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 αparticles 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 αparticle 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 aparticle be taken into account, it seems surprising that some of the αparticles, 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 10^{9} absolute units would be required.
Three different determinations showed that of the incident αparticles about 1 in 8000 was reflected, under the described conditions.
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.
§1The 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 goldfoil about 0.00004 cm. thick, which was equivalent in stoppingpower 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….
§2Consider 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….
§7In 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.
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.
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.
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.
mathematics
classical start  bohr hypothesis  debroglie hypothesis  
F_{c}  = F_{e}  L = mvr  =  nh  C  =  2πr  =  nλ = n  h  
2π  mv  
m_{e}v^{2}  =  1  e^{2}  L^{2} = m^{2}v^{2}r^{2}  =  n^{2}h^{2}  C^{2}  =  4π^{2}r^{2}  =  n^{2}h^{2}  
r  4πε_{0}  r^{2}  4π^{2}  m^{2}v^{2}  
v^{2}  =  1  e^{2}  m_{e}^{2}  ⎛ ⎝ 
1  e^{2}  ⎞ ⎠ 
r^{2}  =  n^{2}h^{2}  4π^{2}r^{2}  =  n^{2}h^{2}  ⎛ ⎝ 
4πε_{0}  m_{e}r  ⎞ ⎠ 

4πε_{0}  m_{e}r  4πε_{0}  m_{e}r  4π^{2}  m_{e}^{2}  1  e^{2}  
r = n^{2}  ε_{0}h^{2}  = n^{2}a_{0}  r = n^{2}  ε_{0}h^{2}  = n^{2}a_{0}  
πe^{2}m_{e}  πe^{2}m_{e} 
Bohr radius, a_{0} …
a_{0} =  ε_{0}h^{2}  =  (8.854 × 10^{−12} C^{2}/Nm^{2}) (6.626 × 10^{−34} Js)^{2}  = 5.293 × 10^{−11} m 
πe^{2}m_{e}  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  m_{e}v^{2} −  1  e^{2}  
2  4πε_{0}  r 
Replace speed with the formula derived earlier for the speed of an electron in a classical circular orbit. Then simplify.
E =  1  m_{e}  ⎛ ⎝ 
1  e^{2}  ⎞ ⎠ 
−  1  e^{2}  = −  1  e^{2}  
2  4πε_{0}  m_{e}r  4πε_{0}  r  4πε_{0}  2r 
Replace radius with the formula derived earlier for the radius of an electron in an allowed orbit. Then simplify.
E_{n} = −  1  e^{2}  ⎛ ⎝ 
πe^{2}m_{e}  ⎞ ⎠ 
= −  e^{4}m_{e}  1  =  E_{1}  
4πε_{0}  2  n^{2}ε_{0}h^{2}  8ε_{0}^{2}h^{2}  n^{2}  n^{2} 
ground state energy, ionization energy of hydrogen
E_{1} = −  e^{ 4}m_{e}  = −  (1.602 × 10^{−19} C)^{4}(9.109 × 10^{−31} kg)  = − 2.179 × 10^{−18} J 
8ε_{0}^{2}h^{2}  8(8.854 × 10^{−12} C^{2}/Nm^{2})^{2}(6.626 × 10^{−34} Js)^{2} 
or in electron volts
E_{1} =  − 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  n^{2}  n_{0}^{2} 
It can be derived from the Bohr model.
c =  ƒλ  
E_{n} =  e^{4}m_{e}  1  ⇒  1  = −  e^{4}m_{e}  ⎛ ⎝ 
1  −  1  ⎞ ⎠ 

8ε_{0}^{2}h^{2}  n^{2}  λ  8ε_{0}^{2}h^{3}c  n^{2}  n_{0}^{2}  
ΔE =  hf 
Rydberg constant, R_{∞}
R_{∞} =  e^{4}m_{e}  
8ε_{0}^{2}h^{3}c  
R_{∞} =  (1.602 × 10^{−19} C)^{4} (9.109 × 10^{−31} kg)  
8(8.854 × 10^{−12} C^{2}/Nm^{2})^{2} (6.626 × 10^{−34} Js)^{3} (2.998 × 10^{8} m/s)  
R_{∞} =  1.097 × 10^{7} m^{−1}  
spectral lines are classified according to the energy level the electron lands on
photochemistry
dilemma
Electrons have threedimensional extent, but the Bohr model assumes the electron to be a onedimensional standing wave wrapped around the nucleus.
major new idea
electron forms a threedimensional standing wave around the nucleus, electron clouds, restricted wavelengths (spherical harmonics)
Erwin Schrödinger (18871961) 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, timedependent form
iℏ  ∂  Ψ(r, t) = −  ℏ^{2}  ∇^{2}Ψ(r, t) + V(r)Ψ(r, t) 
∂t  2m 
can be separated into two halves
Ψ(r, t) = ψ(r)φ(t)
spatial, timeindependent half
Eψ(r) = −  ℏ^{2}  ∇^{2}ψ(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  z^{2}  dumbbell–torus  
−1  d  yz  double dumbbell  
−2  d  x^{2} − y^{2}  double dumbbell  
4, 5 …  3  +3  f  y(3x^{2} − y^{2})  flat triple dumbbell 
+2  f  xyz  quadruple dumbbell  
+1  f  yz^{2}  triple dumbbell  
0  f  z^{3}  dumbbell–double torus  
−1  f  xz^{2}  triple dumbbell  
−2  f  z(x^{2} − y^{2})  quadruple dumbbell  
−3  f  x(3y^{2} − x^{2})  flat triple dumbbell  
5 …  4  −4 to +4  g  manylobed 
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?
⎡ ⎣ 
γ^{μ}c  ⎛ ⎝ 
i  ∂  − eA_{μ}  ⎞ ⎠ 
− m_{e}c^{2}  ⎤ ⎦ 
Ψ (x^{ν}) = 0 
∂x^{μ} 
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 (_{137}Uts) also known informally as feynmanium (_{137}Fy). It's full electron configuration would be something like …
1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} 3d^{10} 4p^{6} 5s^{2} 4d^{10} 5p^{6} 6s^{2} 4f^{14} 5d^{10} 6p^{6} 7s^{2} 5f^{14} 6d^{10} 7p^{6} 8s^{2} 5g^{17}
or is it …
1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} 3d^{10} 4p^{6} 5s^{2} 4d^{10} 5p^{6} 6s^{2} 4f^{14} 5d^{10} 6p^{6} 7s^{2} 5f^{14} 6d^{10} 7p^{6} 8s^{1} 5g^{18}
finish it