Thursday, August 21, 2014

Looking at Basic Atomic Physics (1)


1.The Rutherford Model of the Atom.

 What may be called the first foray into basic atomic physics by which further theory could be built upon, commenced with the Geiger and Marsden experiment – first suggested by Lord Rutherford in 1909. The basic setup is depicted in the rough sketch below:
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Fig. 1: Basic Layout of the Geiger-Marsden Experiment

From the Rutherford experiment design, Geiger and Marsden made use of a source of alpha particles to bombard a thin metal foil, on the other side of which was a detecting zinc sulfide screen. They found that while most alpha particles arrived at A, in the direction shown, a few also scattered to positions at B and C which could be detected when the screen at A was moved to the other positions.  The nature of the scattering and deflections (especially some alpha particles at very large angles) was such that there had to be a highly concentrated charge or “nucleus” at the center of the atom. Since the alpha particles are relatively massive (at about 4.002 amu each) the deflections at wide angles meant nearly all the atomic mass was concentrated in the center of the atom and electrons were in the distant outer regions.

Rutherford thereby proposed a model of the atom in which nearly all the mass was concentrated in a very small nucleus while the electrons were scattered at some distance away. This is depicted below in Fig. 2.

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Fig. 2: The Rutherford Model of the atom

The key consequence was that the Rutherford experiment, carried out by Geiger and Marsden, showed that the “pudding pie” model of J.J. Thomson was incorrect. If Thomson’s model was correct, then the expected deflection could be no larger than 0.0001 radians or less than a degree. Since the observed deflections were in some cases more than 100 degrees, it failed the experimental test.


Despite this success, Rutherford’s model still hadn’t won the day. It was largely accepted because it could quantitatively alpha-scattering by thin foils. His model could not: 1) explain line spectra in atoms, including both absorption and emission lines, 2) account for the stability of atoms and could only account for half the nuclear mass.



2. The Bohr  Model of the Atom.


The Bohr Model of the atom, proposed by Neils Bohr, directly challenged the Rutherford model by showing how the observed emission and absorption lines of spectra could be explained. At the heart of Bohr’s model was simplicity, with the hydrogen atom – for example – configured to a miniature solar system with the nucleus at the center and the electron in orbit around it.


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Fig. 3: The Bohr Model of the atom.



From the diagram the electron (e) orbits at a radius r from the central nucleus of charge Ze. As with the planets, a centripetal (inner directed force) F acts toward the center.


 Bohr’s major concept was to quantize the electron orbits. He proceeded by first quantizing the angular momentum of the orbit:



m vr  = nh/ 2p  = n ħ


where  ħ  =  h/ 2p   is the Planck constant divided by 2p.

The Planck constant, first proposed by Max Planck, is:

h = 6.626069 x 10- 34 J-s

Then the value of ħ  = 1.0546 x 10- 34 J-s

Next: both sides are squared:

(m vr ) 2 =  (n ħ)2

So:  m2 v2 r 2 =  n2 ħ2


And:  v2  =  n2 ħ2     / m2  r 2

Now, Bohr looked at the total energy of the H-atom in terms of it kinetic (K) and potential (V) contributions, so:


E = K + V  =    ½ m v2  -  k e2  / r

E = K + V  =  

k e2  /  2r    - k e2  / r =  - k e2  /  2r    

(Since  ½ m v2   =  k e2  /  2r )


Now solve for r (actually the quantized r n ):


r n  = [  n2 ħ2/ m2 v2 ] ½


But  from the kinetic energy equivalence:


v2 =  k e2  /  mr =  n2 ħ2   m2  r 2


\    r n  = [n2 ħ2   / m k e2  ]

The Bohr radius is just the value when the principal quantum number n = 1, so :

r o  = [ ħ2   / m k e2  ]   = 0.0529 nm = 5.2917 ×10−11 m

This is just the most probable radius, i.e. distance between proton and electron, in the hydrogen ground state.

Now, to obtain the quantized energy (E n) we substitute the value for r n  into the total energy equation:

E =  - k e2  /  2r     =

 - k e2  /  2[n2 ħ2   / m k e2 ]   


E = - m k 2 e4  /  2n2 ħ2   =


 - m k 2 e4  /  2 ħ2     [1/ n2] = - 13.6/ n2   

Where the last quantity  is in eV, or electron volts. Here the n refers to the energy level, ground state is n = 1, so can allow the computation of energy for a given level. Or, the energy for a photon emitted from an atom when an electron makes a transition – say from n = 2 to n = 3. Such a situation is shown below:


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Fig. 4: A few energy transitions made in Hydrogen

    
An important point is that the quantized angular momentum postulate (m vr  = n ħ) restricts the possible circular orbits to defines sizes according to the quantized radii (r n  etc.). Thus the normal state of the atom, say hydrogen, will be that for which it has the least energy or the ground state – corresponding in the case of hydrogen to the Bohr radius. Some transitions for different spectral series are shown below:

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Fig. 5. Some Energy transitions in the Hydrogen Bohr atom

     As shown in Fig. 4, emission occurs when an electron in the atom, say hydrogen, makes a transition from a higher to a lower energy level, accompanied by the emission of a photon with a defined energy E = hf = h (c/ l). Consider for example, a transition from the n = 2 to the n = 1 level, as depicted in the lower right of Fig. 4 and in the first line of the Lyman series of Fig. 5.


The energy at the n= 2 level is:

E(n=2) =  - 13.6/ n2   = - 13.6/ (2)2      =  - 13.6/4  (eV)


Now, 1 eV =  1.6 x 10-19 J  so:


E(n=2) =  - 13.6/4  (eV) =  -(3.4) x 1.6 x 10-19 J  =


 -5.4 x 1.6 x 10-19 J 


The n= 1 level has energy:


E(n=1) =  - 13.6/ n2   = - 13.6/ (1)2      =  - 13.6  (eV)


E(n=1) =  -(13.6)  x 1.6 x 10-19 J  = -21.8 x 10-19 J 


Then the energy difference is:

E2 – E1 = [- 5.4 – (-21.8)]  x 10-19 J  = 16.4 x 10 -19 J 


From this, the wavelength of the photon emitted can be found. Since E = hf = h (c/ l):


l =   hc/ (E2 – E1) 


l =    (6.626069 x 10- 34 J-s)(3 x 10 8 m/s)/ 16.4 x 10-19 J 


l =    1.21 x 10 -7 m 


The frequency can be found from:

f = (c/ l) = (3 x 10 8 m/s) / 1.21 x 10 -7 m  = 2.47 x 10 15 Hz

Insight problem:  Using Fig. 5 as a basis, compute the energies and wavelengths of the photons emitted when the electron in the hydrogen atom makes the 1st, 2nd and 3rd Balmer transitions.

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