dual nature exhibiting its wave characteristics called matter waves.[5,6].The wavelegth associated with an electron moving with an momentum p is given by h/p. According to de Broglie,the motion of an electron can only be stable if the phase wave is tuned with the length of the path [7]. The wave characteristics of the electron in an atom demand that the circumference of any electronic orbits must be equal to some multiples of electron wavelength. i.e., 2πrn = nλ. where n is a quantum number which can have only some integral values. This condition makes the wave to retrace its circular path with same phase at every point. It is required to keep the atomic structure in the same state of stability at all time. The first Bohr's postulate states that the angular momentum of the orbital electron is some integral multiples of h/2π.The wave characteristics of the electron make the electronic orbits to be discrete (Fig.1) Fig.1. Discreteness of electronic orbits in an atom If the orbital electron is represented by de-Broglie wave, they should move together, that is the wave velocity should be very same as that of the electron. For any type of waves, the product of frequency and wavelength must be equal to its velocity of propagation. The phase velocity u = ν λ , energy equivalent of the electron of mass m is mc^2 which gives the frequency of its characteristic wave as mc^2/h and λ = h/p = h/mv .Substituting these values u = (mc^2 /h) (h/mv) = c^2/v. Since the particle velocity v cannot equal or exceed the velocity of light in free space c, the de-Broglie wave velocity u is always greater than c. It is vivid that v and u usually are not equal for a moving body. Since u is greater than c , the de Broglie wave must be different in nature from that of the other known waves electromagnetic and sound. It implies that the wave velocity of matter wave has no simple physical significance at all. To avoid this difficulty, the matter wave associated with a moving body is represented as a wave packet. The group velocity of a system of waves is given by w = u- λ (du/dλ). This can be written as w=-λ^2 d/dλ(u/λ) =- λ^2 dν/dλ or 1/w =- (1/λ2) (dλ/dν) = d/dν(1/λ) If E and Vrepresent the total and potential energies of the system, then(1/2) mv^2 = E-V or v = [2(E-v)/m]^1/2 and E=hν.Using these relations, 1/λ = mv/h=(1/h)[2m(hν- V)]1/2 hence 1/w = (1/h)[2m(hν- V)]^1/2 = 1/v or w = v. Thus the wave packet travels with the same velocity as the body. The conceptual difficulty in the interpretation of the wave velocity exists since a single relation is employed to represent the frequencies of radiation and matter waves. Here an attempt [8] is made to derive a relation for the phase velocity of de-Broglie matter wave, where the matter wave is considered as entirely different from that of radiation. When the matter at rest is completely converted into energy, then according to Einstein's theory of relativity νo = moc^2/h . If the matter is moving, then ν = mc^2/h = moc^2/h + K.E/h or ν > νo where ν and νo represent the frequencies of radiation that comes out of complete destruction of matter in motion and at rest respectively. Hence E= moc^2 = hνowave and E = mc^2 = h ν(wave). When the matter is at rest, there is no matter wave at all, but still it has energy equivalent of mass. Hence it is reasonably supposed that matter wave is related with particle's kinetic energy rather than its total energy. The kinetic energy of matter is (m-mo) c^2 and for its wave nature it is h νmatter. With this knowledge,one can derive an expression for ν (matter) νmatter = (m-mo) c^2/h = mo c^2 /h{[1/ √(1-v^2/c^2)]-1} but ν(wave) = E/h = mc^2/h = [(moc^2)/h√(1-v^2/c^2)] which give ν(matter)/ν(wave) =1- √(1-v^2/c^2). Similarly λ (wave)= c/ν(wave) = hc/E and λ(matter)= h/p and its ratio gives λ(matter)/ λ (wave) = E/pc = mo c^2 / √(1-v^2/c^2) x[1/[mov/√(1-v^2/c^2)]c = c/v . When v = 0, λ(matter) → ∞,it corresponds to a particle at rest.When v = c, λ(matter) = λ wave.It corresponds to a specific case where the matter wavelength is exactly equal to radiation wavelength. This is true only for luxons like photons. Let umatter be the velocity of matter waves. Then umatter/uwave=(λmatter/ λ wave) x (νmatter/ νwave)= (c/v)[1- √(1 - v^2/c^2)].Substituting u(wave) = c , u(matter) = (c^2/v)[1- √(1-v^2/c^2)]. At two extreme specific cases, when v = 0 , um → 0 and v =c, um→ c.Thatis de-Broglie wave velocity u(m) can never be greater than c, the velocity of light in free space. This expression accounts for the relation for group velocity. The angular frequency of the matter wave is given by ω=2πν(m)=2πν(wave)[1- √(1-v^2/c^2)] = [2π moc^2/h] {[1- √(1-v^2/c^2)]/[√(1-v^2/c^2)]} or dω/ dν =2πmov/h(1-v^2/c^2)^3/2 and the propagation constant is given by k = 2π/λ = 2πmov/[h √(1-v^2/c^2)] or dk/dv = 2π mo/[h (1-v^2/c^2)^3/2], hence ω = dω/dk = (dω/dv)/(dk/dv) = v. That is the de-Broglie wave group associated with a moving particle travels with the same velocity as the body. Very same relation for hydrogen spectrum can be obtained by assuming the orbital electrons as matter waves. When an electron jumps from outer orbit to any one the inner orbits, its radius and potential energy are decreased ,orbital velocity and binding energy are increased and these changes result with a decrease in the wavelength of the matter waves associated with the orbital electron. During this jumping the excess energy is radiated out as electromagnetic radiation of wavelength λ(wave),. As they are all the after-effects of the above electronic transition, one may believe that they may be linked indirectly. Based on this assumption, let us determine the wavelength of em radiation emitted in a transition from orbit with quantum number n2 to orbit with quantum number n1 The energy of transition (ΔE) from n(2)to n(1) = E(2)- E(1) , where E(1),E(2) denote the total energy of the orbital electron in the orbit 1 and 2 respectively. E(2)- E(1)= ΔE =hν=hc/λ(wave) or λ(wave) = hc/ΔE(2→1). The wavelengths of matter wave associated with the electron in the orbit with quantum number n(2)and n(1) are h/mv(2) and h/mv(1) respectively. It gives mv(1) = h/λ(matter1) and mv(2) = h/λ(matter2). With this knowledge one can estimate the kinetic and potential energies of the orbital electron in terms of its wavelength of matter wave. Kinetic energy of the electron in orbit n(2) = (1/2) mv2^2= (1/2m)h^2/λ^2(matter2). Potential energy of the electron in orbit n(2) =- e^2/Kr^2=- mv(2)^2 =- (1/m)h^2/λ^2(matter2). Total energy of the electron in the orbit n(2)= Sum of kinetic and potential energies E(2) =- (1/2m) h^2/λ^2(matter2).Total energy of the electron in the orbit n(1)=E(1)=- (1/2m)h^2/λ^2(matter1) ΔE(2→1) =(h^2/2m) [1/λ^2(matter1)- 1/λ^2(matter2)] The radius of the electronic orbit with quantum number n in hydrogen atom is shown as r(n)= n^2ao , where ao is Bohr's radius, the radius of the innermost orbit of the electron in the hydrogen atom. From this one can derive mv^2 = e^2/Kn^2ao and λ^2(matter) = h^2 Kn^2ao/me^2 = (2πnao)^2. This can be verified by substituting the values for matter wavelength in ΔE(2→1). ΔE(2→1) =(h^2/2m)[1/(2πao)2][1/n2^2- 1/n1^2] = [h^2/8π^2mao2][1/n2^2- 1/n1^2] = [e^4m/8h^2εo^2][1/n2^2- 1/n1^2] and λ(2→1)= 8cεo^2h^3/(1/n(f)^2- 1/n(i)^2)me^4= (1/n(f)^2- 1/n(i)^2)/RH where R H= me^4/8cεo^2h^3 called Rydberg constant. By redoing non-relativistic Bohr's theory of hydrogen atom the energy radiatied out during jumping of the electron from one orbit to another is derived.To understand the mutual dependence of various variables such as orbital velocity v(n), orbital radius r(n) and the order of the orbit (n) is derived. (1) Orbital velocity v(n ) Vs orbital radius (rn) The stability of the electron in the orbit requires that the mutual attractive force between the nucleus and the orbital electron must be equal to the cetrifugal force experienced by it. e^2/(4πεo)m = constant = v(n)^2 r(n) ....... (1) It predicts that v(n) is inversely proportional to r(n)^1/2 .The same result is obtained from the Bohr's postulate on the angular momentum of the electron mv(n)r(n) = nh/2π ........ (2) It gives v(n)r(n)/n = v(n)^2r(n)/n v(n) or n v(n) = constant as v(n)^2r(n) = constant . Squaring equ (2) and dividing by equ (1) one can arrive r(n) = n^2 ao . By substituting this value for r(n) we get v(n)^2 n^2 ao = constant, It shows that the radii of higher orbits in hydrogen atom are 4 ao,9 ao,16 ao.... n^2ao and the spacing between the nth and (n+1)th orbits is (2n+1)ao (2) Orbital velocity v(n) Vs Quantum number(n) Dividing equ (1) by equ (2) we have n v(n) = [e^2/2εoh] = constant ,which indicates that v(n) is inversely proportional to n.i.e, as we go away from the center, the orbital electron moves slower. Very same relationship can be derived from the condition introduced for the circumference of the orbit (3) Orbital readius r(n) Vs Quantum number (n) Again by squaring the relation [v(n)r(n)/n]^2 = v(n)^2 r(n)/[n^2 /r(n)] or n^2 /r(n) = constant as v(n)^2 r(n) = constant. It shows that r(n) is directly proportional to n^2. (4) Rate of variation of orbital velocity with respect to radius dv(n)/dr(n) Differentiating v^2 r = constant with respect to r, v^2 + 2vr (dv/dr) = 0, which gives dv(n)/dr(n) =-v(n)/2r(n). By substituting the values of r(n)[r(n) = n^2 ao] and the corresponding v(n)[mv(n) r(n) = mv(n)n^2ao = nh/2π or v(n) = h/2π mnao which states nv(n) = constant] one can find out the rate with which the orbital velocity changes with radius of the orbit, dv(n)/dr(n) =- v(n)/2r(n) =- h/4πmao^2 n^3. i.e., dvn/drn is negative and varies inversely proportional to n^3. The standard Bohr model works well for hydrogen, but the relativistic model is necessary for heavy hydrogen-like ions where the orbital electrons travel at a significant fraction of the speed of light, rendering non-relativistic calculations inaccurate. Objective of the study The objective of the Study is to predict how the non-relativistic theory gets modified when the relativistic increase of mass of the orbital electron is incorporated in the Bohr's theory of hydrogen atom. When Bohr's atomic theory was reviewed with the aim of extending Bohr's atomic theory to atoms other than hydrogen, the first thing that came to mind was taking into the account of the relativistic variation of mass of the orbital electrons moving rapidly in circular paths. This correction is assumed to be necessary because the speed of an electron moving in an atomic orbit is of the order of 10^6 m/s. As the velocity of the orbital electron increases, its mass increases relative to its rest mass. Therefore, the relativistic effects on the core electrons are to give them larger masses that gives more kinetic energy to the particle and this shrinks the Bohr radius. For the expandable utility, the present non-relativistic Bohr's theory of hydrogen atom is studied again by incorporating the relativistic variation of mass of the orbital electron. The present relativistic Bohr's theory of hydrogen atom provides appropriate guidance to minimize its limitations. With this focus in mind the theoretical study based on famous Niels Bohr atom model is undertaken. Related Work/Literature Review The standard Bohr model was studied by many researchers (9-12) by introducing relativistic variation of mass of the orbital electron that improves the Bohr model by giving an account to one or more limitations. The relativistic energy of an electron, moving at a significant fraction of the speed of light c, is defined by its total energy En = moc^2/(1- v(n)^2/c^2)^1/2 , which includes both rest energy moc^2 approximately 0.511 MeV and kinetic energy. The total energy of the electron is E = (p^2c^2 + mo^2c^4)^1/2 , where p is the momentum of the electron. The relativistic energy levels for an electron of charge e and rest mass mo in a hydrogenic atom with atomic number Z are given by: En = [mo^2 c^4 + mo^2c^4α^2Z^2/n^2]^1/2 where α = 2πe^2/hc is the fine-structure constant, which approximates to E(n)= moc^2[1 +α^2Z^2/2m ]. The relativistic model shows that the energy depends not only on the principal quantum number n but also includes relativistic corrections that affect the fine (structure of hydrogenic spectral lines. Again the radius of the orbits [r(n)= n^2(h/2π)^2/meke^2] technically changes because the electron's mass m(e) increases with velocity, making the orbits slightly smaller than predicted by the non-relativistic model, especially for low n. Relativistic Bohr's Theory of Hydrogen atom The hydrogen atom with a central proton and an orbiting electron is a two body problem. When a free electron enters the active space it gets accelerated towards the proton, thereby its kinetic energy is increased gradually due to accelerated motion towards the nucleus and simultaneously it gains some mass due to relativistic variation of its velocity. The energy required for this comes from the source that accelerates the electron [13]. The work done by the electron reduces its potential energy. In this motion of the electron within the atom, the energy must be conserved at every point of its path. It means that the loss of its potential energy must be equal to gain in its kinetic energy and the energy used for the relativistic increase of mass. That is a fraction of the loss of potential energy is converted into its kinetic energy and remaining is stored as its relativistic increase of mass, so that the energy is conserved all along it path. As the accelerated electron is abruptly stopped at any one of the allowed orbits, it takes up a curved path until it attains stability in a stable orbit. It revolves round the proton in circular orbit, where the electrostatic force of attraction is exactly counter-balanced with its centrifugal force The conservation of energy of the electron at all of its position requires that the loss of its potential energy must be equal to gain in its kinetic energy and the energy used for the relativistic increase of mass. e^2/Kr(n) = (1/2)m(n) v(n)^2 + Δmc^2 ,where Δm = (mn- mo) .....(3) When the electron is attracted by the nucleus it gets accelerated. Since the force is inversely proportional to intermediate distance, the acceleration experienced by the electron is not uniform. Let mn, r(n) , v(n) be the mass, radius and velocity of the electron in its n th orbit F = m(n)a(n)=e^2/Kr(n)^2 = m(n)v(n)^2/r(n) or a(n) = v(n)^2/r(n).......(4) It shows that e^2/K = m(n)a(n) r(n)^2 = m(n) v(n)^2 r(n) = [mo/(1- v(n)^2/c^2)^1/2]v(n)^2 r(n) . From the discreteness of the electronic orbits in atom 2πr(n) = nh/m(n)v(n). Squaring both sides and substituting the value for mnvn, 4π2rn2 = n2h2/mn2vn2 = n^2 h^2 K r(n)/m(n)e^2. It gives r(n) = n^2ao(1-v(n)^2/c^2]^1/2. By substituting the value of rn we get e^2/K = mo v(n)^2 n^2 ao = constant or v(n) n = constant irrespective of the relativistic variation of mass of electron.The variable acceleration a(n) = v(n)^2/r(n) = [v(n)^2/(1-v(n)^2/c^2)^1/2][1/n^2ao]. It helps to study how the relativistic approach on the Bohr's theory of hydrogen atom makes changes in the dependency among the various dependent variables (1).Orbital velocity v(n) Vs Orbital radius r(n) Let us suppose that an electron in the hydrogen atom is in its nth orbit having radius rn.The nuclear attractive force is counter-balanced with the centrifugal force. Relativistic mass m(n) increases with velocity v(n) according to Einstein’s Special Theory of Relativity, defined by m(n)= m(o)/ [1- v(n)^2/c^2]^1/2.Incorporating this value in equ (1) e^2/moK = v(n)^2r(n)/(1-v(n)^2/c^2)^1/2 = v(n)^2n^2ao = constant ....... (5) The relativistic variation of mass of the orbital electron makes no correction in the dependency of vn on rn.It implies that the added mass of the electron in the orbital motion is converted into atomic binding energy which holds the orbital electron with the nucleus. (2).Orbital velocity v(n) Vs Quantum number(n) Dividing equ (1) by equ (2) , v(n) = e^2 /2nhεo ........(6) As the orbital velocity vnis independent of mass of the electron, the dependency of v(n) on n is not altered due to the relativistic variation of mass of the orbital electron. (3) Orbital radius r(n)Vs Quantum number (n) From equ (2) m(n)^2v(n)^2r(n)^2 = n^2h^2/4π^2 Dividing one by equ (1), m(n)r(n) = n^2h^2/4π^2x(K/e^2) [mo/(1-v(n)^2/c^2)^1/2]r(n) = n^2h^2εo/πe^2= n^2moao r(n) = [n^2h^2εo/moπe^2][1-v(n)^2/c^2]^1/2= n^2ao(1-v(n)^2/c^2]^1/2 ........(7) The relativistic variation of mass of the orbital electron makes the orbit to shrink about its center. Approximately the change in the radius is h^2/8π^2m0^2aoc^2irrespective of n.Substituting this value of r(n) in m(n)v(n)r(n) = nh/2π [mo/(1-v(n)^2/c^2)^1/2 ] x v(n)xn^2ao(1-v(n)^2/c^2]^1/2 = nh/2π , v(n)= h/n2π moao .........(8) Comparing equs (6) and (8) we get the same result for the Bohr's radius ao = εoh^2/ π moe^2 By estimating the orbital velocity of the electron in the hydrogen atom and the corresponding change in its relativistic mass, one can show that the energy equivalent to the relativistic change of mass is equal to its binding energy with the nucleus. The velocity of the electron in the nth orbit in hydrogen atom is given by v(n)= e^2/2nεoh = (1.602 x 10^-19)^2/2nx(8.85 x 10^-12)x(6.626 x 10^-34) =[2.188 x 106/n] m/s v(n) = h/n 2πmoao= 7x6.626 x10^-34/n x 22 x 9.11 x 10^-31x 5.29 x 10^-11 = [2.187 x 106/n] m/s v(n)^2/c^2 = (2.188 x 10^6)^2/n^2(2.998 x 10^8)^2=4.787 x 10^12/n^2x 8.988 x 10^16 =[0.5326/n^2] x 10^-4 The mass of the orbital electron in the nthorbit mo /[1- v(n)^2/c^2]^1/2 ≃ mo[1 + v(n)^2/2c^2] The relativistic increase of mass Δm = mov(n)^2/2c^2and its equivalent energy Δm c^2= [mov(n)^2/2e] eV Δmc^2 =9.108 x10^-31x(2.188 x 10^6)^2/2 x (1.602 x 10^-19) n^2 = 13.6 /n^2 eV It is in agreement with the practical value of ionization energy of hydrogen atom from its various energy levels. The wavelength of various spectral lines in hydrogen spectrum can be studied from the knowledge of difference in the binding energy of the electron in different orbits, which can be worked out from the relativistic mass of the orbital electron in various orbits. The relativistic increase of mass of the electron in the orbits with orbital quantum number n1 and n2 approximately is given by Δm1 = mov1^2/2c^2 and Δm2 = mov2^2/2c^2 respectively and its equivalent energy is [mov1^2/2] and [mov2^2/2] respectively. When the electron jumps from n(2)to n(1) ,the difference of energy due to relativistic change of its mass is emitted out as em radiation. Its wavelength λ(2→1)= hc/ΔE(2→1) =2 hc/mo(v(2)^2- v(1)^2) where v(n)= e^2/2nεoh. Substituting the values for v(1)and v(2),we get the same relation for λ(2→1) as shown in non-relativistic Bohr's theory λ(2→1) = [8h^3εo^2c]/e^4mo[1/n(2)^2- 1/n(1)^2] . Results The relativistic variation of mass of the electron makes no changes in the Bohr's theory of hydrogen atom except the orbital radius, r(n) = r(n)o (1- v(n)^2/c^2)^1/2 . The revision on Bohr's theory of hydrogen atom provides an acceptable explanation in classical way for the non-existence of hydrogen negative ion and for the limiting reachability of the attracted electron towards the nucleus. The first (1s) orbit can accommodate a maximum of two electrons, but in the hydrogen atom its first orbit cannot be filled with two electrons. Two electrons can stay in the 1s orbit only when one more proton is present in the nucleus. It is confirmed with the existence of Helium atom. If one more electron is allowed in the first orbit of hydrogen, its total energy becomes positive and hence it moves away from the nucleus and the system transforms into a less potentially stable state. Electrostatic attractive force due to the nucleus is e^2/Kr^2. Since both the electrons are in the same orbit there must be mutual interaction between them that is why they attain stability by staying exactly diametrically opposite. Electrostatic repulsive force due to the presence of second electron is e^2/4Kr^2. The resultant force experienced by the electron is (3/4)e^2/Kr^2 . As it is counter-balanced by centrifugal force mv^2/r,the kinetic energy of both the electrons which are identical in all respect is m v^2= (3/4) e^2/Kr. The potential energy of the first electron due to the presence of the nucleus only is- e^2/Kr. When the second electron moves towards the neutral hydrogen atom, its electrostatic potential energy is zero until it reaches the electronic orbit of the first electron. Due to additional electron-electron interaction the potential energy is increased by e^2/2Kr. The total energy associated with the system = (3/4) e^2/Kr- e^2/Kr + e^2/2Kr = (e^2/4Kr) . Since there is no binding energy, the system gets transformed into normal hydrogen atom by expelling out the additional electron. Same result is obtained in relativistic approach of Bohr's theory. Let us suppose that both the electrons are placed in the 1s orbit to make negative hydrogen ion. The electrostatic attractive force experienced by an electron due to the presence of nucleus is e^2/Kr^2 and the electrostatic repulsive force experienced by it due to the presence of other electron is e^2/4Kr^2.The resultant force is balanced by centrifugal force required for its orbital motion .It gives the kinetic energy acquired by the electron as (3/8) e^2/Kr. When a negative hydrogen ion is formed, there is a drop of potential energy e^2/Kr due to nucleus only in the case of first electron as it is moved in potential field and it is zero for the second electron as it is moved in potential free field outside the orbit .When it reaches the orbit there is an increase in its potential energy by an amount e^2/2Kr due to the additional electron in the same orbit. Both the electrons in the orbit are identical in all respect and hence they have its own kinetic energy.Total kinetic energy acquired by the electrons is (3/4) e^2/Kr . Conservation of energy gives e^2/Kr - e^2/2Kr = e^2/2Kr = (3/4)e^2/Kr + dm c^2 where dm = 2(m - mo) dm c^2 = - (1/4) e^2/Kr Since the binding energy is negative, the negative hydrogen is practically impossible. The revised Bohr's theory of hydrogen atom gives an acceptable explanation for the question 'Why the atomic electron under normal circumstances cannot reach beyond the innermost orbit?'. When an electron is attracted by the proton in hydrogen atom, it gets accelerated towards the nucleus until it reaches a point where its velocity is exactly equal to the orbital velocity required to keep the electron stable in the orbit. It is predetermined by the conservation of energy, loss of potential energy = gain in kinetic energy + relativistic increase of mass. Since the gain in kinetic energy of electron is half of the loss of its potential energy, the energy equivalent of relativistic increase of mass of the electron comes from the remaining half of the loss of potential energy (1/2) e^2/Kr(n) = (1/2)m(n)v(n)^2 or v(n)^2 = e^2/Km(n)r(n), where r(n) = n^2ao(1-v(n)^2/c^2)^1/2 and m(n)= mo(1-v(n)^2/c^2)^-1/2 or v(n)^2 =(1/n^2) [e^2/Kmoao] The highest value of v(n) is v(1) with n = 1; v(1)=[e^2/Kmoao] For stability of the electron in the orbit, electrostatic force = centrifugal force e^2/Kr(n)^2 = m(n)v(n)^2/r(n) or v(n)^2= e^2/Kr(n)m(n) = (1/n^2) [e^2/ Kmoao] Whenanaccelerated electron moves towards the nucleus, usually it will not make a straight line motion, If so, it cannot be stopped abruptly in an allowed orbit The accelerated electron moves along a curved path and ultimately it attains stable orbital motion with uniform velocity. The innermost orbit has lowest possible radius in hydrogen atom and is equal to ao, Bohr's radius. That is when hydrogen atom is formed, the orbital electron cannot be brought closer to nucleus with an intermediate distance less than ao. This can be proved form the Bohr's postulates. From the first Bohr's postulate on angular momentum of the orbital electron m(n)^2v(n)^2 r(n)^2 = n^2h^2/4π^2 and from the second Bohr's postulate stating the equivalence of nuclear attractive force with the centrifugal force mn v(n)^2 r(n) = e^2/K. Dividing one by the other, m(n)r(n) = n^2h^2εo/πe^2 or m(n)= n^2h^2εo/π e^2 r(n).From m(n)r(n)= [mo/(1- v(n)^2/c^2)^1/2][n^2ao(1- v(n)^2/c^2)^1/2] = n^2 mo ao or m(n) = n^2 moao/r(n) and from Bohr radius ao=h^2εo/moπ e^2, or mo = h^2εo/πe^2ao. The relativistic increase of mass cannot be less than zero. i.e., dm = m(n)- mo ≃(1/2)mo v(n)^2 /c^2 = [h^2 εo/π e^2] [n^2/r(n)- 1/ao] . The highest possible value of n^2/r(n) = 1/r1 , since dm cannot be negative r1= ao . The lowest possible radius of the orbital electron in hydrogen atom is ao. Conclusion and Future Scope This theoretical work based on non-relativistic Bohr's theory of hydrogen atom is independently carried out without any financial assistance from any funding agencies. I take this opportunity to acknowledge the love and patience of my family and friends. Their belief in me and constant support provided me the creative strength required to persevere. The present work introduces a new line of thought about the source of relativistic increase of mass of the electron, and its atomic binding energy with the nucleus. During an electron jump from outer orbit to any one of the inner orbits there is a mutual conversion between its potential energy with kinetic energy and relativistic mass energy in accordance with the law of conservation energy. It avoids the continuous release of energy in jumping from its initial to final position. When the electron is fixed in an orbit, the orbital electron binds with the nucleus where the required binding energy comes from the mass of the bound system. Since the nucleons are bound more tightly, its share to the atomic binding energy is negligible. This approach gives a potential way to study the spectral features of atoms with higher atomic numbers, if not very heavy atleast light elements like helium, lithium,berllium and boron. References: [1].N.Bohr.,"On the Constitution of Atoms and Molecules,Part-I"Phil.Mag, Vol.26,1913,1-24 [2].N.Bohr.,"On the Constitution of Atoms and Molecules,Part-II" Phil.Mag, Vol.26,1913,476-502, [3].N.Bohr.,"On the Constitution of Atoms and Molecules,Part-III" Phil.Mag, Vol. 26,1913,857-875. [4]. Yuan Lixin,"Discussion on the Problems of Bohr’s Hydrogen Atom Theory in Basic Theory" Applied Science and Innovative Research , 857-875. Vol7(2), 2023 pp 18,DOI:10.22158/asir.v7n2p18 [5].L.de Broglie., "A Tentative Theory of Light Quanta ", Phil.Mag, Vol.47,1924 ,446-458 [6}.L.de Broglie., "Recherches sur la théorie des quanta ", Ann.de.Physique, Vol.3,1925, pp 22 [7].P.Weinberger,"Revisiting Louis de Broglie's famous 1924 paper in the Philosophical Magazine" Phy.Mag, 86,(7), 2006,405-410, http://doi.org/10.1080/09500830600876565. [8].M Meyyappan, "On the phase velocity of matter waves" Acta Ciencia. Indica, Vol.11 (P) 3, 1985.147-149, [9].J.T.Gushing, "Relativistic Bohr model with finite-mass nucleus" Am. J. Phys. Vol.38.1970,1145–50. [10].D.W.Kraft, "Relativistic corrections to the Bohr model of the atom" Am.J.Phys. Vol.42, 1974,837–9. [11].Andreas.F.Terzis, "A simple relativistic Bohr atom" , Eur.J.Phys., Vol.29(4),2008,735, Doi-10, 1088/0143 0807/29/4/1008 [12].L.Nanni, " Relativistic Bohr model", available Doi:10.13140/RG 2.2.35770 85443.2018 [13].Albert Einstein," Eistein's original papers on Special Relativity"Nutty Physics, Translated Harry Yoon, 2026 [14]M.Meyyappan, "On the optimization of Bohr's Theory of Hydrogen atom" available DOI:https://doi.org/10.56975/ijcsp.v16i1.303905 2025 ....................................

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