0
$\begingroup$

I'm reading about measuring radiation and trying to come up with a more intuitive retelling for an audience familiar with math but not necessarily chemistry or any physical science. Is any or all of the following inaccurate or oversimplified?

  • Radioactivity (Ci, Bq) - Radiation from the material's point of view. Measured as a rate of disintegration.

  • Exposure (R, C/kg) - Radiation from the air's point of view. Measured as a sum of radiation that's passed through.

  • Absorbed dose (rad, Gy) - Radiation from the absorbing target's point of view. Measured as a sum of energy deposited.

  • Dose equivalent (rem, Sv) - Radiation also in terms of sum energy-absorption, but adjusted for a medical point of view.

As you can tell, the 2 pieces I want to highlight are (1) where-it's-measured and (2) time dimensionality (rate versus summable quantity).

My objective question is, where have I got it wrong? What has potential to cause confusion?

But of course, interested in complete reformulations.

$\endgroup$
1
  • 1
    $\begingroup$ Exposure rather as Measured as an air ionization effect of passed radiation. Dose equivalent as adjusted for biological sensitivity $\endgroup$
    – Poutnik
    Nov 16, 2020 at 8:23

1 Answer 1

1
$\begingroup$

Maybe some numerical values may help.

  1. Radioactivity. The activity is the number of disintegrations occurring per second in a given source. The unit is the Becquerel : $1 Bq = 1 s^{-1}$. For example, the experience shows that $3.7·10^7$ nucleus are decomposed per second in $1$ g $\ce{Ra-226}$. The activity of $1$ g $\ce{Ra-226}$ is $3.7·10^7 Bq$ which was previously called $1$ Curie ($1$ $Ci$). One Curie is a huge activity for usual measurements in the laboratory.

  2. Exposure dose. If a gamma-ray radioactive source is in air, the gamma rays ionize the air around. As a consequence, they deposited some energy in a small volume of air situated at a distance from the source. If the small air sample is far away from the source, the amount of energy is smaller than if it is near the source. The amount of energy so absorbed by the air sample is called the dose (or exposure dose) and is determined with dosimeters, and it is measured in $J/kg$ air. The unit of dose is röntgen ($1~r$), equal to $0.01 J/kg$.

The experience shows that $1 Ci$ of a gamma-ray emitter delivers a dose of $1$ röntgen per hour ($1 r/h$) in air at $1 m$ distance from the source. This value is somewhat dependent on the nature of the radioactive source. Let's forget about this difficulty for the present purpose. Most important, the dose decreases with the square of the distance. So, to be consistent with the units, the dose delivered by a gamma-ray source is $1 r·m^2 Ci^{-1} h^{-1}$. A small volume of air (some milliliters) situated at $1 m$ distance from a $1 Ci$ gamma source, absorbs in one day ($24 h$) a dose equal to $(1~r·m^2·Ci^{-1} h^{-1})·1 Ci·24 h/(1 m)^2 = 24~ r = 0.24 J/kg $. This energy is finally transformed into heat.

  1. If the irradiated matter is not air, but another substance, like water, the amount of energy is called absorbed dose and is expressed in Gray, abbr. $Gy$ ($1 Gy = 1 J/kg$). The experience shows that an exposure dose of $1$ röntgen delivers an absorbed dose of $0.01 Gy$ in water, like in air.

  2. Living tissues are mostly made of water. But they don't react the same way to a given absorbed dose. Some molecules like ADN are much more sensitive to a given adsorbed dose than water. To compare the biological effects of a given dose expressed in Gray, it is necessary to define a new parameter, the effective dose, which is expressed in Sievert, abbr. $Sv$. $1 Sv = 1 Gy$ in water. But $1 Sv$ = up to $20~Gy$ in sensitive tissues.

The present explanation is just a first approach of the domain of dosimetry and radioprotection. As quickly mentioned in the text, the energy deposited in a target depends also on the energy of the gamma ray, which was forgot here. And we have not spoken about alpha and beta rays, whose dosimetry is not similar to gamma rays. Dealing with these problems should take hours, and cannot be done in the present message.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.