It deals with the methods and calculations to measure the masses of photons with different wavelengths. where I was able to create two experimental calculations to explain the measurements of the masses of the photons. and I hope that this thesis competes with others, in order to obtain a physics prize.

1. Theoretical Physicist Wilson Vitelio Hidalgo Suarez, is accredited through
Senescyt, as one of the researchers of Ecuador and Ecuadorian Author,
registered in the Ecuadorian book chamber. He has written a total of 10
books for more than 25 years.
Project name: Calculation and measurement method, to measure the
masses of photons with different wavelengths.
Sole author Wilson Vitelio Hidalgo Suarez
NATIONAL COPYRIGHT REGISTRY No. 025512
Email: Wilson.hdlg@gmail.com Cell: 0990895071
January 3, 2024 Guayaquil-Ecuador

2. EXPERIMENT TO DETECT THE MASSES OF PHOTONS WITH
DIFFERENT WAVELENGTHS.
Scattering of electromagnetic radiation
When the electromagnetic radiation that has passed through a region in
which there are free electrons is analyzed, it is observed that in addition to
the incident radiation, there is another of lower frequency. The frequency
or wavelength of the scattered radiation depends on the direction of
scattering.
Let be the wavelength of the incident radiation, and be the wavelength of
the scattered radiation. Compton found that the difference between both
wavelengths was determined solely by the dispersion angle, as follows
Where c is a constant that is equal to 2.4262 10-12 m
The Compton effect is explained in terms of the interaction of
electromagnetic radiation with free electrons, which we assume initially at
rest in the observer's reference system.
Linear momentum of photons
In the photoelectric effect we have only considered that the photon has an
energy E=hf. Now, a photon also has a linear momentum p=E/c.
This relationship is not new, but rather arises when posing the equations
that describe electromagnetic waves. Electromagnetic radiation has
momentum and energy. When we analyze any process in which
electromagnetic radiation interacts with charged particles, we must apply
the laws of conservation of energy and linear momentum.

3. In the case of the photoelectric effect, the law of conservation of linear
momentum was not applied, because the electron was bound to an atom,
a molecule or a solid, the energy and momentum absorbed are shared by
the electron and the atom, the molecule or the solid with which it is
bound.
We are going to obtain the formula for the Compton effect from the study of an
elastic collision between a photon and an electron initially at rest.
Principle of conservation of linear momentum
Let p be the linear momentum of the incident photon.
Let p' be the linear momentum of the diffused photon,
Let pe be the linear momentum of the electron after the collision. P = p' + pe
Energy conservation principle
The energy of the incident photon is E=hf.
The energy of the scattered photon is E'=hf '.
We cannot write the kinetic energy of the electron after the collision as mev2/2
since the recoiling electron reaches speeds close to that of light; we have to
replace it with the equivalent relativistic formula

4. Where me is the rest mass of the electron 9.1•10-31 kg
The principle of conservation of energy is written
Solving the system of equations, we arrive at the following expression
Taking into account the relationship between frequency and wavelength become
the equivalent expression.
We have obtained the value of the proportionality constant from the
fundamental constants h, c and me.
We then come to the conclusion that we can explain the scattering of
electromagnetic radiation by free electrons as an elastic collision between a
photon and an electron at rest in the observer's reference frame.
From the linear momentum and energy conservation equations, we arrive at
the equation that relates the wavelength of the incident radiation to the
wavelength of the scattered radiation and to the dispersion angle

5. Espectro electromagnético a diferentes longitudes de onda
Domain of the electromagnetic spectrum

6. X-RAYS AND GAMMA RAYS
X-rays. Wavelengths between 10-8 and 10-17 m; frequencies between 1016
and 1025 Hz.
Atoms emit X radiation when electrons undergo transitions between the
innermost shells of atoms. These types of transitions can occur when a
beam of electrons bounces off or is stopped by a metallic target.
The frequency and, therefore, the maximum energy of the radiation that is
generated is determined by the energy with which the electrons reach the
target. In turn, this energy depends on the voltage by which the electrons
have been accelerated. Therefore, the maximum frequency of X-rays
depends on the voltage at the electron accelerator.
X-rays have several applications in technology, in the medical field and
industry.
Gamma rays. Wavelength equal to or less than 10-17 m. Frequency equal
to or greater than 10 25 Hz.
The gamma ray region of the electromagnetic spectrum overlaps with that
of They are also a component of the so-called cosmic radiation, radiation
that bathes the Earth from outer space.
Gamma rays are the most energetic radiation known and, in the case of
cosmic radiation, are produced by the most energetic events in the
universe, such as supernova explosions.
Many of the cosmic events that produce the observed gamma rays are not
fully understood and an entire gamma ray astronomy is being developed
to try to understand them.

7. Summarizing
We can express the photon scattering effect using free electrons. For the energy of the
incident photon and the energy of the free electron with rest mass, we obtain ε +
moc^2 = ε` + E` with the elastic collision without energy loss after the collision, the
sum of the incident photon + the sum of the electron before the collision, is equal to
the sum of the scattered photon + the electron after the collision.
The total energy of the electron E = mc^2 + k, where k is the kinetic energy of the
photon after the collision. ε` = ε + k it is observed that the energy of the incident
photon is relatively greater than the energy of the scattered photon.
In the case of the photon, k is the differential of the energy of the incident photon
minus the energy of the scattered photon. k = ε - ε` the prime energy of the scattered
photon can be explained in the following way ε` = hc / λ`, where λ` is the wavelength
of the photon after the collision.
We can express k in the following way k = ε - ε` = or k = ε - hc / λ` therefore we say
that hc / λ` = ε – k solving for λ` we obtain λ` = hc / ε – k
The constant hc has a value of 12.4 KeV and a wavelength of 2.43 x 10 - 12 m. The
wavelength λ of the incident photon is expressed as a function of the wavelength λc
and we have λ` - λ = λc (1 – cos θ)
Solving for the Dispersion Angle, we have cos θ = 1 - λ` - λ / λc therefore
θ = 〖cos〗^(- 1) λc - λ` + λ / λc

8. MEASUREMENT OF THE MASS OF THE PHOTON EXPERIMENTALLY
It is a fact that photons of different wavelengths have mass, which depends exclusively
on their wavelength and frequency. The method of measuring the mass of photons
with different wavelengths can be seen in the two figures below of this text.
Compton effect
Angle formed by the photon after the collision, where the detector
analyzes it.

9. Photon dispersion, the incident photon and the scattered photon are
observed, and the angle that the scattered photon formed in relation to the
electron bound to the atom.
MEASURES TO TAKE BEFORE CARRYING OUT THIS EXPERIMENT.
1) The experiment is not done in a vacuum, so after passing through the
magnetic field the air absorbs them quickly.
2) The Curies confirmed the negative charge of gamma rays in 1900 using
an electroscope.
3) Electromagnetic radiation is neutral, as are particles that carry equal
amounts of positive and negative charge.
CALCULATIONS OBTAINED IN THE EXPERIMENT
Scattering occurs whenever the propagation of light depends on the wavelength.
Sunlight, considered white, actually appears a little yellow, due to its mixture of
wavelengths, but it contains all visible wavelengths. This implies that white light is
distributed in a rainbow depending on the wavelength.
According to the experiment observable in the previous figures, the calculations relate
the prime wavelength of the scattered photon, the mass of the electron at rest and the
dispersion angle of the photon in relation to the strongly bound orbital electron.
These two equations were found that give the same result
1 ) Mf = 2(mo. λ`) / (Cos θ¿
2
2) Mf = (hc / λ`) / c
2
Where hc is a constant, whose value is 12.4 KeV and the wavelength λc has a constant
value of 2.43 x 10^ - 12 m.
I will give technical and practical examples

10. A photon with a wavelength of 400nm is scattered by electrons bound to its atom,
assuming elastic collision and after the collision the photon is scattered forming an
angle that will depend on the wavelength of the incident photon.
We can say that a photon with a wavelength of 400 nm has an energy of
ε = (hc/λ). the energy of the electron cannot be mev2/2, since the electron is moving at
a speed relative to light, then we can describe the kinetic energy of the electron:
To then calculate the wavelength of the prime photon, the wavelength of the photon after
the collision.
ε = 6.63 x 10
−34
J.s X 3 x 10
8
m/s / 4 x 10
−7
m = 4.9725 x 10
−19
J.
which is the energy of the incident photon. Now we will calculate the kinetic energy of
the electron
Ek = √me2
.c2
+ p2
- me
2
.c
2
Ek = √(9.11e−31Kg)
2
¿¿ + (9.11 e−31Kg .)
2
X ¿ =
Ek = √7.5 x 10
−44
¿ = 2.7 x 10
−22
Kg.m/s
Ek = 2.7 x 10
−22
Kg.m/s + 7.5 x 10
−44
¿ = 2,7 x 10
−22
J.
Ek = 2.7 x 10
−22
– 7.5 x 10
−44
= 2.7 x 10
−22
J.
Therefore the prime wavelength λ` or of the scattered photon is
λ` = hc / Ek – ε
λ` = (6.63 x 10−34
J.s X 3 x 108
m/s) / 4.9725 x 10−19
J – 2.7 x 10−22
J. =
λ` = 1.989 x 10−25
J . / 4.9698 x 10−19
J. = 4.0021731 x 10−7
m.

11. Therefore the mass of the photon would be equal to
Mf = (hc / λ`) /c
2
= (6.63 x 10^ - 34) J.s X 3 x 10^8 m/s) / 4.0021731 x 10
−7
m =
Mf = / 4.9698 x 10
−19
J / (3 x 10
8
m/s ¿2
= 5.522 x 10
−36
Kg.
Another more practical and experimental way to obtain the masses of photons with
different wavelengths is the following equation.
Mf = 2(mo. λ`) / (Cos θ¿2
This is the experimental equation and on which the experiment of measuring the
masses of photons with different wavelengths is based.
As we see, what makes up the Compton effect experiment is involved in the equation,
such as the rest mass of the electron, the wavelength of the scattered or prime photon
and the angle of dispersion of the photons, which will depend on the wavelength of
each photon...
So the wavelengths of the photons vary the scattering angle, likewise the different
angles vary the wavelengths.
We will test my experimental equation and measure the masses of photons with
different wavelengths…
We observed a photon with a wavelength of 400 nm, this photon was scattered, where
the photon varied its wavelength after the collision λ` of 4.0021731 x 〖10〗^(-7) m,
which is called prime, wavelength cousin. This dispersion and the variation of the
wavelength were observed to depend on the dispersion angle. Find the scattering angle,
in order to collect the experimental data and find or find the mass of the photon in this
experiment, which is something that appears to the Compton effect...
To find the scattering angle of the incident photon, we have. cos θ = 1 - λ` - λ / λc
Cos θ = 1 - (4.0021731 x 10−7
m. – 4 x 10−7
m.) / 2.43 x 10−12
m. = Cos 89 =
0.51.

12. Already knowing the scattering angle in the experiment, we can easily find the mass of
this photon!
The mass of this photon would be defined in the following way
Mf = 2(mo. λ`) / (Cos θ)^2
Mf = 2 x (9.11 x 10
−31
Kg. X 4.0021731 x 10
−7
m. ) / (Cos 89¿
2
Mf = 7.29 x 10
−37
/ (0.51¿2
=
Mf = 7.29 x 10
−37
/ 0.26 = 2.8 x 10
−36
Kg.
So we see in the experiment that the variation of the scattering angle of a photon not
only varies the wavelength of the photon, but also the mass of the photon in a
relativistic way... this calculated mass is of the scattered photon, relative to the incident
photon .
COMPARISON WITH THE THEORETICAL CALCULATIONS OBTAINED, TO
FIND THE MASS OF THE PHOTONS
INDISCUTABLELY, PHOTONS HAVE MASS
PHOTONS WITH DIFFERENT WAVELENGTHS HAVE INVARIANT MASSES.
Photons from infrared to high-frequency gamma rays have masses that never vary.
It is the discovery that I made more than 30 years ago and published it in my first book
in 2006 in the Ecuadorian book chamber.
TWO EQUATIONS THAT MEASURE THE MASS OF PHOTONS WITH
DIFFERENT WAVE LENGTH
This direct proof explains that photons have mass at different wavelengths. That is,
each photon with different wavelengths has different masses that never vary. These
different masses of the photons depend on their wavelengths. It is a mass domain, as
are the frequency and length domains throughout the electromagnetic spectrum.

13. Well, the equations to directly calculate the mass of photons with different
wavelengths are:
Mf = (h/λ) x C / C^2 = (hC / λ) / C^2
These two equations give exactly the same results. I used this method before to find the
mass of the photon, and it is
E = hv, and then find the mass of the photon M = hv / C^2, the same result remains. I
will give a practical example just knowing the wavelength of the photon.
A photon is observed in the laboratory and measured, giving a wavelength of 400 nm.
Calculate the mass of this photon.
Mf = (6,63 x 10^ - 34 J.s x 4.0 x 10^ - 7 m.) x 3.0 x 10^ 8 m/s /
(3 x 10^ 8 m/s)^2 =
Mf = 4, 9725 x 10^ - 19 Joule /(3 x 10^ 8 m/s) = 5,525 x 10^ - 36 Kg. Where we see
the mass of this photon at this wavelength.
Well now we will try the second equation
Mf = 6,63 x 10^ - 34 J.s X 3 x 10^ 8 m/s / (3 x 10^ 8 m/s)^2 =
Mf = 4,9725 x 10^ -19 Joule / (3 x 10^ 8 m/s)^2 = 5,525 x 10^ - 36 Kg.
If we look closely, the mass of photons depends absolutely on their frequency and
wavelength. This equation is very accurate for measuring the masses of different
photons with different wavelengths.

14. THESE EQUATIONS ARE RELATIVIST AND PREDICT THAT PHOTONS OF
DIFFERENT WAVELENGTHS HAVE MASS AND THAT THOSE MASSES AT
CERTAIN DOMAINS OF WAVELENGTHS NEVER VARY.
EINSTEIN WAS COMPLETELY RIGHT WITH HIS FAMOUS EQUATION
Einstein expressed energy in a way that assumed that nothing in the universe can fall
to absolute zero, nothing is absolute, least of all the masses of photons! Einstein
expressed the energy E = mc^2 for photons and particles traveling at the speed of light.
For particles traveling at a speed less than or relative to light E = mv^2
Therefore we can now with great confidence relate the following equations
E = h/λ x c = hc / λ = hv = mc
2
and we can associate it with this equation for
particles with charge E = e∆V. The mass of the photons is related to these equations:
Mf = (h/λ) x c / c^2 = (hc / λ) / c^2 = hv / c^2 = 2(mo λ`) / (Cos θ)^2 =
We see that Einstein was absolutely right E = mc
2
By. Wilson Vitelio Hidalgo Suarez
Author.