Radioactive Decay - working out activity

AI Thread Summary
The discussion revolves around calculating the initial activity of 500g of Thorium-234 and its activity after two weeks, using the half-life of 24 days. The initial activity was calculated to be approximately 4.29 x 10^17 Bq, and the activity after two weeks was found to be around 2.86 x 10^17 Bq. Participants expressed confusion regarding the units of measurement, particularly between becquerels (Bq) and curies (Cu), clarifying that a curie is significantly larger than a becquerel. The consensus is that the calculated values are plausible, but the large mass of Thorium-234 raises questions about practical scenarios. The conversation highlights the importance of understanding radioactive decay units in the context of radioactivity measurements.
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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