Average Dose Rate for 1.11x108 Bq of 131I Injection

AI Thread Summary
The discussion focuses on calculating the average dose rate for a patient injected with 1.11×10^8 Bq of 131I, with 30% uptake by a 20 g thyroid. The calculated dose rate is presented as 51 μGy s–1, derived from specific activity and energy values. The formula used includes parameters such as the activity concentration and energy emitted. Participants are asked to identify any errors in the calculation. The conversation emphasizes the importance of accuracy in radiological dose assessments.
xamy
Messages
26
Reaction score
0
A patient receives an injection of 1.11×108 Bq of 131I,30% of which goes to the thyroid, having a mass of 20 g.What is the average dose rate in the organ?

Solution:
D=1.60×10–10AE Gy s–1

A=(1.11x108x(30/100))/20=1.665x106Bqg-1

E=0.192Mev

=1.60x10-10 (1.665)(0.192Mev)

= 51μGy s–1

This is my solution. Can anyone point out the mistake here.
 
Physics news on Phys.org
This isn't nuclear engineering. I have moved your thread.
 
Thread 'Confusion regarding a chemical kinetics problem'
TL;DR Summary: cannot find out error in solution proposed. [![question with rate laws][1]][1] Now the rate law for the reaction (i.e reaction rate) can be written as: $$ R= k[N_2O_5] $$ my main question is, WHAT is this reaction equal to? what I mean here is, whether $$k[N_2O_5]= -d[N_2O_5]/dt$$ or is it $$k[N_2O_5]= -1/2 \frac{d}{dt} [N_2O_5] $$ ? The latter seems to be more apt, as the reaction rate must be -1/2 (disappearance rate of N2O5), which adheres to the stoichiometry of the...
Back
Top