1.
The Wave Model of the Atom.
Consider the following dynamical picture: as the electrons whir about the nucleus they
ought to be losing energy, in the
context of Bohr’s orbital model. If they lose kinetic energy over time they
must spiral into the nucleus, and the atom then ceases to exist. This ought to
happen in a very short time, so that most atoms in the universe cease to exist
and hence the whole universe. But this isn’t observed. Why?
The
only explanation is that Bohr’s orbital model can’t be correct. Thus was born the theoretical basis for the wave
model which we mostly accept today in modern quantum mechanics. Unlike the Bohr model, electrons don’t follow
defined orbital paths but instead are
referenced to regions or volumes in
which they will be more or less probable.
The basic allocation of electrons, say for the hydrogen atom, is then
confined to “orbitals” or regions of higher probability.
We
now look at the experimental basis provided for this model.
Around
1926, a young French physicist named Louis
de Broglie actually postulated the basis for material particles, such as
electrons, acting as waves. This was
experimentally verified in the (1927) Davisson and Germer electron diffraction
experiment sketched below:
Fig.
1: The Basic Davisson and Germer experiment
From
the experiment, with electrons moving through a potential difference V = 4,000
volts, the kinetic energy gained should be equal to the work done, or:
½ m v^{2 } = eV
Where
m the mass of the electron is: 9.1 x 10 ^{31} kg
And
the electron charge e = 1.6 x 10^{19} C
The
velocity then is:
v
= Ö (2eV/m)
The
momentum p = mv = m Ö
(2eV/m) = Ö (2eVm)
And
the de Broglie wavelength is:
l_{D}= h/p = h/Ö (2eVm)
For
a voltage V = 3,000 V one would find:
l_{D} =
(6.626
x 10^{34} Js) / [(3.2 x 10^{19} C) (3000V) ( 9.1 x 10^{31}
kg)]^{1/2}
l_{D} = 2 x 10^{11} m
Which
is the de Broglie wavelength of the electron in this experiment.
A first step to uncovering the wave model
from Bohr’s is to examine his quantized relationship:
m
vr = nh/ 2p = L
where
L is the angular momentum. Rearranging:
h/
mv = 2p r/ n = l_{D } or 2p r = n l_{D}
Showing
the circumference is scaled into n (standing) waves of wavelength l_{D }as shown below:_{}
Fig.
2. Standing wave model for the Bohr atom.
This
waveorbiting electron atom still has a radius r, but with waves each separated by one de
Broglie wavelength , l_{D.}
Thereby an integral number of such wavelengths form
the circumference
of the atomic orbit, as required by the condition:
2p r = n l_{D.}
For
example, in the case of hydrogen the first three of these cloudwave regions are shown in Fig. 3.
Fig.
3: Electron (“orbital”) cloudsregions in Hydrogen
Let
us ‘zoom in’ on the more spherical n= 1 configuration, and the probability for
the electron in this space as depicted in Fig. 4 below:
Fig.
4: The n = 1 electron orbital for hydrogen
This
diagram more than any other dispenses with the notion that hydrogen electron
occupies a definite position. Instead, it’s confined someplace within a “cloud”
or probability space (b) but that probability can be computed as a function of the
Bohr radius (a_{o} = 0.0529 nm).
The probability P_{1s} for the 1s orbital is itself a result of
squaring the “wave function” for the orbital. If the wave function is defined y (1s) = 1/Öp (Z/ a_{o}) exp
(Zr/ a_{o}), and the probability function is expressed:
P
= ½y (1s) y (1s) *½
Where
y (1s) * is the complex
conjugate, then the graph shown in Fig. 4 is obtained. Inspection shows the
probability of finding the electron at the Bohr radius is the greatest, but it
can also be found at distances less than or greater than 0.0529 nm.
We
thereby see from Figs. 34 that Bohr’s original quantizing number, n (the "principal quantum number"), has far
more meaning than simply to parse the number of standing waves for a given
atom. We already see that it determines the energy of the atom, viz.
E
_{n} =  13.6/ n^{2 } ^{ }
But
it also indicates the average distance of the electron from the nucleus. Thus, de Broglie’s wavelength provides the
basis for the waveparticle duality that lies at the basis of the “smeared”
probabilistic atoms peculiar to modern atomic theory.
At
the bottom of Fig. 3 are the quantum numbers: n and ℓ, which are identified
as: the principal quantum number, and the angular momentum quantum number,
respectively.
There are two physical meanings attendant on n: i) it determines the energy of an orbital (specifically in the Hatom), and (ii) it indicates the average distance of an electron in a particular orbital, to the nucleus, To fix ideas, I show in the accompanying diagram (Fig. 5) a sketch of one lobe for an electron orbital associated with the (3, 2, +2) state in the Hydrogen atom. The key point is the orbital denotes an electron density associated with a probability of finding the electron in some defined space.
There are two physical meanings attendant on n: i) it determines the energy of an orbital (specifically in the Hatom), and (ii) it indicates the average distance of an electron in a particular orbital, to the nucleus, To fix ideas, I show in the accompanying diagram (Fig. 5) a sketch of one lobe for an electron orbital associated with the (3, 2, +2) state in the Hydrogen atom. The key point is the orbital denotes an electron density associated with a probability of finding the electron in some defined space.
Fig. 5: one lobe for an electron
orbital associated with the (3, 2, +2) state in hydrogen
In
the case shown one must also visualize a symmetrical lobe on the other side
(making the whole orbital resemble a dumbbell) to make it complete. As one
alters the set of quantum numbers the electron densities change and so do the
probabilities associated with the orbit.(See Fig.6)
Fig. 6: Further hydrogen orbitals with higher quantum numbers
Fig. 6: Further hydrogen orbitals with higher quantum numbers
Describing
orbitals using the set of given quantum numbers means knowing the numbering
rules applied to each. In the case of the principal quantum number, n, we allow
it to have integral (nonzero) values: 1, 2, 3, 4 etc.
The physical significance of the angular momentum quantum number (ℓ) is to convey the shape of the probability density cloud or orbital. The numbering rule for l is directly contingent on the value for n. Thus, for any given n, then ℓ must be such that it has integral values from 0 to (n 1). This means if n = 2, then ℓ can have (n 1) = (2 1) = 1. But if n =1, then ℓ = (n  1) = 1  1 = 0.
Note that the ℓ quantum numbers appear more than once for any orbital with n >1. Thus, for the n = 2 case, we have two values of ℓ occurring: one for l = 0, the other for ℓ = 1. If we go on to n= 3 there are three values of ℓ, for n = 4, four values and so on. One also finds the ℓ value specified for lettered orbitals: s, p, d, f, g, h. The sorbital is for ℓ =0, the p for ℓ =1, the d for ℓ = 2 and so on.
The physical significance of the angular momentum quantum number (ℓ) is to convey the shape of the probability density cloud or orbital. The numbering rule for l is directly contingent on the value for n. Thus, for any given n, then ℓ must be such that it has integral values from 0 to (n 1). This means if n = 2, then ℓ can have (n 1) = (2 1) = 1. But if n =1, then ℓ = (n  1) = 1  1 = 0.
Note that the ℓ quantum numbers appear more than once for any orbital with n >1. Thus, for the n = 2 case, we have two values of ℓ occurring: one for l = 0, the other for ℓ = 1. If we go on to n= 3 there are three values of ℓ, for n = 4, four values and so on. One also finds the ℓ value specified for lettered orbitals: s, p, d, f, g, h. The sorbital is for ℓ =0, the p for ℓ =1, the d for ℓ = 2 and so on.
There
is no special significance to the letters (apart from the physical meaning we
already gave for the angular momentum quantum number, ℓ, and they are mainly of
historical import though still retained, for example, in chemistry. (By
extension, one also often hears the term "atomic shell" used in
chemistry). A collection of orbitals under the same value of n is called a
"shell". Thus, for n = 4, we have ℓ =0, ℓ = 1, ℓ =2, ℓ = 3 so comprising the
collection of orbitals: s, p, d and f.
Lastly, there is the magnetic quantum number,
usually designated m _{ℓ} (subscript the same as the angular momentum
quantum number) because it is contingent upon it. This quantum number describes
the orientation of the orbital in 3D space. For a given angular momentum quantum
number, ℓ,
we have integral values specified as follows:
m _{ℓ} =  ℓ, (ℓ +1)....0......( ℓ  1), + ℓ
Note the above set of m _{ℓ} numbers is given as a SERIES, e.g. starting with (ℓ) and terminating at + ℓ. Look at the simplest example for ℓ = 0, then:
m _{ℓ} = 0. (Since all terms are zero)
m _{ℓ} =  ℓ, (ℓ +1)....0......( ℓ  1), + ℓ
Note the above set of m _{ℓ} numbers is given as a SERIES, e.g. starting with (ℓ) and terminating at + ℓ. Look at the simplest example for ℓ = 0, then:
m _{ℓ} = 0. (Since all terms are zero)
What
about ℓ =
1?
Then: m _{ℓ} = 1, 0, 1
What about ℓ = 2?
We have:
m _{ℓ} = 2, 1, 0, 1, 2
As a general rule then, we can use the formula:
N(m _{ℓ}) = {(2 x l) + 1}
Then: m _{ℓ} = 1, 0, 1
What about ℓ = 2?
We have:
m _{ℓ} = 2, 1, 0, 1, 2
As a general rule then, we can use the formula:
N(m _{ℓ}) = {(2 x l) + 1}
to
give the total number of m _{ℓ} numbers.
In
any problems to do with identifying electron shell structure configuration, it
is well to bear in mind the Pauli
Exclusion Principle to make sure the electrons are distributed so that no
two electrons have the same set of quantum numbers. It is instructive
to study the table below to see how the principal quantum number, n, changes
with ℓ, and
the subshell as well as the symbol for the primary electron shell. The
Table shown below shows the respective Shell and Subshell Symbols and
associated Quantum Numbers.
n =

SHELL

ℓ =

Subshell

1

K

0

s

2

L

1

p

3

M

2

d

4

N

3

f

5

O

4

g

6

P

5

h

In working out assorted problems, preparation of a schematic energy level diagram associated with the state of the system can also be of immense value.
Problems:
(1) Find the de Broglie wavelength l_{D } of a proton subject to 4000 V in a DavissonGermer type experiment.
(Take m _{p} = 1.7 x 10 ^{27 } kg)
(Hint: The m _{s} or spin quantum number, which we will encounter in the next section, has either +1/2 or 1/2 value. )
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