How Long Until 24Na Decays from 5μg to 1μg?

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Homework Help Overview

The discussion revolves around the decay of a 5μg sample of 24Na to 1μg, utilizing its half-life of 234 hours. Participants are exploring the calculations related to radioactive decay and the application of the decay constant.

Discussion Character

  • Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of the radioactive decay formula, N = N0e-λt, and how to manipulate it to find the time required for the sample to decay to the desired amount. There are questions about the correctness of the calculations and the interpretation of the decay constant.

Discussion Status

There is an ongoing exploration of the mathematical approach to the problem, with one participant presenting their calculations and seeking validation. Another participant confirms the correctness of the final answer while providing a slightly different result, indicating a productive exchange of ideas.

Contextual Notes

Participants are working under the constraints of homework rules, focusing on understanding the decay process without providing complete solutions. The discussion reflects varying interpretations of the calculations involved.

kkid
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TA sample of 24Na has a half-life of 234 hours, How much time elapses before a 5μg sample contains 1μg of undecayed atoms?



I have calculate the decay constant (8.23x10-7)


What do I do now?
 
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kkid said:
TA sample of 24Na has a half-life of 234 hours, How much time elapses before a 5μg sample contains 1μg of undecayed atoms?

I have calculate the decay constant (8.23x10-7)

What do I do now?
Hello kkid. Welcome to PF !

What's the formula for radioactive decay?
 


I eventually did this by altering this equation:

N = N0e-λt

if one divides by N0 then the left hand side is just a faction of the start and end amounts (whether this is in terms of amount of atoms or in terms of mass it will be the same.

In this case the left hand side of the equation (the fraction) will be 1/5 as 1 is 1/5 of 5.


This gives me 1/5 = e-λt

taking ln of both sides to eliminate e gives:

ln(1/5) = -λt

t = ln(1/5)/-λ


I substitute the value of decay constant (λ) to get my overall answer of 543.241 hours


Is this correct?
(It took me about 3 hours to get this)
 


kkid said:
I eventually did this by altering this equation:

N = N0e-λt

if one divides by N0 then the left hand side is just a faction of the start and end amounts (whether this is in terms of amount of atoms or in terms of mass it will be the same.

In this case the left hand side of the equation (the fraction) will be 1/5 as 1 is 1/5 of 5.


This gives me 1/5 = e-λt

taking ln of both sides to eliminate e gives:

ln(1/5) = -λt

t = ln(1/5)/-λ

I substitute the value of decay constant (λ) to get my overall answer of 543.241 hours

Is this correct?
(It took me about 3 hours to get this)
Yes, that's correct.

Without rounding until the final answer, I get t1/5 = 543.331 hours .
 

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