Structure of the atom
Bohr atom
1. Classical model of the atom
After the work of Rutherford on 1911, it is stated that the atom is composed of a nucleus and electrons moving around. In this study, It is considered the simplest atom, hydrogen. The force between the electron and the proton (nucleus) is given by the Coulom's law:
Fe = - (1/4πε0)(e2/r2) er (1.1) Where: . Fe is the Coulombian electrostatic force exerted by the proton on the electron. . - e is the charge of the electron (+ e is the charge of the proton), . r is the distance between the two particles, and . er is the unit vector in the direction from the proton to the electron. This Fe needs to be balanced by the centripetal force Fc in order to maitain the electron of mass m moving (rotating) at constant tangential speed v in a circular orbit of radius r. This phrase is written as: Fe = Fc, that is : (1/4πε0)(e2/r2) = m v2/r (1.2) We have then:v = e/[4πε0mr]1/2 (1.3) The kinetic energy of the system is the kinetic energy of the electron because the proton is so massive regarding the electron (mp ≈ 1836 mee). We can then consider that the proton is at rest. Then KE = (1/2)mv2 . The Coulomb potential of the electron (Origin at the proton) is PE = - e2/(4πε0 r) Usingthe equation (1.2), we obtain: PE = mv2 = 2 KE The total energy of the atom is E = KE + PE = KE - 2 KE = - KE = PE/2Total Energy E = - e2/(8πε0 r) (1.4) In this expression, energy is negative, just to sho that it cannot given by the system (atom); instead it must be received in order to move the electron. That is the electron is bounded. When r decreases, - E incereases, so E decreases.
2. Bohr atom
Bohr's general assumptions (Postulates) in order to derive the Rydberg equation. 1. It exists for the orbiting electrons in atoms some stationary states in which they do not radiate electromagnetic energy. 2. The emission or absorption of electromagnetic radiation for an atom occurs only relatively to the transition of electrons between two stationary states. The frequency ν of the related (absorbed or emitted) radiation obey the equation: ΔE = (h/2π) ν. ΔE is the energy difference between two stationary states and h is the Planck's constant. 3. The classical laws of Physics govern the dynamical equilibrium of the system in the stationary states, but they are not applicable to transitions between states. 4. For circular motion, the angular momentum of the system (electron-nucleus) in a stationary state is an integral multiple of (h/2π) For circular motion, the angular momentum L of th electron is: L = r x p. Its magnitude is L = mvr. This angular momentum should be (Assumption 4) as follows: mvr = n (h/2π), or: n is called the principal quantum number. It quantizes all the related physical values such as velocities, radii and enrgies.v = nh / 2πmr (2.1) Equating (1.3) and (2.1), we get:rn = (4πε0)n2(h/2π)2 /m e2 (2.2) Only certain values of the radius r are allowed. The equation (2.2) can be written as: rn = n2 a0, where a0 is called the Bohr radius: Where:a0 = r1 = (4πε0)(h/2π)2 /m e2 (2.3) a0= 0.53 x 10 - 10 m 1/4πε0 = 8.99 x 10 9 N.m2/C2 h/2π = 1.055 x 10 -34 J.s m = 9.11 x 10 -31 kg e = 1.6 x 10 -19 C The value n = 1 gives r1 = a0, the radius of the hydrogen atom in its lowest energy state, called the ground state. The other values for which n> 1, correspond to the excited states of the atom. The diameter of the atom is about 2a0 ≈ 10 -10 m; that is 1 Angstrom (Å). The total energy in the relationship (1.4) is written: En = - e2/(8πε0 rn). Also only some values of enrgy are pemitted. En is the energy of an electron in the stationary state n. Using the equation (2.2), we get: En = - e2/(8πε0) [1/a0 n2] = - E0/n2 (2.4) Where E0 is the lowest energy state: - E0 = E1, the lowest energy ( state n = 1). E0 = e2/(8πε0) (m e2/(4πε0)(h/2π)2) = [m e4/2(h/2π)2][1/4πε0]2 = 13.6 eV (2.5)En = - E0/n2 (2.6) E0 = [m e4/2(h/2π)2][1/4πε0]2 = 13.6 eV (2.7) The atom can then exist only in the stationary states with definite and quantized energies En. When an electron undergoes a transition from En to Em, the radiation emitted (or absorbed) of frequency ν and wavelength λ occurs the following way: En - Em = h ν Using the equation (2.6) and ν = c/λ yield: c/λ = (1/h)(- E0/n2 + E0/m2) = (E0/h)(1/m2 - 1/n2) Then:1/λ = (E0/hc)(1/m2 - 1/n2)= R∞(1/m2 - 1/n2) (2.8) Where :R∞ = E0/hc (2.9) R∞ = 1.097 x 107 m- 1 is called the Rydberg Constant. The subscript ∞ stands for the infinite nuclear mass, regarding the mass of the orbiting electron. This value agrees with the experimental value of Rydberg. The emitted radiation is called is called a photon or a light quatum. Using the equation (2.1), we get the speed of the electron moving around its nucleus:vn = nh / 2πmrn = nh / 2πm a0 n 2 = h / 2πm n a0 (2.10) v1 = h / 2πm a0 = 2.2 x 106 m/s The ratio: α = v1/c ≈ 1/137 is called the fine structure constant.
3. The Bohr's Correspondence Principle:
It states: In the limits where classical and quantum theories should agree, the quantum theory must reduce to the classical result. Example: Classical electodynamics gives the frequency of the electron in its orbit around the nucleus: νClassical = 1/T = ω/2π = (1/2π) v/r (3.1) Using the new quantized values of rn and vn (2.2) and (2.10), we find: v/r = (h/2π)(1/n3 m a02). the equation (3.1) becomes: νClassical = [h/1/(2π)2][1/n3 m a02] (3.2) The Bohr's model gives the frequency νBohr of the transition from the state n + 1 to the nearest state n. The equation (2.9) yields: νBohr = [E0/h][(1/n2) - (1/(n + 1)2)] = [E0/h][(2n + 1)/n2)(n + 1)2]. (3.3) The approximation for large orbits, that is for large values of the quatum number n, we have: νBohr = 2E0/hn3. The two frequencies are equal and the Bohr's correspondance is verified.4. Remark:
Let's say that there is nothing extraordinary in this model. Bohr was just striving to build an ad hoc theory to account for the hydrogen spectral lines experimented by Balmer. He said it himself: " As soon as I saw the Rydberg formula, everything became clear to me". The set hypothesis mvr = nh/2π, matched the predictions and became then the found key formula; that is the angular momentum is quantized. The quatization of energy was already thought before by Max Planck. The introduction of the integral number n gives the discrete values for the related values such as radius, velocities and energies. This hidden quantization feature appears in the experimental (empirical) Rydberg formula. Plus, this model that consider the circular orbiting motion of the electron around the nucleus is wrong. The correct motion of the electron is somehow everywhere near the nucleus. It can be correctly described by the probability of presence concept derived from the the Schrodinger equation.
5. Improvement of Bohr atom
5.1. Reduced mass correction
In fact, the mass of the nucleus is not infinite. the system (nucleus, electron) rotate about the center of mass. The mass in the previous equations would be the reduced mass μ of the system. The more accurate Rydberg constant becomes then , for the hydrogen atom: RH= R∞ μ/m = R∞ / (1 + m/M) (5.1) We have RH = 1.096776 x 10R7 - 1 which fit the measuremements.5.2. Model for only the hydrogeneic
The Bohr model may be applied to the hydrogen-like, that is the sigle-electron atom such as H+, Li++; the e2 in Bohr's formulas becomes, because of he Coulomb interaction, Ze2. Z is the Atomic number of the atom. The Rydberg constant R changes into Z2R and the Rydberg equation becomes: 1/λ = Z2R[(1/ni2) - (1/nf2)] (5.2)6. Frank & Hertz Experiment
The German physicits James Franck and Gustav Hertz ( nephew of Heinrich Hetz), in 1914, wanted to study the ionization phenomenun. They used an electron bomardment of gaseous vapors. The above figure describes the experiment. Electrons extracted from a heated filament ( cathode) are acccelerated by a varying voltage (0 - 45 V)controlled by a voltmeter. these electrons ionize the monoatomic gas Hg. The resultant electrons arrive at the grid and then decerated between the grid a nd the electrometer( sensitive ammeter). As the potential difference between the electrometer is only 1.5 V, that means only the electrons with energy equal to 1.5 eV can be detected by the electrometer device.The results are as follows: If the accelerating power supplies less than 5 V ( then the energy of the electrons are 5 eV), a current is detected (all the elctrons with energy between 1.5 and about 5 eV). The electrons do not lose energy, the collisions are elastic. When the voltage is icreased to about 5 V ( precisely 4.88 eV), the current drops suddenly. The collisions becomes inelastic. That is all the electrons with 4.88 eV do not go to the grid, they excite the gas atoms. The same process occurs with electrons at energy about 10 eV ( precisely about 4.88 x 2 = 9.76 eV). The expected ionization process (wich would occurs with very high energy)turns out to be excitation. The electromagnetic radiation (photon) were not seen because they are not in the visible spectra. Nevertheless, an emission line of wavelength 254 nm (Ultraviolet) was measured. Setting E = 4.88 eV, and used the Planck relationship E = hν Franck and Hertz showed that the Planck constant was in good agreement with the values determined by others. They are other highly states in Hg that could be excited ( from L to M , N to L , ...) in enalstic collisions; but the probabilty of that occur is more smaller that the first ( from L to K , M to K , N to K) excited states. This quatization explanation accounts for the results of the experiment. ---------- Due its circular motion about the nucleus, the electron is accelerated. An accelerating charged particle looses (gives off or radiates)energy in the form of electromagnetic radiation ( as light, xray, ...). If energy of the electron decreases continusly, then its radius deccreases continusly, spiralling inward until it crashes into the nucleus, so the atome collapses!Hydrogen Spectra
BalmerS formula: ---------------- Johann Jakob Balmer, a Swiss mathematician and an honorary physicist, in 1885, heated a Hydrogen tube and analyzed the lines emitted. He established an empirical formula for the spectral lines of the Hydrogen atom he obtained: λ = hn2/(n2 - 22) This relationship gives the value of the wavelength emitted by the hydrogen atom. h is a constatnt equal to 3.65 x 10- 7m n = 3, 4, 5, 6, and so forth. n = 3 4 5 6 7 λ(nm)= 656 486 434 410 397 red green blue indigo violet Rydberg Formula: --------------- In 1888,Johannes Robert Rydberg, Swedish physicist inverted both sides of Balmer's formula and gave: 1/λ = RH[(1/22) - (1/n2)] RH is the Rydberg constant = 0.010972 (nm)- 1 --------- ******************* Failures of the Bohr Model While the Bohr model was a major step toward understanding the quantum theory of the atom, it is not in fact a correct description of the nature of electron orbits. Some of the shortcomings of the model are: 1. It fails to provide any understanding of why certain spectral lines are brighter than others. There is no mechanism for the calculation of transition probabilities. 2. The Bohr model treats the electron as if it were a miniature planet, with definite radius and momentum. This is in direct violation of the uncertainty principle which dictates that position and momentum cannot be simultaneously determined. The Bohr model gives us a basic conceptual model of electrons orbits and energies. The precise details of spectra and charge distribution must be left to quantum mechanical calculations, as with the Schrodinger equation. ********************