FAMILIARITY IN FUNDAMENTAL RELATIONS AND UNITS
One of the difficulties which someone, who is not a physicist encounters when attempts to
understand the relation of matter to energy is what units should be used so as they give the correct result. Also, the concept of energy not immediately
reveals clearly in our minds, what phenomenon summarizes exactly. But the original ambiguity, indeed should not prevent the proper use of this term and
calculations. We have an ability to we use a concept properly, which is hard to define it in the begin or because it is general and not clearly defined.
We have this ability with many words and concepts and not only with the energy. A provocative example and with history is the concepts of "truth" and "existence".
An example of calculation of energy (E) and rest mass (M) of the electron with the most famous
equation: E = Μοc2 If we insert mass in kilograms (kg) and the speed of light (c2) in
meters per second at the square (m2/s2) at this simple formula, then the result will be energy in Joule units. If a quantity of mass multiplied with
speed of light at square thus result an equivalent quantity of energy E=mc2. If the quantity of energy is divided by speed of light at
square thus result an equivalent mass m=E/c2
Remind:
1 Joule is equal to 1kgr m2/s2. The
1eV =1,602176462 ×10-19 Joule and 1Joule =
0,62415097445 × 1019 eV (or 1,6 ×10-12 erg).
That is, according to the units of Physics, 1Joule is the force of
1Newton × displacement of 1m. The 1Newton is defined as the force
where is should to be apply on a body of 1kg mass so as is accelerated one meter per second at squared (1 m/s2).
EQUIVALENT ENERGY OF THE ELECTRON
Eο=Μο
c2 → Ee=Μe
c2
Ee= (9,109389 ×10-31 kg) × (8,987551 ×1016
m2/s2) = 8,187110 ×10-14 J
In eV: 8,187110 ×10-14 J × 6,2415097445 ×1018
eV = 510,9993 ×103 eV/s (= 0,5109993 MeV/s) |
1eV =1,602176462 ×10-19
= 1,6021764 ×10-12 erg → 1Joule = 6,2415097445
×1018 eV
MASS OF AN ELECTRON Me
BY ITS EQUIVALENT OF ENERGY
Mο
=Eο/c2 → Me=
8,18711 × 10-14 J / 8,987551 ×1016 m2/s2 = 9,109389 × 10-31 kg
Me= (510,9993 ×103 eV/s) × (1,6021764 × 10-19
J) / 8,987551 ×1016 m2/s2 |
HOW MANY TIMES "FIT" THE CONSTANT h IN THE ENERGY of an electron.
THIS CALCULATION GIVES FREQUENCY (f) That is, how
many times the energy of the numerator is greater than the constant h in the denominator (in time 1sec).
f=
E/h
fe= Ee / h = 8,18711 × 10-14 J /
6,626069 ×10-34 J·s = 1,2355909 × 1020 Hz
In units eV: Ee
/ h = 510,9993 ×103 eV/s / 4,1356676 × 10-15 eV sec = 1,2355908 × 1020
Hz |
EQUIVALENT
MASS OF ELECTRON BY FREQUENCY
h=E/f
=Μc2/f →Μe
= h fe/c2 = 6,626069
×10-34 J s × 1,2355908 ×1020 Hz / 8,987551 ×1016 m2/s2 = 8,187110 × 10-14 J / 8,987551
×1016 m2/s2 = 9,109389 ×10-31 kg
* In physics, the formula c=f·λ for calculation of the frequency
and wavelength applies only for electromagnetic waves and photons, which are considered massless. |
EQUIVALENT LENGTH "WAVE" of
electron λe which called Compton's length.
Since frequency units we can find a length (corresponding)
wavelength or a radius (r=c/ω), according to the known relation: λ=c / f → λe = c / fe
λe=2,997924 5 ×108 m/s / 1,2355908 ×1020
Hz = 0,2426308 ×10-11m (=2,426308 ×10-12 m)
The same result is found with formula: λ=h/Μc
(Compton's length)
λe = (6,626069 ×10-34 J·s) / Me c
→
λe = (6,626069 ×10-34 J·s) / (2,730926 ×10-22)
= 0,24263085 ×10-11 m
1m =109 nm | 1nm =10-9
m
0,24263085 ×10-11 (×109 nm) = 0,24263085 ×10-3
nm
The visible light wavelength: 400 nm to 700 nm (or 4000 Å - 7000
Å ≈ 4 ×10-7 - 7 ×10-7 m). A comparison Compton electron length to the wavelength of visible light:10-7
m/10-12 m. For the electron is about 105 times less.
The length measuring unit Angstrom:
Å = 0,1nm = 1 ×10-10 m and micron µ = 1
×10−6 m |
MASS ELECTRON SINCE RESPECTIVE WAVELENGTH
λ=h/Μc
→Μ=h/cλ =
6,626069
×10-34 /
(2,997924 ×108
) (0,2426308 ×10-11
)→
Me=
6,626069 ×10-34
/ 0,7273886 ×10-3
= 9,1094 ×10-31
kg |
THE CONSTANT h IN eV UNITS
6,6260693 ×10-34 J sec
×
0,62415097445 × 1019 eV = 4,1356676 × 10-15 eV sec (or eV/Hz) |
THE CONSTANT h
EQUIVALENT TO FREQUENCY IN Hz
If 4,1356676 × 10-15 eV corresponds per Hz (or for sec) then for 1 eV :
1Hz ·1eV / 4,1356676 ×10-15
eV = 2,41798929 ×1014 Hz (times the quantity h).
1 eV = 2,41798929 ×1014
Hz (times the quantity h)
1eV correspond at frequency n=241,8 THz (where 1THz
=1012 Hz)
h =4,1356 ×10-15 eV/Hz = 1eV / 241,8 THz
1012 Hz=1THz and 1Hz=10-12 THz
|
Frequency fe
of electron in THz: 1,2355908 × 1020 Hz × 10-12
THz =
123559,08 ×103 THz
Equivalent mass of 1eV energy
1Joule = 0,62415097445 × 1019 eV
→ 150,919037 × 1031 Hz
1eV = 1,602176462 ×10-19 J
→ 241,798929 ×1012 Hz
(or 241,798929 THz)
m=E/c2 → 1ev × (1,602176 × 10-19 J) /c2 =
0,178266131 ×10-35
kg
The minimal quantity h multiplied by a frequency f is equivalent with a quantity of
energy (energy that are bring electromagnetic waves with " null " mass of photons). Mass on the square in speed of light
amounts also with energy, that is to say:
h f = Μ c2 = E → Μ = h f / c2
= h /cλ = h / f λ2 or Μ = Mmin f |
We can find a frequency f and reversely period for inertia M. When we multiply the Planck's
constant h by a minimal frequency 1Hz then results a minimal quantity of energy Emin = h 1Hz. The minimal quantity h on
each other frequency f gives us multiples quantity of energy Ε=hf. When on the contrary, we divide the quantity of energy with a certain
frequency then results the minimal quantity of energy h=E/f .
Because the reverse of the frequency is the (periodic) time
t = 1/f is also in effect t = h/E.
(!) We do not forget,
a mass M in microscopic dimensions does not exist in (stable) situation of rest and we use the measurement in kg for comprehension and simplification
reasons. The formula c= f λ is valid for the calculation of frequency and length of electromagnetic waves and photons only, that are considered
in physics without mass. However, this consideration does not prevent to we speak with term of
frequency, when we observe phenomena of repetition and similar quantities that change multiple or sub-multiple.
We observe, the first useful mathematic relations for the description of structure of matter and shaping of mass.
A minimal quantity of energy Emin = h 1Hz is connected with a minimal frequency fmin and the largest length λmax=c/fmin.
As long as the frequency f increases, also increases the quantity of energy h f and it minimizes the length λ
(since c/f and λ=h/M c). In increased frequency and quantity of energy amounts more mass (h fmax / c2
= Mmax). In more much quantity of mass corresponds more quantity of energy and mass is connected with higher frequency and smaller
length. Thus in the theoretical mass of unification (Mplanck =√(h c/G) we find a maximum high frequency fmax,
a most quantity of energy Emax and a minimal length λmin.
Without well knowledge of physics we observe:
Emax / c2
= h fmax / c2 = h / c λmin
= Mmax
The minimum quantity of energy that express by constant h
(×1Hz) can be considered as a product of a minimum mass by c2 (that is Mmin c2). Therefore, if the
minimum quantity of energy of the constant h is divided by speed of light at square c2 then result a minimum quantity of mass h/c2 =Mmin
sec. (If we want a mathematical result for clear mass in unit of kg then will multiply the constant h in numerator by the unit of
the minimum frequency f =1Hz).
That is, if energy E = h*1Hz then :
h
1Hz/c2 =Mmin→6,6260693 ×10-34
J / 8,9875513 ×1016 m2/sec2 =0,737249 ×10-50
kg
► As multiple quantities of energy (E=hf) result from the minimum quantity h,
equivalently multiple quantities of mass (M=Mmin f) result from the minimum quantity Mmin.
Therefore, if we divide a quantity of mass m by the minimum quantity Mmin results a frequency of mass, as correspondingly if we divide a
quantity of energy E by the minimum quantity h results a frequency of energy quantity.
f =Μ / Mmin
or Μ / (h/c2 ) and f
=E/h = Μc2 / h
For example f =Μe/Mmin→9,1093897
×10-31
kg / 0,73724967 ×10-50kg =12,355908 ×1019
Hz. We find the same result as from the formula f = E/h
The h in multiplication with a frequency f give us the same result as much as mass m
in multiplication with the c2, that is the equal quantity of energy:
Μ
c2 = h f = E
→
Μ = h f / c2
→
Μ = h /cλ
or Μ =
Mmin
· f
|
* The mark "→" in formulas is used as "implies".
* Attention to the decimal point. This is a different mark for many countries


|