Atomic Models

Discussion

rutherford

Rutherford's assistants did all the work. Rutherford's idea was, "well let's see if these guys (Geiger, Marsden etc.) are good at detecting alpha particles scattered from gold foil. Make them look for them at large angles. That sounds like a really difficult task." But then, Geiger and Marsden actually found particles scattered at extreme angles.

Experiment — alpha particles bombarding gold foil (polonium a source)

A small fraction of the α-particles falling upon a metal plate have their directions changed to such an extent that they emerge again at the side of incidence.

Compared, however, with the thickness of gold which an α-particle can penetrate, the effect is confined to a relatively thin layer. In our experiment, about half of the reflected particles were reflected from a layer equivalent to about 2 mm of air. If the high velocity and mass of the a-particle be taken into account, it seems surprising that some of the α-particles, as the experiment shows, can be tamed within a layer of 6 × 10−5 cm. of gold through an angle of 90°, and even more. To produce a similar effect by a magnetic field, the enormous field of 109 absolute units would be required.

Three different determinations showed that of the incident α-particles about 1 in 8000 was reflected, under the described conditions.

Geiger, Fellow, Marsden, 1909


[slide]

It was left to Rutherford to make conclusions from their observations

observation  ⇒  conclusion
most particles
did not deviate
  atoms are mostly empty space
some
deviated slightly
  there's something positive inside the atom
a tiny fraction
deviated 90°–180°
  a small positively charged region (nucleus)
contains most of the atom's mass
    electrons orbit the nucleus like a planet
orbiting the sun

Rutherford's own words.

The observations, however, of Geiger and Marsden on the scattering of α rays indicate that some of the α particles, about 1 in 20,000 were turned through an average angle of 90 degrees in passing though a layer of gold-foil about 0.00004 cm. thick, which was equivalent in stopping-power of the a particle to 1.6 millimetres of air…. It seems reasonable to suppose that the deflexion through a large angle is due to a single atomic encounter, for the chance of a second encounter of a kind to produce a large deflexion must in most cases be exceedingly small. A simple calculation shows that the atom must be a seat of an intense electric field in order to produce such a large deflexion at a single encounter….

Consider an atom which contains a charge ±Ne at its centre surrounded by a sphere of electrification containing a charge ∓Ne supposed uniformly distributed throughout a sphere of radius R. e is the fundamental unit of charge, which in this paper is taken as 4.65 x 10−10 E.S. unit. We shall suppose that for distances less than 10−12  cm. the central charge and also the charge on the alpha particle may be supposed to be concentrated at a point. It will be shown that the main deductions from the theory are independent of whether the central charge is supposed to be positive or negative. For convenience, the sign will be assumed to be positive. The question of the stability of the atom proposed need not be considered at this stage, for this will obviously depend upon the minute structure of the atom, and on the motion of the constituent charged parts….

In comparing the theory outlined in this paper with the experimental results, it has been supposed that the atom consists of a central charge supposed concentrated at a point, and that the large single deflexions of the α and β particles are mainly due to their passage through the strong central field.

Ernest Rutherford, 1911

bohr


[slide]

Niels Bohr (1885–1962) Denmark

Let us at first assume that there is no energy radiation. In this case the electron will describe stationary elliptical orbits….

The circumstance that the frequency can be written as a difference between two functions of entire numbers [whole numbers] suggests an origin of the lines in the spectra in question similar to the one we have assumed for hydrogen; i.e. that the lines correspond to a radiation emitted during the passing of the system between two different stationary states.

Neils Bohr, 1913

For this it will be necessary to assume that the orbit of the electron can not take on all values, and in any event, the line spectrum clearly indicates that the oscillations of the electron cannot vary continuously between wide limits….

Let us now try to overcome these difficulties by applying Planck's theory to the problem….

The subject of direct observation is the distribution of radiant energy over oscillations of the various wave lengths. Even though we may assume that this energy comes from systems of oscillating particles, we know little or nothing about these systems. No one has ever seen a Planck's resonator, nor indeed even measured its frequency of oscillation; we can observe only the period of oscillation of the radiation which is emitted. It is therefore very convenient that it is possible to show that to obtain the laws of temperature radiation it is not necessary to make any assumptions about the systems which emit the radiation except that the amount of energy emitted each time shall be equal to hν, where h is Planck's constant and ν [ƒ] is the frequency of the radiation….

During the emission of the radiation the system may be regarded as passing from one state to another; in order to introduce a name for these states we shall call them "stationary" states, simply indicating thereby that they form some kind of waiting places between which occurs the emission of the energy corresponding to the various spectral lines….

Under ordinary circumstances a hydrogen atom will probably exist only in the state corresponding to n = 1. For this state W will have its greatest value and, consequently, the atom will have emitted the largest amount of energy possible; this will therefore represent the most stable state of the atom from which the system cannot be transferred except by adding energy to it from without.

Niels Bohr, 1913

In a letter to  …

There appears to me one grave difficulty in your hypothesis, which I have no doubt you fully realize, namely, how does an electron decide at what frequency it is going to vibrate at when it passes from one stationary state to the other? It seems to me that you would have to assume that the electron knows beforehand where it is going to stop.

Ernest Rutherford, 1913


Only an integral number of wavelengths fit in an allowed electron orbit. [slide]

mathematics

classical start   bohr hypothesis   debroglie hypothesis
Fc  = Fe   L = mvr  =  nh   C  =  r  =   = n  h  
mv
         
mev2  =  1   e2   L2 = m2v2r2  =  n2h2   C2  =  2r2  =  n2h2  
r 4πε0 r2 2 m2v2
         
v2  =  1   e2   me2 
1   e2
 r2  =  n2h2   2r2  =  n2h2  
4πε0   mer
4πε0 mer 4πε0 mer 2 me2 1 e2
         
    r = n2  ε0h2  = n2a0   r = n2  ε0h2  = n2a0
πe2me πe2me

Bohr radius, a0 …

a0 =  ε0h2  =  (8.854 × 10−12 C2/Nm2) (6.626 × 10−34 Js)2  = 5.293 × 10−11 m
πe2me π(1.602 × 10−19 C)2 (9.109 × 10−31 kg)

Thus the diameter of a hydrogen atom in its ground state is approximately 10−10 m, a unit also known as an ångstrom and represented with the symbol Å.

energy levels of hydrogen: total energy is the sum of the kinetic and electric potential energy of the electron

E = K + U =  1  mev2 −  1   e2
2 4πε0 r

Replace speed with the formula derived earlier for the speed of an electron in a classical circular orbit. Then simplify.

E =  1  me 
1   e2
 −  1   e2  = −  1   e2
2 4πε0 mer 4πε0 r 4πε0 2r

Replace radius with the formula derived earlier for the radius of an electron in an allowed orbit. Then simplify.

En = −  1   e2  
πe2me
 = −  e4me   1  =  E1
4πε0 2 n2ε0h2 02h2 n2 n2

ground state energy, ionization energy of hydrogen

E1 = −  e 4me  = −  (1.602 × 10−19 C)4(9.109 × 10−31 kg)  = − 2.179 × 10−18 J
02h2 8(8.854 × 10−12 C2/Nm2)2(6.626 × 10−34 Js)2

or in electron volts

E1 =  − 2.179 × 10−18 J  = − 13.6 eV
1.602 × 10−19 C/e

energy level changes are followed by the emission of a photon

ΔE = hf

Spectroscopists like wavelengths, which leads to the following funky formula. The Rydberg formula for hydrogen.

1  = −R 
1  −  1
λ n2 n02

It can be derived from the Bohr model.

c =  ƒλ    
En =  e4me   1  ⇒  1  = −  e4me
1  −  1
02h2 n2 λ 02h3c n2 n02
ΔE =  hf    

Rydberg constant, R

R =  e4me  
02h3c  
R =  (1.602 × 10−19 C)4 (9.109 × 10−31 kg)  
8(8.854 × 10−12 C2/Nm2)2 (6.626 × 10−34 Js)3 (2.998 × 108 m/s)  
R =  1.097 × 107 m−1  
 

spectral lines are classified according to the energy level the electron lands on


[slide]


[slide]

photochemistry

  1. Grotthuss-Draper law: Light must be absorbed by a chemical substance in order for a photochemical reaction to take place. Molecules that do not absorb light of a particular frequency will not undergo a photochemical reaction when irradiated at that frequency
  2. Stark-Einstein law (photoequivalence law): Each photon of light can cause a photochemical reaction of only one light-absorbing molecule.
  3. The amount of photoreaction that takes place is directly proportional to the product of the light intensity and the time of illumination. In other words, more light produces more photoproduct.

schrödingder

dilemma
Electrons have three-dimensional extent, but the Bohr model assumes the electron to be a one-dimensional standing wave wrapped around the nucleus.

major new idea
electron forms a three-dimensional standing wave around the nucleus, electron clouds, restricted wavelengths (spherical harmonics)

Erwin Schrödinger (1887–1961) Austria, Abhandlungen zur Wellenmechanik. Wave equation for matter reminiscent of Maxwell's equations for electromagnetic waves. The story I heard is that Schrödinger went to Switzerland with two goals: to keep his mistress happy and to derive a wave equation for matter. How successful he was with the former is open to speculation.

full, time-dependent form

iℏ  Ψ(rt) = − 2 ∇2Ψ(rt) + V(r)Ψ(rt)
∂t2m

can be separated into two halves

Ψ(rt) = ψ(r)φ(t)

spatial, time-independent half

Eψ(r) = − 2 ∇2ψ(r) + V(r)ψ(r)
2m

The electrons around an atom are standing, probability waves. It's their interference, when atoms bond and form molecules, that determine molecular structures.

     Probability density of a ground state electron
| 1, 0, 0 > in a hydrogen atom

The four quantum numbers

Wolfgang Pauli (1900–1958) Austria — exclusion principle

The ground states of all elements follow the pattern of the excited states in the hydrogen atom. The structure of the periodic table, which was determined empirically, can be derived theoretically from first principles. Chemistry is the bastard child of physics.

Some Probability Densities for Excited Electrons in a Hydrogen Atom

| 2, 0, 0 >

| 2, 1, 0 >

| 2, 1, 1 >

| 3, 0, 0 >

| 3, 1, 0 >

| 3, 1, 1 >

| 3, 2, 0 >

| 3, 2, 1 >

| 3, 2, 2 >
n
shells
l
subshell
m
orbital
orbital
name
orbital
shape
1, 2, 3, 4, 5, 6, 7, 8 … 0 0 s   spherical
2, 3, 4, 5, 6, 7 … 1 +1 p x dumbbell
0 p y dumbbell
−1 p z dumbbell
3, 4, 5, 6 … 2 +2 d xy double dumbbell
+1 d xz double dumbbell
0 d z2 dumbbell–torus
−1 d yz double dumbbell
−2 d x2 − y2 double dumbbell
4, 5 … 3 +3 f y(3x2 − y2) flat triple dumbbell
+2 f xyz quadruple dumbbell
+1 f yz2 triple dumbbell
0 f z3 dumbbell–double torus
−1 f xz2 triple dumbbell
−2 f z(x2 − y2) quadruple dumbbell
−3 f x(3y2 − x2) flat triple dumbbell
5 … 4 −4 to +4 g   many-lobed
This Table Needs to Be Checked for Accuracy

Dirac

1928: Dirac developed the relativistic quantum theory. Paul Dirac states his relativistic electron quantum wave equation. Charles G. Darwin and Walter Gordon solve the Dirac equation for a Coulomb potential. Paul Dirac combines quantum mechanics and special relativity to describe the electron.

Is this right?


γμc 
i  − eAμ
 − mec2 
 Ψ (xν) = 0
xμ

This equation …

Dirac showed that there are no stable electron orbits for more than 137 electrons, therefore the last chemical element on the periodic table will be untriseptium (137Uts) also known informally as feynmanium (137Fy). It's full electron configuration would be something like …

1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f14 5d10 6p6 7s2 5f14 6d10 7p6 8s2 5g17

or is it …

1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f14 5d10 6p6 7s2 5f14 6d10 7p6 8s1 5g18

finish it