Understand why electronic clouds of atoms do not collapse on the nucleus due to electrostatic attraction at a first-year university level

1. 1
ELECTRON CLAUSTROPHOBIA AND STABILITY OF ATOMS
1. The objective of this file:
Understand why electronic clouds of atoms do not
collapseon thenucleus dueto electrostaticattraction.
A roughreasoningaccessible tostudentsintheirfirstyearof universitymakesitpossible toestablish
the "claustrophobia"of electrons(and,ingeneral,of anyquantumparticle).
(https://readingpenrose.files.wordpress.com/2015/08/bohr-radius.gif)
Thispropertyleadstorationalize several propertiesof the matterwhich,amongothers :
 The stability of the hydrogen atom, its dimensions, the existence of quantum numbers and
energy levels, its spectroscopy ...
 The contributionof this"claustrophobia"tochemical linkageswithparticularreference tothe
general trend observed to establish delocalized molecular orbitals.
2. Prerequisite,the de Broglie'srelationship :
In an attempt to generalize the duality nature of light (wave versus photons), de Broglie suggests
(1924) to unite the properties of waves and particles through their energy.
Thus,the special case of the lightservesas the basis for the establishment of Broglie's relationship.
 The energyof an electromagneticwave isproportional toitsfrequency(E= h.).
 The energyof a particle isproportional toitsmass(E = m.c2
).
And so:
𝐸 = 𝑚 ∙ 𝑐2 = ℎ ∙  = ℎ ∙
𝑐
𝜆
→ 𝑚 ∙ 𝑐 = 𝑝 =
ℎ
𝜆
→ 𝝀 ∙ 𝒑 = 𝒉
The photon-specific characteristics no longer appear explicitly in this relationship
which isthereforesupposed to apply to anyobject.In thisinterpretationof quantum
mechanics, any impulse particle (p = m.v) would be associated with a wave
wavelength λ.

2. 2
3. The "claustrophobia" of electrons :
Thispropertyappearsasa consequenceof de Broglie'sequationwhenanelectronisconsideredtobe
locked into a box. Two conditions must be met for the electron to "survive" in the box.
 A condition on the particle: The box must be large enough.
 A condition on the wave: The wave must be able to be stationary in the box.
If, at the level of the atoms,the conditiononthe particle is not binding,the one on the wave, on the
otherhand,requiresthatthe length(L)of the box(supposedtobe linear) isequaltoanintegernumber
(n) of half-wavelengths.
𝐿 = 𝑛 ∙
𝜆
2
→ 𝜆 =
2 ∙ 𝐿
𝑛
It comes as follows :
𝑝 =
ℎ
𝜆
=
𝑛 ∙ ℎ
2 ∙ 𝐿
And, if one is interested in the kinetic energy of the particle :
𝐸 𝑐 =
𝑚 ∙ 𝑣2
2
=
𝑚2 ∙ 𝑣2
2 ∙ 𝑚
=
𝑝2
2 ∙ 𝑚
→ 𝐸 𝑐 =
𝑛2 ∙ ℎ2
8 ∙ 𝑚 ∙ 𝐿2
The kinetic energy of a particle enclosed in a
(linear) box increases (rapidly) when the size of
the box decreases.
The particles will therefore always seek to occupy a box as large as possible in order to reduce their
content in energy. They will run away from little boxes as a claustrophobic person would do it.
4. Application to the stability of the hydrogen atom :
During its fall to the nucleus the electron decreasesits potential energy(electrostatic attraction) but
increasingly limits its accessible domain, which causes an increase in its kinetic energy
("claustrophobia"). A balance situation is reached when any reduction in potential energy is
compensated by the concomitant increase in kinetic energy.
The stability of the hydrogen atom can be understood as a
compromisebetween a potential fall and a kinetic blockage.
The Coulombattractionisacentral force forwhichthe classical solutionsof the movementare planar,
circular or other. To maintain a simple mathematical formalism, the electron-accessible domainwill
be a circle and any stationary wave must overlap (L = n.) after traversing a perimeter :
2 ∙ 𝜋 ∙ 𝑟 = 𝑛 ∙ 𝜆 𝑒𝑡 𝜆 ∙ 𝑝 = ℎ → (
2 ∙ 𝜋 ∙ 𝑟
𝑛
) ∙ 𝑝 = ℎ → 𝑝 =
𝑛 ∙ ℎ
2 ∙ 𝜋 ∙ 𝑟

3. 3
Thus, for the total energy of the (proton + electron) system :
𝐸 = 𝐸 𝑘𝑖𝑛𝑒𝑡𝑖𝑐 + 𝐸 𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑒𝑙 =
𝑝2
2 ∙ 𝑚
−
𝑒2
4 ∙ 𝜋 ∙ 𝜀0 ∙ 𝑟
𝐸 𝑛 =
𝑛2 ∙ ℎ2
8 ∙ 𝜋2 ∙ 𝑚 ∙ 𝑟2 −
𝑒2
4 ∙ 𝜋 ∙ 𝜀0 ∙ 𝑟
The energy drop (coulomb) will be blocked by the rise in kinetics energy when the total energy is
minimal, i.e. when the derivative in relation to the radius cancels.
𝑑𝐸 𝑛
𝑑𝑟
= 0 =
−2 ∙ 𝑛2 ∙ ℎ2
8 ∙ 𝜋2 ∙ 𝑚 ∙ 𝑟3 +
𝑒2
4 ∙ 𝜋 ∙ 𝜀0 ∙ 𝑟2 → 𝑟𝑛 = (
𝜀0 ∙ ℎ2
𝜋 ∙ 𝑚 ∙ 𝑒2
) ∙ 𝑛2
The solutionsare organizedaccordingtothe numberncalled"quantumnumber".The smallestsize is
n = 1.
𝑟1 = (
𝜀0 ∙ ℎ2
𝜋 ∙ 𝑚 ∙ 𝑒2
) = (
8,854.10−12 ∙ (6,626.10−34)2
3.1416 ∙ 9,109.10−31 ∙ (1,602.10−19)2
) = 52,9.10−12 𝑚 = 52,9 𝑝𝑚
That's Bohr's value for the fundamental state of the hydrogen atom !
It now remains to express the total energy for the values of Rn.
𝐸 𝑛 =
𝑛2 ∙ ℎ2
8 ∙ 𝜋2 ∙ 𝑚 ∙ (
𝜀0 ∙ ℎ2
𝜋 ∙ 𝑚 ∙ 𝑒2)
2
∙ 𝑛4
−
𝑒2
4 ∙ 𝜋 ∙ 𝜀0 ∙ (
𝜀0 ∙ ℎ2
𝜋 ∙ 𝑚 ∙ 𝑒2) ∙ 𝑛2
𝐸 𝑛 =
𝑚 ∙ 𝑒4
8 ∙ 𝜀0
2 ∙ ℎ2 ∙
1
𝑛2 −
𝑚 ∙ 𝑒4
4 ∙ 𝜀0
2 ∙ ℎ2 ∙
1
𝑛2
Note that the virial theorem (EPot = - 2.ECin) is satisfied.
𝐸 𝑛 = −
𝑚 ∙ 𝑒4
8 ∙ 𝜀0
2 ∙ ℎ2 ∙
1
𝑛2 = − 2,18.10−18 ∙
1
𝑛2 𝐽 = −13,6 ∙
1
𝑛2 𝑒𝑉
Note : It is remarkable toobtainthe exactexpressions of Rn and En inspite of the approximations
necessary to meet the program of a first year of university course.
5. Attempt to improve the model of the hydrogen atom :
At the previouspoint,the electronisconfinedinalinearbox (circumference) whilethe systemclearly
has spherical symmetry. One can try to extend the calculation for a spherical three-dimensional
domain(volumeof asphere of radiusR) whilelimitingitselftothe mathematicaltoolsof afirstyearof
university.

4. 4
If the condition of stationarity carries, for example, on the diameter (D) of the sphere it comes:
𝐷 = 𝑛.(
𝜆
2
) = 2. 𝑅 → 𝜆 =
4. 𝑅
𝑛
=
ℎ
𝑝
→ 𝑝 =
𝑛.ℎ
4. 𝑅
→ 𝐸 𝑐 =
𝑛2 ∙ ℎ2
32 ∙ 𝑚 ∙ 𝑅2
As the electron has access to all the volume of the sphere, the radius r, its distance (mean,efficient,
most probable...) to the nucleus will be required less than R:
𝐸 𝑝 = −
𝑒2
4 ∙ 𝜋 ∙ 𝜀0 ∙ 𝑟
( 𝑟 < 𝑅) → 𝑬 𝒏 =
𝒏 𝟐 ∙ 𝒉 𝟐
𝟑𝟐 ∙ 𝒎 ∙ 𝑹 𝟐 −
𝒆 𝟐
𝟒 ∙ 𝝅 ∙ 𝜺 𝟎 ∙ 𝒓
𝑑𝐸 𝑛
𝑑𝑅
= 0 =
−2 ∙ 𝑛2 ∙ ℎ2
32 ∙ 𝑚 ∙ 𝑅3 +
𝑒2
4 ∙ 𝜋 ∙ 𝜀0 ∙ 𝑟2 → 𝑅 𝑛
3 = (
𝜀0 ∙ ℎ2 ∙ 𝜋 ∙ 𝑟2
4 ∙ 𝑚 ∙ 𝑒2
) ∙ 𝑛2
For r = Rn/2, for example, it comes:
𝑅 𝑛
3 = (
𝜀0 ∙ ℎ2 ∙ 𝜋 ∙ 𝑅 𝑛
2
4 ∙ 𝑚 ∙ 𝑒2 ∙ 4
) ∙ 𝑛2 → 𝑅 𝑛 = (
𝜀0 ∙ ℎ2 ∙ 𝜋
16 ∙ 𝑚 ∙ 𝑒2
) ∙ 𝑛2 = (
𝜀0 ∙ ℎ2
𝜋 ∙ 𝑚 ∙ 𝑒2
) ∙ (
𝜋2 ∙ 𝑛2
16
)
→ 𝑅1 = (
𝜀0 ∙ ℎ2
𝜋 ∙ 𝑚 ∙ 𝑒2
) ∙ (
𝜋2
16
) = 0,62. 𝑟( 𝐵𝑜ℎ𝑟) 𝑇ℎ𝑒 𝑎𝑡𝑜𝑚 𝑖𝑠 𝑓𝑜𝑢𝑛𝑑 𝑎 𝑙𝑖𝑡𝑡𝑙𝑒 𝑡𝑜 𝑠𝑚𝑎𝑙𝑙.
6. Some consequences in chemistry :
The pooling of electrons from several atoms to form molecules (chemical bonds) also leads to an
enlargement of the electron-accessible domains. This results in a decrease in potential energy that
contributes to the bond energies.
Example (1)
The hydrogen molecule (H2) results from the union of two atoms that only use 1s orbitals to justify
their fundamental state chemical properties (other solutions obviously exist but are too costly in
energy).
Example (2)
All the chemistry of butadiene (H2C=CH-CH=CH2) indicates that the single (too short) bond has a
significantdouble character. Furthermore, 1,4additions are regularlyobserved.Thisisrationalizedby
admittingadistributionof electrons on the 4 carbon atoms (= a large box) in the molecular orbitals.
According to the theory of molecular orbitals
Energy
H H
H H
a bonding orbital (0 node)
an anti-bonding orbital (1 node)
The biggest box has the lowest energy

5. 5
Where it is necessary to associate two forms to describe reality. The right-hand limit form shows a
charge separation, unfavourable from an energetic point of view but compensated by the expansion
of the electron-accessible domain.
Example (3)
Benzene does not contain alternating singles and double bonds.
Where it is necessary to invoke two limit-forms to
justifythe physico-chemical propertiesof benzene.
According to Lewis theory (out of quantum mechanics)
According to the theory of molecular orbitals
3 nodes.
It is an anti-bonding orbital.
Energy
2 nodes.
It is an anti-bonding orbital.
1 node.
It is a bonding orbital.
0 node, a large box over 4 atomes.
It is a bonding orbital.
According to the theory of molecular orbitals.
0 node.
The box extends over 6 atoms.
This is the lowest energy orbital.
1 node.
Two boxes over 3 atoms.
This is a next-to-the lowest orbital.
etc ...
According to Lewis theory (out of quantum mechanics)
or