Re: Once in a lifetime opportunity

[LUDWIK KOWALSKI <KOWALSKIL@alpha.montclair.edu>]

> Date: Wed, 03 Dec 1997 11:23:32 -0800
> From: Leigh Palmer <palmer@sfu.ca>
> Subject: Re: A once in a lifetime opportinity
>
> Since the source is localized to the heart, the dose is, too. Having
> said that I decline to do the calculation. 

Ask people in the hospital, they should know what a typical total "whole
body" dose is. I would really like to know how it compares, for example,
with 5 rads per year which is a maximum permissible dose for anybody
working in a nuclear plant. Or with 500 rads which kills about 50% of
rats, or so,

The upper limit is easy to calculate but this can differ from the real
dose by a large factor. We know that 20 mCi corresponds to 2*10^13 Tc99m 
atoms. Each of them emits one 140 KeV photon --> total energy of 
2*10^13*0.14=2.8*10^12 MeV=0.45 J. One rad, as I remember, is close to 
100 ergs/gram or 0.01 J/kg. Suppose the upper 1/2 of the body (30 kg) 
absorbs all that energy. 0.45/30 --> 0.015 J/kg or 1.5 rads. 

My guess is that the real dose, for the 140 KeV photons, is betwee 1/5 
and 1/10  of the maximum dose, perhaps 0.3 rads. How close is this
guess from what we will hopefuly find from medical people?

After all many photons escape without interacting and those which do
interact with body atoms produce secondary photons (Compton effect). 
Many secondary photons also escape. And metabolic elimination during 
the first day also helps to reduced the dose (not much, according to
Leigh). The distribution of the dose must of course be non-uniform
within the body.
                                              Ludwik Kowalski
P.S.
Brian, by decay rate I ment the number of desintegrations/second, not 
the percentage value you used in the comment. My prediction is much
less than your "0.8 rads/hr, initially".