From key@physics.utoronto.ca Wed Apr  2 19:33:33 2008
Date: Wed, 02 Apr 2008 19:33:30 -0400
From: Tony Key <key@physics.utoronto.ca>
To: xiaoou.he@utoronto.ca
Subject: Re: Phy138 Question

Hi Xiaou - no, this is just a  matter of language.  It just means that of
all the radioisotope administered, only 70% actually reaches the liver.
Presumably the rest  gets distributed around the body.  But your math is
correct - multiply the 20mCi by 0.7 to get the starting activity in the
liver.

TonyK

xiaoou.he@utoronto.ca wrote:
> Dear Professor.
>
> I have a quick question regarding one of the suggested problems.It is #1
> of SNV
>
> Committed Dose from a Tracer. 20 mCi of 99mTc is administered to a
> patient. The uptake in the liver, which weighs 1.8 kg, is known to be 70%.
> The average energy of the gamma rays from the decay of the 99mTc is 140
> keV and the nuclear half life of 99mTc is 6.0 hours. The biological half
> life in the liver is 6.0 days, and about 25% of the gamma ray energy is
> deposited in the liver. A) Calculate the absorbed dose to the liver in
> mGy.
>
> Is the "uptake" essentially the same thing as an isotopic dilution?,
> except dealing with masses instead. Meaning you multiply 0.7 by 20mCi to
> find out your activity at time 0, for the liver?
>
> Thanks
>