  # 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/Mmin9,1093897 ×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 Go to Top