Electric charge movement and gas tube discharge
Albert Einstein paraphrase:
The most beautiful experience we can have is the
stupidity. It is the fundamental emotion that stands at the cradle of modern
science.
The problem is to have enough supporters,
eventually in key positions to promote the stupidity, and the success is
guaranteed.
This is, in short, the history of modern science
…….
Background and actual interpretation
A gas tube,
in principle, is a very simple device. It consists of an evacuated glass tube
fitted at each end with a metal terminal called electrode. The simplest tube
contains inside, a small amount of highly purified inert gas or a mixture of
compounds for more complex one. Connected to the two electrodes is a source of
high-voltage electrical power able to deliver a limited current. When the
current is turned on, the tube glows.
In the
absence of a potential difference, the gas inside the tube is formed by neutral
particles and the current does not pass through. In the presence of a high
voltage potential difference, a process of ionization takes place and the
current can flow through tube.
It is
admitted the existence of a small quantity of charge in a specific medium (gas)
due to cosmic radiation background. These charges are not enough to insure a
flow of an electric current. Besides this charge, when a high enough voltage is
applied on gas, the electrons of gas medium are stripped away from their
respective nuclei, creating conditions of ionization. In these conditions, with
a higher concentration of free electrons able to migrate from a region to
another, gas becomes a relatively good conductor of electricity.
In a
simplified description, the electrons coming from electric current, having a
minimum specific energy for every gas, are powerfully enough to hit the gas
molecules, to generate new charges between cathode and anode and the low
rarefied gas become conductor of electricity.
Because
like electrical charges repel one another and unlike charges attract each
other, a free electron is strongly attracted to any nearby positive ion. This
attraction leads to the rapid combination of the positive ions and electrons
into neutral particle, and this process is responsible of light production and
its color depends upon on gas type.
Thus, to
produce a gas discharge and tube glows, electrons must be removed from neutral
molecules and recombined with positive ions to form other neutral molecules.
The practical way of producing this ionization is by passing a current through
the gas.
When
voltage is applied to the electrodes, cathode emits electrons and anode
attracts these electrons. If the voltage is high enough, the electrons will be
attracted with tremendous force and will accelerate toward the positive
electrode reaching speeds of tenth of km per second.
Figure 1 Action of
electrons and positive ions in gas-filled tube.
During this trip toward anode,
electrons collide with a neutral gas molecule that lies in its path. It hits
this molecule with such force that one or more (usually more) electrons are
liberated. These secondary electrons, once free, starts moving toward anode.
They soon collide with other neutral molecules, generating in a cascade process
more charge into gas. Shortly after high
voltage is applied to the tube, there is a general process of ionization in
entire volume of gas. Electrons are liberated from molecules,
free electrons combine with positive ions, giving off light, and then are
blasted apart again.
For our discussion it is necessary
to remind the fact that gas current that flow through the tube depends largely
on the pressure. In the low-pressure gas, the atoms are fewer in number and
farther apart from one another; therefore, the free electron has a longer
distance in which to get up speed before it hits a neutral atom. As a
consequence, when the free electron hits, it hits much harder, and more free
electrons are liberated than in case of higher pressure. Hence, as the pressure
is lowered, the current will increase.
If, however, the pressure is reduced
so much that a nearly perfect vacuum forms inside the tube, then there are so
few atoms available that the current will decrease due to lack of electrons and
ions, even though the electron speeds are very high.
Depending on the gas composition, it
is known from practice that gas tubes present specific colors:
Gas
|
Color
|
Helium
|
Whitish orange
|
Neon
|
Red-orange
|
Argon
|
Violetish pale - lavender blue
|
Krypton
|
Grayish dim
|
Xenon
|
Grayish
|
|
|
|
EXPERIMENTAL PART
EXPERIMENT 1
Long time
ago, I’ve found that a gas tube approached by a charged Van der
Graff device produce light. There was only a curiosity at that time, but after that
I was thinking of the mechanism of tube discharge. At that time, the tube was
moved toward VDG great sphere (charged positively) with metallic electrode
directed to sphere. At a distance of about 1 cm between VDG sphere and tube
electrode, a spark appear in the space between, and simultaneously an electric
discharge appear in the tube (fig.1).
When the
gas tube is moved toward small sphere of Van der
Graff device (charged negatively), thee same electrical discharge is observed
as in fig. 2.
Figure 1.
Glows of a gas tube near a positively charged sphere of VDG device
Figure 2.
Glows of a gas tube near a negatively charged sphere of VDG device
For the
beginning, in a trial to elucidate the mechanism of tube lightening, a gas tube
in form of U is used. The tube is moved close to the positively charged VDG
sphere, keeping the metallic electrodes outside, as in fig. 3.
Figure 3.
Lightening of a U form gas tube near a positively charged sphere of device
Contrary to normal expectations, the tube
starts lightning even there is no metallic electrode able to be charged and to
emit light. The tube is lightening for minutes (with my small VDG device) in
the absence of an apparent source of electrons. In order to block a eventually
electron charge displacement from air, both tube electrodes are put in short
circuit and covered with a insulating material as in fig. 4 and the experiment
repeated.
Figure 4.
U form gas tube variant
When this modified U tube is moved closed to
negatively charged VDG sphere, the tube is lightening as in precedent case as
in fig. 5.
The tube glow again, when the U tube
is moved toward positive or negative sphere, with both electrodes, maintaining
the same distance between sphere and electrodes as in fig. 6,
Figure 5.
Lightening of a U form gas tube near a negatively charged sphere of device
Figure 6. Lightening of a U form gas tube
with both electrodes at the same potential
Where is the absurdity
in actual explanation …..
An old Van der
Graaf device and a simple neon tube can be used as
cut off experiment to
the charge movement during tube glow.
It is accepted (see the previous post related
to the absurdity of VDG working principle) by actual orthodox science, that a
VDG device produce as result of friction a charge separation, and consequently
a sphere will become positively charged and the other negatively charged. For
the purpose of experiment it is not important which one is positively charged
and which remain negatively.
As was up described, in order to have a electric current passing inside a gas tube, there is
necessary for a flux of electron to produce a gas ionization.
Therefore, in our experiment it should appear a
clear difference between the comportament of a gas tube
closed to a negatively charged sphere and the same gas tube closed to a
positively charged sphere.
In first case it is possible to ,,admitʼʼ that a negatively charged object can
somehow produce a gas discharge. Of course a ,,common sense mindʼʼ,
based on actual classical electrodynamics,
expects to have a null result (no light emission) when gas tube is moved
close to the positively charged sphere, because in this case there is no
electron flux able to excite the gas in the tube.
The experiments contradict these theoretical
predictions and actual mediocre theoreticians should choose what is wrong in
their explanations.
I doubt that their minds are able to resolve
this simple and schoolboy problem, but in case someone is able to do that, the
second part of experiment using a U gas tube will indicate him that his
explanation is completely out of sense.
So, when a U form gas tube is used, with
metallic electrodes outside, the positively or negatively charge from VDG
spheres act directly on glass material and indirectly on the gas from tube.
There is no flux of electrons, there is no breakdown
for the glass material. How is possible to have a charge moving inside an
insulator (glass) in these conditions? How is possible to have a gas tube glow
when a positive or negative potential is applied on a insulator
?
When the U gas tube is used with both
electrodes inside, at the same distance related to VDG sphere, there is no
difference of potential between electrodes and independent on the charge
present on the VDG sphere, and according to actual electrodynamics the tube
must not glow. The reality is completely opposite …..
I doubt that actual theoreticians minds are able to understand this simple cut
off experiment, but I donʼt write for the
present, I write for the future.
EXPERIMENT 2.
A tube of gas is connected to a high voltage
source, as in fig 7. For commodity, in the experiment as high voltage source
the same VDG device is used with tube connected simoultaneously
at positive and negatively charged sphere.
Figure 7. Lightening of a gas tube with both
electrodes connected at VDG device spheres
Depending on the gas composition, it is known
from practice that gas tubes present specific colors. When this light is
analyzed with a spectrograph, a discontinuous spectra formed by one line or few
lines is observed. According to actual electrodynamics in order to have a
current inside gas tube an ionization process must take place. In the same
time, the quanta hypothesis must be respected during elementary processes of
light generation.
Why the actual explanation for this spectra is
absurd….
There is a simple problem with actual
explanation: in order to have a consistent explanation one of up presented
ideas must be ruled out. Or a gas tube permit an electric current to pass
inside in absence of a gas ionization process, or the quanta hypothesis is
false. For proposed explanation, presented in detail in the book, both ideas
are absurd.
Let’s analyze a little bit these ideas from
actual orthodox theory point of view.
Considering a simple gas tube with a single gas
inside, let’s analyze the process of ionization taken in discussion the quantum
hypothesis and the experiment Franck –Hertz (an experiment fundamental for
quantum theory).
In Franck Hertz experiment, it was
demonstrating that, in a tube filled with mercury vapors (the results are valid
for every other gas, only the values of currents are different), atoms could absorb (and are excited) only by specific
amounts of energy (quanta).
At low potentials, the accelerated electrons
suffer purely elastic collisions with mercury atoms in the tube. This is due to
the prediction of quantum mechanics that an atom can absorb no energy until the
electron kinetic energy is equal with energy necessary to lift an electron into
a higher energy state.
The fundamental question is: An electron lift
in a higher energy state means an ionization process?
In case of Franck Hertz experiment there isn’t
an ionization process effectively, because an ionization process means for
colliding electron to have enough energy able to lift the electron from excited
state to other higher excited states until the electrons becomes free as
indicated in fig.8.
Figure 8.
Details about excitation and emission
But, if the collision between electrons and
atom gas does not produce ionization how can be interpreted the increasing of
electric current in case of Franck Hertz experiment?
I leave for the mediocre theoreticians to
resolve this problem in the frame of actual theory.
In experiment the ionization process does not
take place, and emission spectra of excited mercury give exact indication about
type of excitation. The spectra of mercury vapors emission gives only a single
line in ultraviolet (l = 254 nm) in case of Franck Hertz experiment, and this means only
excitation and not ionization ( fig. 8-a details).
If the actual mediocrity think are able to
solve the up presented gap, a new problem can be formulated:
In Franck Hertz experiment with mercury, it was found that mercury atoms
emit radiation of 253.6 nm, in ultraviolet, if and only if electrons having at least the
excitation energy of 4.88 eV collide with them as in fig. 9.
Figure 9. Comparison between mercury and neon
emission lines
In Franck Hertz experiment with neon
gas-fig. 9, the process of absorbing energy from electron collisions produces
visible line and this is caused by some levels situated between 18.3 to 19.5 eV .
How is possible to have for mercury a line
emission in UV with a excitation of 4,9 eV, and for example in case of neon a line in visible with a excitation of 18 eV or more?
Maybe after actual theoreticians, it is
possible for a excitation of 1 eV
to produce a X-ray photon?
In few scientific texts a new stupidity is formulated
for Ne tube working principle. So the neon atom
becomes excited by a energy of electron beam equal with 18,4
eV. From this excited state atoms falls on a lower
state at 16,57 and 16,79 eV
as in fig. 10. The energy difference, in the range of few electrons volts,
gives light in the visible range. Of course, no actual theoreticians give an
explanation of what’s happened with these intermediate levels. I suggest to
those theoreticians who believe in this mechanism to put a hand on a neon tube to
feel its temperature and the other hand on their scalp. One of these two
objects must boil. If the gas tube does not boil when every atom inside release
a energy of about 16,5 eV,
they must be sure that their minds are boiling (if there is something to boil!).
Figure 10. Variant of neon emission lines in Franck Hertz
experiment
By analogy, let’s analyze the case of simple
glowing gas tubes. If an ionization process take
place, as in fig. 11, a positive cation and a free
electron are produced in the first stage, when the energy of incident particle
is greater then ionization energy. For
the simplicity of interpretation, lets consider only the simple ionization
process, and the resulting charge are not moving under the influence of
electric field. After ionization, at least a part of these charge recombine
again, and light is produced. From a free electron and a cation,
during the recombination process, an multi-line, with
line in UV,
Figure 11.
Theoretical emission spectra of a gas tube
But the reality is completely different, and a
gas tube is very specific. In normal condition, the emission spectrum is composed
by a single line or combinations of very few lines.
In the same time it must be highlighted that a
direct collision between a free electron with high energy and a positive ion,
should determine the generation of primary X-ray photons and in an indirect
manner a visible photons.
Despite this prediction, most of common used
tubes (mercury is an exception) works directly with visible photons and this
means the impossibility of existence of an ionized gas, but only an excited
gas. Even in case of mercury the line transition in UV does not mean
ionization, it is only excitation, with a greater gap between ground level and
excited level. In case it is admitted that mercury line produced in UV is
characteristic for an ionized medium, the Franck Hertz experiment must receive
a new more fantastic explanation…..
A discussion similar to one presented already
at photoelectric effect is further presented, where the generation and
extinction of electric charge is unable to explain the observed current
variation.
Why the actual explanation is absurd
Let’s consider a gas tube having only five
neutral gas particle (neon or mercury atoms for example).
The cathode emits only 3 electrons which hit,
if they meet on their trajectories the gas particles.
If the
kinetic energy of emitted electrons is lower then ionization energy, no
supplementary charge can be produced so the electric current into circuit must
have a uniform variation and surely must be lower then the current of same tube
without filling gas.
When the energy of incoming electron is 4,9 eV for mercury or 18,4 for
neon, the ionization process does not take place. For ionization of these elements are
necessary 10,4 eV in case of mercury and 21,5 eV for neon.
Consequently a energy of 4,9 or 18,4 eV can’t produce other free electrons or other cations into gas tube. The actual explanation for Franck
Hertz experiment and for gas tube glowing deserves a monument of science
stupidity and not a Nobel price.
If the electron with a 4,9 or 18,4 eV hit a mercury or a neon atom, once or many times, the
intensity of electric current should decrease, because the electron is in
retard on its way between cathode and anode. The observed variation of electric
current in case of Franck Hertz experiment should be considered a new enigma of
modern physics like entanglement and wave corpuscle duality.
Let’s consider that energy of
electron fascicle is increased and the ionization conditions are obtained, more
precisely, 10,4 eV in case
of mercury and 21,5 eV for neon. For simplicity, it
can be considered that all emitted electrons meet ideal condition and produce
ionization for 3 gas particle as in fig. 12.
Figure 12.
Gas ionization produced by electron beam
According to actual interpretation, these
secondary electrons are accelerated toward anode and they participate to
electric current increasing into circuit.
But can be admitted as true this
interpretation?
If secondary electrons are accelerated and
produce an increase current into circuit, where is the mechanism of light
production?
What’s happened with
remaining positive ions?
There are more possibilities and actual
theoreticians should choose which is less absurd.
1. The
positive ions remain somewhere in the space between anode and cathode. As
consequence after few seconds of electric current increasing, there will be a
current decreasing because cathode emitted electrons will be attracted by
positive ions instead of anode. So a sinusoidal current should pass through a
gas tube in this case.
2. There is a movement of cations
toward cathode, in the same time with a movement of electrons toward anode. The
cations are neutralized due to collision with
electrons or with cathode. But in this case, there is only a local generation
and extinction of electric charge, and this does not affect the electric
current into external circuit. As result, the electric current into external
circuit should be the same like in case of vacuum tube. In this case only the
yield of process is affected due to conversion of a part of system energy into
other forms of energy.
Maybe actual theoreticians have invented a new
method of charge counting? Or maybe there is a spatial temporal distortion and
because electrons have greater speeds then cations,
after a ionization process, a electron is able due to ,,temporal
distortion” to make a tour of external
circuit and ,,young and vigorous” is
able to extinct the cation charge. In this case … of
course, any unskilled person counts an increased current into external circuit.
3. Maybe some new processes are taken place
into the space between anode and cathode. The actual physics should clarify
these phenomena.
For a common sense mind, admitting as true the
actual definition of electric current, the actual explanation of gas tube glowing
must be ruled out. In proposed theory, the ionization phenomenon is not a
dominant factor responsible for gas tube glowing.