|Message: Re: Discrepancy between definition of materials using effective atomic number and true atomic number||Not Logged In (login)|
Click on the Forum title, e.g. on the "Forums by Category" page, to read a sequence of postings to the Forum and its threads all in one page. If you are only interested in one thread or the thread following a specific posting, click the thread or the posting, which takes you to a smaller page, which contains only the part you are interested in and may be easier to navigate.
Messages are "chained" if there are only replies at the first level, i.e. 1/1.html, 1/1/1.html etc. In case of "chained" messages the message number is replaced by the icon and there is no indentation.
Inline: Display the subject line only or also the text of the posting(s); for the choice "All" the "Outline" choices are switched off.
|1||0||1||no text / full text of posting|
|2||1||All||text for level 1 only / text for All postings|
Outline: Choose the depth of the posting thread, successive toggle controls provide increasing detail.
|1||2||1||2 levels / 1 level (original posting)|
|2||3||2||3 levels / 2 levels|
|3||3||All||3 levels / all levels (all postings)|
On Fri, 01 Oct 2010 15:38:48 GMT, David Oxley wrote:
> Dear all, > > I am trying to define materials in G4 using their effective atomic > number. Later I also want to look at effective electron density, but > that will not be the subject of this post. > > This is because I want to know how significantly an error in the > effective atomic number effects the range of protons. So I want to > define the material in two different ways, add an uncertainty to one and > observe the difference in the Bragg peak. > > As a starting point, I have defined water using both means and hoped to > get the same answer, but I dont.
There is no theoritical foundation to choose Zeff = Zmean.
1- dE/dx is proportional to electron density. This leads to the definition of (Z/A)eff = w1 Z1/A1 + w2 Z2/A2 Then one can choose Zeff and Aeff according to this ratio
2- if one choose Aeff = Amol, then one find Zeff = Zmol (=10)
3- Ionisation cross section is proportional to atom density (not electron). This gives a definition of Aeff : 1/Aeff = w1/A1 + w2/A2 (in case of water one find Aeff = 18/3). Hence Zeff = 10/3
I let you play with TestEm0 and TestEm7 (in examples/extended/electromagnetic) to see the effects of these various choices.