4 ms·

Like UV light, alpha radiation is easily shielded by most kinds of material. Even after skimming the linked publication, it does not seem clear to me how the al

Like UV light, alpha radiation is easily shielded by most kinds of material. Even after skimming the linked publication, it does not seem clear to me how the alpha radiation (and UV) would reach the devices sensor. Probably something behind "... and measure alpha and beta radiation from contaminated surfaces".

thyristan 1y ago
"Shielding" is a relatively crude term. Alpha radiation is "shielded" by a sheet of paper in the sense that the attenuation is so high, and the attenuation length is so short when compared to the thickness of that paper, that less than 1/1000th of the original particles arrive on the other side of the paper. However, this means that there are still very few but not zero alpha particles arriving on the other side. Plus, the energy of those alpha particles doesn't just go away, it will be either transformed into highly energetic photons (in the hard UV to gamma range) or it will ionize the attenuating medium, leading to photons from line emissions (infrared to visible to UV) or other ionization-related effects (e.g. an ionisation current in semiconductor diodes). All those things can be detected, and you can calibrate your detector for the relative sizes of those effects you are seeing when compared to the incident alpha rays.
- mppm 1y ago
  This is somewhat inaccurate from a physical perspective. What you say would be applicable to the attenuation of gamma rays, which is governed by low-probability interactions and is therefore exponential. But alpha particles, being charged an heavy, lose energy continuously through electron interactions and have a relatively fixed range in matter, beyond which the incidence is practically zero.
  - thyristan 1y ago
    No, that is just plain wrong. All attenuation is exponential (except very special stuff like energy resonances at certain particle energies). You just get very different exponents for rarely vs. frequently interacting particles. The charge of an alpha particle is also only twice the charge of a beta particle with a far higher (orders of magnitude, not just a factor of 2) penetration depth. The relevant difference is actually the mass and size (i.e. cross-section) of an alpha particle that makes it far more frequently interacting. There are some more differences that govern the cross-section like conservation of momentum and angular momentum in certain interactions where there are differences between gamma and alpha particles. Or magic number nuclei that are relevant for neutron absorption. But in general, for a given material and particle, attenuation is always always always exponential over the penetration depth vs number of particles, until you hit a low enough energy that some other mechanism for energy transfer can come into play or become unavailable, where the exponent will change and you will see a change in slope.
    - mppm 1y ago
      Please take a look at slide 11 of this presentation [1]. Stopping power and Bragg peak (slide 30) are also relevant concepts. What you are describing applies to photons and neutrons, but not to charged particles. 1. https://indico.cern.ch/event/975141/contributions/4137563/attachments/2156087/3646193/2020-Lecture-3-Interactions%20of%20Particles%20with%20Matter.pdf https://indico.cern.ch/event/975141/contributions/4137563/at...
      - thyristan 1y ago
        Well then, please read what those slides say. Then read what I wrote. The Bragg peak is about energy deposition, not the number of particles arriving at some point. When the energy of a particle changes going through the medium, the absorption changes. Still an exponential curve, just piece-wise exponential. Look at slide 11, that cutoff. Look at the logistic formula for it and its limit going right: still an exponential.
- theglocksaint 1y ago
  Shielding is not a crude term, it is indeed used as technical language in the world of radiation safety. You are also using the term attenuation incorrectly, as sibling comment points out.
  - thyristan 1y ago
    Shielding in radiation safety means attenuation to a safe level. But for detection, that is irrelevant because radiation detectors can be very sensitive, such that even very minute quantities making it through a "shielding" are still sufficient. When in doubt, you just increase integration time to almost forever and you will still get a sufficient number of particles to get a measurement.
- tzs 1y ago
  > Plus, the energy of those alpha particles doesn't just go away, it will be either transformed into highly energetic photons (in the hard UV to gamma range) or it will ionize the attenuating medium, leading to photons from line emissions (infrared to visible to UV) or other ionization-related effects (e.g. an ionisation current in semiconductor diodes). Why can't it end up as thermal energy in the attenuating medium instead?
  - thyristan 1y ago
    Because thermal energies are far lower than what those particles are depositing. The thermal movement at those energies would still be ionizing because it would some nucleus away leaving a few stray electrons in the lattice. But of course, sooner or later the secondary, tertiary, ... particles will end up as thermal energy. And of course any ionization always has a thermal component, the lattice will always get a transfer of momentum.
jjtheblunt 1y ago
maybe it infers alpha radiation from detection of decay chain particles / photons at energies indicative of alpha presence.

thyristan 1y ago

"Shielding" is a relatively crude term. Alpha radiation is "shielded" by a sheet of paper in the sense that the attenuation is so high, and the attenuation length is so short when compared to the thickness of that paper, that less than 1/1000th of the original particles arrive on the other side of the paper. However, this means that there are still very few but not zero alpha particles arriving on the other side. Plus, the energy of those alpha particles doesn't just go away, it will be either transformed into highly energetic photons (in the hard UV to gamma range) or it will ionize the attenuating medium, leading to photons from line emissions (infrared to visible to UV) or other ionization-related effects (e.g. an ionisation current in semiconductor diodes). All those things can be detected, and you can calibrate your detector for the relative sizes of those effects you are seeing when compared to the incident alpha rays.

mppm 1y ago
This is somewhat inaccurate from a physical perspective. What you say would be applicable to the attenuation of gamma rays, which is governed by low-probability interactions and is therefore exponential. But alpha particles, being charged an heavy, lose energy continuously through electron interactions and have a relatively fixed range in matter, beyond which the incidence is practically zero.
- thyristan 1y ago
  No, that is just plain wrong. All attenuation is exponential (except very special stuff like energy resonances at certain particle energies). You just get very different exponents for rarely vs. frequently interacting particles. The charge of an alpha particle is also only twice the charge of a beta particle with a far higher (orders of magnitude, not just a factor of 2) penetration depth. The relevant difference is actually the mass and size (i.e. cross-section) of an alpha particle that makes it far more frequently interacting. There are some more differences that govern the cross-section like conservation of momentum and angular momentum in certain interactions where there are differences between gamma and alpha particles. Or magic number nuclei that are relevant for neutron absorption. But in general, for a given material and particle, attenuation is always always always exponential over the penetration depth vs number of particles, until you hit a low enough energy that some other mechanism for energy transfer can come into play or become unavailable, where the exponent will change and you will see a change in slope.
  - mppm 1y ago
    Please take a look at slide 11 of this presentation [1]. Stopping power and Bragg peak (slide 30) are also relevant concepts. What you are describing applies to photons and neutrons, but not to charged particles. 1. https://indico.cern.ch/event/975141/contributions/4137563/attachments/2156087/3646193/2020-Lecture-3-Interactions%20of%20Particles%20with%20Matter.pdf https://indico.cern.ch/event/975141/contributions/4137563/at...
    - thyristan 1y ago
      Well then, please read what those slides say. Then read what I wrote. The Bragg peak is about energy deposition, not the number of particles arriving at some point. When the energy of a particle changes going through the medium, the absorption changes. Still an exponential curve, just piece-wise exponential. Look at slide 11, that cutoff. Look at the logistic formula for it and its limit going right: still an exponential.
theglocksaint 1y ago
Shielding is not a crude term, it is indeed used as technical language in the world of radiation safety. You are also using the term attenuation incorrectly, as sibling comment points out.
- thyristan 1y ago
  Shielding in radiation safety means attenuation to a safe level. But for detection, that is irrelevant because radiation detectors can be very sensitive, such that even very minute quantities making it through a "shielding" are still sufficient. When in doubt, you just increase integration time to almost forever and you will still get a sufficient number of particles to get a measurement.
tzs 1y ago
> Plus, the energy of those alpha particles doesn't just go away, it will be either transformed into highly energetic photons (in the hard UV to gamma range) or it will ionize the attenuating medium, leading to photons from line emissions (infrared to visible to UV) or other ionization-related effects (e.g. an ionisation current in semiconductor diodes). Why can't it end up as thermal energy in the attenuating medium instead?
- thyristan 1y ago
  Because thermal energies are far lower than what those particles are depositing. The thermal movement at those energies would still be ionizing because it would some nucleus away leaving a few stray electrons in the lattice. But of course, sooner or later the secondary, tertiary, ... particles will end up as thermal energy. And of course any ionization always has a thermal component, the lattice will always get a transfer of momentum.

jjtheblunt 1y ago

maybe it infers alpha radiation from detection of decay chain particles / photons at energies indicative of alpha presence.