Radioactive Decay - working out activity

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Homework Help Overview

The discussion revolves around calculating the activity of Thorium-234, given an initial mass and its half-life. The original poster attempts to determine both the initial activity and the activity after two weeks, using relevant equations related to radioactive decay.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the calculations of decay constants and initial activity, with some questioning the validity of the results based on their understanding of radioactivity units. There is a focus on the relationship between becquerels (Bq) and curies (Cu), and whether the original poster's results are reasonable.

Discussion Status

There is ongoing exploration of the calculations presented, with participants providing insights into the units of measurement and their implications. Some guidance has been offered regarding the relationship between Bq and Cu, but no consensus has been reached on the correctness of the original poster's answers.

Contextual Notes

Participants note potential confusion regarding the units of radioactivity, specifically the differences between becquerels and curies, which may affect the interpretation of the calculated activity values. The original poster's use of large quantities of Thorium-234 is also questioned.

FaraDazed
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Homework Statement


If you have 500g of Thorium-234 at t=0. Calculate the initial activity and the activity after 2 weeks. The half life of thorium-234 is 24 days.

Homework Equations


A=A_0e^{-\lambda t} \\<br /> A_0=\lambda N_0 \\<br /> \lambda=\frac{ln2}{t_{\frac{1}{2}}}

The Attempt at a Solution


For the intial activity I first found lambda by converting 24 days into seconds.

<br /> \lambda=\frac{ln2}{60 \times 60 \times 24 \times 24}=\frac{ln2}{2073600}=3.34 \times 10^{-7} \\

Then used the second relevant equation listed above by calculating N_0 with the known data (converting to number of moles and then times by Avogadros number).
<br /> A_0=\lambda N_0=(3.34 \times 10^{-7})(\frac{500}{234})(6.02 \times 10^{23})=4.29 \times 10^{17} Bq<br />

Then after knowing A_0 i can calculate the activity after 2 weeks, with the first relevant equation listed but converting 2 weeks into seconds.
<br /> A=A_0e^{-\lambda t}=(4.29 \times 10^{17})e^{(3.34 \times 10^{-7})(1209600)}=2.86 \times 10^{17} Bq<br />

My problem is with my answers, they seem way way too high. I was told in a lecture it is very unusual to ever go above a micro Bq.
 
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Haven't checked the math, but I think you have it the other way around, a Bq is a very small unit of radioactivity. It's true that background counts should be less than a Bq, but when you're talking about radioisotopes, you're definitely talking about much more than a Bq.
 
Curious3141 said:
Haven't checked the math, but I think you have it the other way around, a Bq is a very small unit of radioactivity. It's true that background counts should be less than a Bq, but when you're talking about radioisotopes, you're definitely talking about much more than a Bq.

Right, that could be it. He was talking about a unit called the curie too at the same time but didnt quite understand what he said as he went over it very quickly but from what he said I got the impression that 1Bq=1Cu but that they usually say Bq these days.

It may be that they are not equal and that he meant that its unusual to ever get a reading of over 1Cu and not 1Bq.
 
FaraDazed said:
Right, that could be it. He was talking about a unit called the curie too at the same time but didnt quite understand what he said as he went over it very quickly but from what he said I got the impression that 1Bq=1Cu but that they usually say Bq these days.

It may be that they are not equal and that he meant that its unusual to ever get a reading of over 1Cu and not 1Bq.

A Cu is a MUCH bigger unit than a Bq.

27pCu (that's picoCurie)= 1Bq
 
Curious3141 said:
A Cu is a MUCH bigger unit than a Bq.

27pCu (that's picoCurie)= 1Bq

That makes sense then and why my answers are so high, I was getting Bq and Cu mixed up. Do you think my answers are correct?
 
yes, they sound ok ... if you would ever have 500 grams of 234Th!
(why do you use "Bq" if it is confusing, rather than the straightforward "decay/sec")
 

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