Scoring Dose Equivalent at 100 cm form the siled surface

Scoring
1

Dear Experts,

With reference to my previous post: Shift in the maximum dose rate point, I extend my ques.:

I have scored Dose Equivalent at 100 cm from the surface of the shield with an isotropic source at centre - please find attached the input and flair files.

I am getting different outputs for different volume of the Cartesian bins used (Cartesian binning in USRBIN) - which seems logical as it is averaged over the volume of the bin - but the question is - how to decide the correct values ? The statistical errors in all the cases (scores with different volume bins) are less than 5 % - so all are acceptable. Is there any other check to decide the correctness of the output - that which result is more reliable out of many ? Please find attached the .lis files for the scores obtained at 100 cm Left and 100 am Right from the shield.

I also found that near surface the outputs are obtained for 8 cc and 16 cc bin volumes with acceptable statistical error, but at 100 cm the volume of scoring bin has to be bigger, say upto 96cc /128 cc / 400 cc to obtain the expected output. Does the isotropic nature of the source has a role to play here ? Scattering is air outside the shield ?

Please explain.

Thanks and regards,
Raksha.
LHS_1m_27.bnn.lis (2.2 KB)
RHS_1m__26.bnn.lis (2.2 KB)
test1.flair (11.9 KB)
test1.inp (13.9 KB)

3

Dear Raksha,

I see from the input and flair files you provided that some of the scoring meshes for dose equivalent don’t include a lot of bins. For example the USRBIN “DELHS5cm” only contains one bin along the X and Y directions. A useful check to determine the correctness of the output is for example to include more bins in the X and Y directions and check if the distribution of values you obtain along each of the axes is smooth, i.e. no large fluctuations in values and acceptable errors. In the same way you could also define a higher amount of bins in the same scoring volume, at the expense of required computation time to achieve good statistics as you already correctly noted.

When moving further away from the shielding it indeed makes sense that you require larger bins to obtain the same statistical error since the fluence drops as 1/distance^2 for an isotropic source. In your case with the shielding geometry and surrounding air this still holds: moving further away requires more statistics.

Hope this helps,

Andreas