# 11C-Carfentanil dosimetry

1. Jun 11, 2015

### Dauthi

1. The problem statement, all variables and given/known data
Hello everyone, I'm designing a PET study with 11C-Carfentanil and my task is to discover the mean effective dose absorbed in a patient, when given 250 MBq of 11C-Carfentanil (about 0,22 µg).

2. The attempt at a solution
My own attempt was done with high school knowledge, so the result is a gross overestimation of the total radiation dose. I first calculated the amount of radioactive nuclei in a 250 MBq dose:
N = α / λ = 250 MBq / 5,66852 x 10-4 = 4,410 x 1011 nuclei. (λ was calculated from ln2/T½, and T½ for 11C was 20,38 minutes).

Next I reasoned that when 1x 11C nuclei degrades via β+, it yields 2 gamma quantums, each having 511 KeV energy. So the total amount of radiation energy produced would be:
E = N x 2 x E(gamma quantums) x 1,602 x 10-19 eV = 4,410 x 1011 x 2 x 511 KeV x 1,602 x 10-19 eV = 0,07220J.

Thus the equivalent dose would be H = Wr x Dr, where Dr = E/m (m being the approximate mass of the patient, 70 kg):
H = 1 x 0,07220J / 70 kg = 1,03 mSv.

So what I need to know here is how to do I make my calculations more accurate. To my information I need to know the following:
- How does the kinetic energy of the 11C divide between the daughter nuclei: 11B, positron and neutrino.
- To my knowledge, 11B does not degrade radioactively, so do I need to include it in any way in my calculations?

Thank you very much in advance, whoever replies :)

2. Jun 11, 2015

### DEvens

http://en.wikipedia.org/wiki/Isotopes_of_boron#Boron-11

B11 is stable. So you can ignore that.

http://en.wikipedia.org/wiki/Isotopes_of_carbon#Carbon-11

Two decay modes for C11. One produces a positron, one absorbs an electron. But the electron capture is a small fraction of the total so can probably be ignored.

The positron decay releases 0.96 MeV. Some of this will go with the neutrino. Only a very small amount of the energy goes into the nucleus produced. Remember the relative masses, and remember that the nucleus only gets recoil momentum. You can probably Google up the spectrum. Probably there is a rule-of-thumb for such cases that gets close enough. Maybe your health physics text will tell you?

Then the positron finds an electron PDQ, and produces the two gammas you included.

You say you got a "gross over estimation." How do you know that?

Your dose seems to be simply the total energy deposited. You should be including some kind of estimate of the dose produced by a specific energy gamma. This includes the fact that higher energy gammas will penetrate farther.

3. Jun 11, 2015

### Dauthi

Hello, and thank you for your input. I was told previously, when I returned my answer, that "Well done, except that it's a gross overestimation of the radiation, because it expects that every gamma quantum is absorbed by the tissue, which is not the case, because we can detect the gamma quantums with PET detectors".

So I can just ignore the fact that 11B has 0,00009 MeV and the positron 0,959907 MeV of kinetic energy? Or how do I incorporate them to the total amount of radiation? I was given these numbers by a former physicist, who did an iteration of the particle energies, but is stuck pretty much with the same question as I am: "Why do we need this?"

Last edited: Jun 11, 2015
4. Jun 11, 2015

### DEvens

Yes, you can ignore the tiny fraction of energy the recoiling B nucleus gets. The chances are the rest of your calculation won't be more than 2 digits accurate. So the tiny fraction that goes to the B nucleus will be less than the round-off on the remainder of your calculation.

No you cannot ignore the energy the positron starts with, the 0.96 MeV. That will be included in the gamma energy produced by the positron. But not all of that 0.96 MeV goes with the positron. Some goes with the positron and some with the neutrino. And of course that tiny bit with the B nucleus. That is what I was referring to when I talked about the spectrum. You need to know how much goes each way.

When I talked about dose produced by a specific energy gamma, and penetration, I was referring to the point you got in the response to your work. Higher energy gammas tend to go farther through matter. So higher energy gammas will do damage along the way, but not leave all their energy behind. The higher the energy the more the penetration. When you have 0.511 MeV gammas you probably see a large fraction of the energy get out, as indicated in the evaluation of your work.

To make your answer more accurate you need to look up the damage produced by a given gamma as a function of energy. If your health physics text does not have some indication of this, then you might find it by Google. But you likely will find, at best, a "rule of thumb" to convert exposure to dose. The wiki page only talks about the conversion of absorbed energy to dose. It does not talk about converting exposure to absorbed energy.

http://en.wikipedia.org/wiki/Equivalent_dose

To get really accurate you would have to model the location of the radiation source in the body, then the path of the gammas after they were emitted. You could use that to calculate what fraction were absorbed and what fraction escaped. That is a major task. You might find somebody has done exactly this and published it in a health physics text. Such tracer isotopes as you are talking about here are fairly standard diagnostic tools these days. So do some Google work and see what you can find.