Re: Ambient dose equivalent H(10) (ICRU 39)

["Robert Gunter" <rjgunter@icnpharm.com>]

Try the mass _absorption_ coefficients.  You are really interested in the dose
deposited, not the relative transmission of uncollided photons.  You will note
that the absorption is smaller.  Check out Cember, there is a graph in Ch 5 or 6
and a quick explanation of the conversion in the section on converting exposure
to dose.

Rob


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Robert J. Gunter         Tel: (714) 545-0100

Sr. Technical Specialist      Tel: (800) 548-5100 Ext. 2414

ICN Biomedicals, Inc.         Fax: (714) 668-3149

Dosimetry Division       Email: rjgunter@icnpharm.com

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Michael McNaughton <mcnaught@lanl.gov> on 06/15/99 09:35:40 AM

Please respond to radsafe@romulus.ehs.uiuc.edu

To:   Multiple recipients of list <radsafe@romulus.ehs.uiuc.edu>
cc:    (bcc: Robert Gunter/HQ/ICN)
Subject:  Ambient dose equivalent H(10) (ICRU 39)




Dear RadSafers

Could anyone help me to understand the physical basis for Figure 3 in ICRU39.
This is the conversion of exposure (roentgen) to ambient dose equivalent
(cSv).
Why is it not equal to the ratio of the mass attenuation coefficients for
tissue and air? For example, near 80 keV, Figure 3 shows about 1.7, whereas
the ratio of the mass attenuation coefficients is about 1.1. What is the
physical basis used to obtain Figure 3?

Thank you, mike

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