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:
Fig. 1: Basic Layout of the Geiger-Marsden Experiment
|Fig. 2: The Rutherford Model of the atom
The key consequence was that the
2. The Bohr Model of the Atom.
The Bohr Model of the atom, proposed by Neils Bohr, directly challenged the
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
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:
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:
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.