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8.5. Calculation of the Spectrum of
the Lithium Atom
The lithium atom has three
electrons. Lithium containing one electron is considered to be a hydrogenlike
atom. We have already shown how the spectra of the hydrogenlike atoms are
calculated, including the hydrogenlike lithium atom (Table 9). Let us
calculate the spectrum of the second electron of this atom.
Ionization energy of the second
electron of the lithium atom is equal to =75.638 eV. According to PauliFermi principle, the second
electron cannot occupy the first energy level , because it has already been
occupied by the electron of the hydrogenlike atom (the third electron). That’s
why it is necessary to find binding energy of the second electron of the
lithium atom corresponding to the second energy level. For this purpose, let us
write from the reference book a row of experimental values of excitation
energies corresponding to stationary energy levels of this electron [5]: 62.41;
69.65; 72.26; 73.48; …. eV.
As the second electron of the
lithium atom cannot occupy the first energy level, the first excitation energy
of 62.41 eV in a variety of excitation energies corresponding to the stationary
energy levels should belong to the second energy level of this electron. Let us
find a difference between ionization energy =75.638 eV of this electron and excitation energy corresponding
to the second energy level =62.41 eV.
(221)
Now let us multiply the difference DE by square of the main quantum number
corresponding to the second energy level: . The result being obtained will correspond to binding energy
of the second electron of the lithium atom with the atomic nucleus at the time
of its stay on the first energy level. In accordance with PauliFermi
principle, this electron cannot occupy the first energy level, that’s why the
energy being obtained, which corresponds to this level, will be fictitious.
Here is its value =13.538×4=54.152 eV.
Thus, ionization energy of =75.638 eV of the second electron of the lithium atom is not equal
to energy of =54.152 eV of its binding with the atomic nucleus
corresponding to the first energy level. If we insert these data into the
formulas (206) and (207), we’ll get (Table 14).
Table 14. Spectrum of the second electron of the lithium atom
Valumes 
n 
2 
3 
4 
5 
6 
(exper) 
eV 
62.41 
69.65 
72.26 
73.48 
 
(theor.) 
eV 
62.41 
69.62 
72.25 
73.47 
74.13 
(theor.) 
eV 
13.54 
6.02 
3.38 
2.17 
1.50 
Let us calculate the spectrum of the first electron of the lithium atom.
Its ionization energy is =5.392 eV, and a row of excitation energies corresponding to
the stationary energy levels is as follows [5]: 3.83; 4.52; 4.84; 5.01; 5.11;
5.18; 5.22; 5.25; 5.28; 5.30; 5.31 eV.
The difference between ionization
energy of this electron and excitation energy corresponding to the third
stationary energy level will be as follows: =5.393.83=1.5 eV. Then we’ll find binding energy of this
electron with the atomic nucleus corresponding to the first fictitious energy
level.
. (222)
Thus, ionization energy of the first
electron of the lithium atom is =5.392 eV, and fictitious binding energy with the nucleus
corresponding to the first energy level is =14.05 eV. If we insert these data into the mathematical
model of formation of the spectra of the atoms and the ions (207) and into the
formula (206) of the calculation of
binding energies of this
electron corresponding to the stationary energy levels, we’ll get a spectrum of
this electron (Table 15).
Table 15. Spectrum of the first electron of the lithium atom
Volumes 
N 
2 
3 
4 
5 
6 
(exper) 
eV 
 
3.83 
4.52 
4.84 
5.01 
(theor.) 
eV 
62.41 
3.83 
4.51 
4.83 
5.00 
(theor.) 
eV 
3.51 
1.56 
0.88 
0.56 
0.39 
8.6. Calculation of the Spectrum of
the Beryllium Atom
The beryllium atom has four
electrons. The fourth electron has the largest ionization energy, the first
electron has the smallest ionization energy. We’ll not give the calculation of
the spectrum of the fourth electron of this atom, because its results are given
in Table 6 as a spectrum of the hydrogenlike atom. We’ll not repeat completely
the details of the calculation method of the spectra of the third, the second
and the first electrons of this atom, we’ll give only the key points of this
method.
Ionization energy of the third
electron of the beryllium atom is equal to =153.893 eV. Excitation energies of this electron corresponding
to the stationary energy levels make the following row [5]: 123.67; 140.39;
146.28; 149.01; 150.50; 151.40 eV. The difference between ionization energy and
the value of the first energy in this row will be equal to
. (223)
Binding energy of the third electron
with the atomic nucleus corresponding to the first energy level is determined
in such a way
(224)
If we insert the values =153.893 eV and =120.892 eV into the formulas (206) and (207), well find
(Table 16).
Table 16. Spectrum of the third
electron of the beryllium atom
Valumes 
n 
2 
3 
4 
5 
6 
(exper) 
eV 
123.7 
140.4 
146.3 
149.0 
150.5 
(theor.) 
eV 
123.7 
140.5 
146.3 
149.0 
150.5 
(theor.) 
eV 
30.22 
13.43 
7.56 
4.84 
3.36 
The second electron of the beryllium
atom has ionization energy =18.211 eV and the following row of excitation energies
corresponding to the stationary energy levels [5]: 3.96; 11.96; 14.7; 15.99;
16.67; 17.08 eV.
Let us pay attention to the fact
that the energy value of 3.96 eV exceeds the limits of the excitation energy
formation regularity supposed by us. In the reference book [25] this spectral
line is given as a bright line, that’s why we have every reason to take it into
consideration. We have one opportunity: we can suppose that the second electron
of the beryllium atom can have two positions in the atom, and it is connected
with the structure of its nucleus. Later on, we’ll analyze the structures of
the atomic nuclei and try to find the answer to this ambiguity. Now we have
only one possibility: we should suppose that excitation energy 3.96 eV and the
remaining energies 11.96; 14.7; 15.99; 16.67; 17.08 eV correspond to various
positions of the second electron in the atom, that’s; why we try to get
theoretically only the row 11.96; 14.7; 15.99; 16.67; 17.08 eV. For this
purpose, we’ll find the difference between ionization energy =18.211 eV and the energy 11.96 eV.
. (225)
The binding energy of the second
electron of the beryllium atom corresponding to the first fictitious energy
level will be: =6.25×9=56.259 eV. If we insert this
value and ionization energy =18.211 eV into the formulas (206) and (207), we’ll find
(Table 17).
Table 17. Spectrum of the second electron of the beryllium atom
Values 
n 
2 
3 
4 
5 
6 
(exper) 
eV 
 
11.96 
14.72 
15.99 
16.67 
(theor.) 
eV 
4.15 
11.96 
14.70 
15.96 
16.65 
(theor.) 
eV 
14.81 
6.25 
3.52 
2.25 
1.56 
The theory predicts (Table 17) the
existence of excitation energy of 4.15 eV corresponding to the second energy
level, but it seems that its is fictitious value of energy.
The first electron of the beryllium
atom has ionization energy of =9.322 eV and the following row of excitation energies [5]:
2.73; 5.28; 7.46; 8.31; 8.69 eV. We’d like to note that there is no energy of
2.73 eV in the reference book [25], and in the reference book [5] it is given
without specification of its brightness. It allows us not to take it into consideration.
Thus, difference of energies will be equal to =9.3225.28=4.04 eV, and energy corresponding to the first
fictitious energy level will be as follows: . If we insert =9.322 eV and =16.17 eV in the formulas (206) and (207), we’ll find (Table
18).
Table 18. Spectrum of the first
electron of the beryllium atom
Values 
n 
2 
3 
4 
5 
6 
7 
8 
(exper) 
eV 
5.28 
7.46 
8.31 
8.69 
8.86 
8.98 
9.07 
(theor.) 
eV 
5.28 
7.53 
8.31 
8.67 
8.87 
8.99 
9.07 
(theor.) 
eV 
4.04 
1.80 
1.01 
0.65 
0.45 
0.33 
0.25 
Note: experimental values of
excitation energies corresponding to the 6th, 7th and 8th energy levels are
taken from the reference book [25].
The mathematical models (206) and
(207) gave satisfactory results. But they were the spectra of the atoms and the
ions of the first four elements of the periodic law of the chemical elements.
They are the simples atoms.
If the electrons actually presses on
the atomic nuclei, they begin to interact when their quantity is increased in
the atom, and the mathematical models fail to take it into consideration (206)
and (207). There is every reason to suppose that in the atoms with a large
number of electrons in these models the correction factors or trigonometric
functions will appear, which can characterize the arrangement of a cell of the
electron in the atom. The cell is considered to be a cavity of a conic form, in
which basis the electron is situated, and the vertex is directed to the atomic
nucleus.
A question arises: what aim can be
when the spectra of the atoms and the ions are calculated? The fist aim is to
get the information for the determination of the structure of the atom and its
nucleus. The second aim is to calculate bindings of valency electrons with the
atomic nuclei in order to use them for the analysis of energy balance in
various chemical reactions. The first aim is a remote one; nevertheless, we
make the first steps to this aim. The second aim is nearer to practice, that’s
why it deserves a priority attention. Taking these facts into consideration
we’ll try to calculate the spectra of valency electrons, which have the
smallest ionization energies.
Later on during the analysis of the structure
of the atomic nuclei and the atoms of chemical elements, we’ll see that all
electrons are in the atom, their binding energies with the protons of the
nuclei are nearly the same.
We have already agreed that we call
an electron with the smallest ionization potential the first electron. This
electron is called valency electron. Later on we’ll see that the atomic nuclei
have equal potential possibilities to be valency electrons. That’s why the
enumeration of the electrons in the atom is a conditional thing. Let us try to
calculate the spectrum of the electron of boron atom, which has the smallest
ionization potential. Let us call this electron the first electron.
8.7. Calculation of the Spectrum of
the First Boron Atom
The boron atom has five electrons.
Let us call the electron, which has the least ionization energy =8.298 eV, the first electron. It has the following row of
excitation energies: 4.96; 5.93; 6.79; 6.82; 7.44; 7.46;
7.75; 7.88; 7.92; 7.95; 8.02; 8.03; 8.08; 8.09; 8.13;
8.16; 8.18; 8.20; 8.22; 8.23; 8.24; 8.25; 8.26; 8.27 eV. The row is rather
long. Let us pay attention to the underlined close values of energies. It seems
they are doublets and triplets, i.e. the slip atoms. That’s why the calculation
should give one of the underlined values or their average values. Let us see if
it is so or not. The difference of energies =8.2984.96=3.34 eV. Binding energy of this electron with the
atomic nucleus corresponding to the first fictitious energy level is determined
according to the formula =DE×2^{2}=3.34× 4=13.35 eV. If we insert =8.298 eV and =13.35 eV in the formulas (206) and (207), we’ll find (Table
19).
Table 19. Spectrum of the first
electron of the boron atom
Valumes 
n 
2 
3 
4 
5 
6 
7 
(exper) 
eV 
4.96 
6.82 
7.46 
7.75 
7.92 
8.02 
(theor.) 
eV 
4.96 
6.81 
7.46 
7.76 
7.93 
8.02 








Valumes 
n 
8 
9 
10 
11 
12 
13 
(exper) 
eV 
8.09 
8.13 
8.16 
8.18 
8.20 
8.22 
(theor.) 
eV 
8.09 
8.13 
8.16 
8.18 
8.20 
8.22 








Valumes 
n 
14 
15 
16 
17 
18 
19 
(exper) 
eV 
8.23 
8.24 
8.25 
8.25 
8.26 
... 
(theor.) 
eV 
8.23 
8.24 
8.25 
8.25 
8.26 
... 
If we analyse the given experimental
row of excitation energies and the results of its calculation given in Table
19, we can see a good convergence of theoretical and experimental data.
Later on we’ll not try to calculate
the spectra of all electrons and all atoms, we’ll give only the calculations of
the spectra of the atoms and those valency electrons, which we have used during
the analysis of the results of our investigations.
The
Foundations of Physchemistry of Microworld
Copyright Ó2003 Kanarev Ph.
M.
Internet Version  http://book.physchemistry.innoplaza.net
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