Re: Positrons

Den 09.07.2025 21:53, skrev Stefan Ram:

ram@zedat.fu-berlin.de (Stefan Ram) wrote or quoted:

To get into this mindset just think about how the mass of an atom
of hydrogen is smaller than the mass of the proton and electron
taken in isolation. This would be another example where the
mass of a combined system differs from the sum of the masses of its
components.

Yes of course.
An atom is a closed system. Its mass includes the mass of its constituents and the binding energy between its constituents.
You can measure the mass of an atom.
The same goes for all closed systems.
The mass of the Earth includes the binding energy.
The gravitational acceleration on the Earth Moon system
is as the mass includes the mass of the Earth + mass of Moon
+ kinetic energy due to rotation and orbiting.
In H the energy of the electron in ground state is -13.6 eV.
That means that the binding energy, the energy to split
the electron from the proton is 13.6 eV = 1.0735e-9 u
Proton   mass   = 1.00727647 u
electron mass   = 0.00054858 u
binding energy  = 1.0735e-9  u = 13.6 eV
Mass of H atom: = 1.007825051 u
The binding energy is negligible in a H-atom.
In the Deuterium nucleus it is much higher:
   Proton  mass   1.007276 u
+ Neutron mass   1.008666 u
- Atomic weight  2.014102 u
------------------------------
= binding energy 0.001840 u  = 1.714 MeV
The binding energy between the proton and neutron
is part of the mass of the closed system Deuterium nucleus.
-----------------
But all this is irrelevant to the issue of this thread,
which is the e⁺ e⁻ annihilation that occurs in a
Positron Emission Tomography scanner.
The only sensible frame of reference is the rest frame of
the gamma photon detectors and the patient's body.
In the patient a F-18 nucleus is decaying and a positron is
ejected at high speed. It will hit an electron in some atom.
It is nonsensical to insist that these two leptons is a system
with mass > than the mass of two leptons.
The mass before the annihilation is 2m where m is the mass of a lepton.
However, the kinetic energy of the positron will most probable be
higher than the kinetic energy of the electron.
The energy released in the annihilation is the energy equivalent
of 2m + the kinetic energy of the positron and electron.
Since the latter is different for the two leptons, the energy
of the two gamma photons will be slightly different.
(conservation of momentum)
It is nonsensical to insist that these two photons constitutes
a system with mass.
After the annihilation we have two mass-less photons, zero mass.
Since the kinetic energy of the leptons will be small
compared to the energy equivalent of their masses,
an approximation which is good enough for our purpose
is to ignore it; the exact energy of the gamma photons
is not critical and is not detected.
The energy equivalence of an electron and a positron is 0.511 MeV,
so the energy of each of the gamma particles is ~0.511 MeV. (0.024 Å)
The whole point with the PET is to observe the gamma photons
individually. There are two detectors which make it possible
to "see" the point in the body where the annihilation occurred.
A 3D picture is created.  Because cancer cells have a higher
metabolic rate than do typical cells, they will appear as bright
spots in the image.

    The binding energy of hydrogen is "-13.6 eV".
    It is the sum of the potential energy and the kinetic energy
   of the electron and is part of the mass energy of hydrogen.
    I take this calculation from the World-Wide Web, but might
   have added typos when converting it to plain text:
 At the mean radius a_B = 0.528 × 10^{−10} m, the potential energy
of the pair is:
          1    e^2
U = − -----  --- = −4.36 × 10^{−18} J = −27.2 eV.
       4πϵ_o  a_B
 The kinetic energy is half this value, and agrees with the classical
result for a particle in uniform circular motion:
 mv^2     1     e^2     1          1    e^2
---- = -----  ----- => - mv^2 = -----  ---.
a_B    4πϵ_o  a_B^2    2        8πϵ_o  a_B
  So the binding energy is: E = K + U = −13.6 eV.
 

--
Paul
https://paulba.no/