Re: Positrons

Paul.B.Andersen:> Den 06.07.2025 06:27, skrev Bertietaylor:

>
In the case of cancer diagnosis a drug containing Fluorodeoxyglucose
is injected into the bloodstream of the patience.
>
Fluorodeoxyglucose contains F-18, which decay via β+ decay.
>
F-18 with 9 protons and 9 neutrons →
O-18 with 8 protons and 8 neutrons + positron + neutrino
>
When the ejected positron hits an electron somewhere in the body,
they annihilate and two gamma particles (photons) are created.
  e⁺ + e⁻  →  γ + γ
Energy and momentum are preserved.
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)
(Mass is converted to pure kinetic energy of photons)
>
These gamma particles are detected by two detectors so a 3D image
is created.

I like to think that the e⁺ and  e⁻  don't actually annihilate, but
instead form a tight ball. Since this ball has no charge, it is for all
practical purposes undetectable. But then, the evidence that e⁺ and  e⁻
completely annihilate into 2 γ photons is compelling, since after all:

m_{e⁻}c^2 == 9.1094e-31 * 299792458^2 / 1.6022e-19 == 511 keV

A reasonable argument against this is that m_{e⁻} measurement is
controversial, and gamma ray detectors are calibrated assuming that the
e⁺ e⁻ annihilation radiation is 511 keV.

https://en.wikipedia.org/wiki/Annihilation_radiation
"
Because of their well-defined energy (511 keV) and characteristic,
Doppler-broadened shape, annihilation radiation can often be useful in
defining the energy calibration of a gamma ray spectrum.
"

https://upload.wikimedia.org/wikipedia/commons/7/78/Annihilation_Radiation.JPG

An independent measurement of the frequency/wavelength/wavenumber would
be more convincing, though I don't know how this can be done for
annihilation gamma radiation.