Bohr Model

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Discussion

Niels Henrik David Bohr (1885 - 1962) Denmark

For this it will be necessary to assume that the orbit of the electron cannot take on all values, and in any event the line spectrum clearly indicates that the oscillations of the electron cannot vary continuously between 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 hn, where h is Planck's constant and n 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.

(Bohr 1913)

In a letter to Bohr dated 20 March 1913 …

There appears to me one grave difficulty in your hypothesis, which I have no doubt you fully realize, namely, how does an electron decide what frequency it is going to vibrate at when it passes from one stationary state to the other? It seems to pre 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. [magnify]

mathematics

classical start   bohr hypothesis   debroglie hypothesis
Fc  = Fe   L = mvr  =  nh   C  =  2πr  =  nλ = n  h  
2π mv
         
mev2  =  1   e2   L2 = m2v2r2  =  n2h2   C2  =  4π2r2  =  n2h2      
r 4πε0 r2 4π2 m2v2
         
v2  =  1   e2   me2 ⎛
⎜
⎝		 1   e2 ⎞
⎟
⎠		  r2  =  n2h2   4π2r2  =  n2h2 ⎛
⎜
⎝		 4πε0   mer ⎞
⎟
⎠
4πε0 mer 4πε0 mer 4π2 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 × 10−10 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 angstrom 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 8ε02h2 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
8ε02h2 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 =hƒ

Spectroscopists like wavelengths, which leads to the following funky formula.

1  = − R∞ ⎛
⎜
⎝		 1  −  1 ⎞
⎟
⎠
λ n2 n02

It can be derived from the Bohr model.

c =  ƒλ    
En =  e4me   1 ⇒ 1  = −  e4me ⎛
⎜
⎝		 1  −  1 ⎞
⎟
⎠
8ε02h2 n2 λ 8ε02h3c n2 n02
ΔE =  hƒ    

Rydberg constant, R∞

R∞ =  e4me  =  (1.602 × 10−19 C)4 (9.109 × 10−31 kg)  = 1.097 × 107 m−1
8ε02h3c 8(8.854 × 10−12 C2/Nm2)2 (6.626 × 10−34 Js)3 (2.998 × 108 m/s)

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


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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.

Summary

Problems

practice

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numerical

  1. problems

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