Substance decrease by factor of 100? (radioactivity)

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Homework Help Overview

The discussion revolves around the decay of radioactive iodine (131-I) and the calculation of the time required for its quantity to decrease by a factor of 100, given its half-life of 8 days.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between the initial and final quantities of iodine, questioning the correct formulation of the decay constant and the implications of the half-life. There is discussion about taking the natural logarithm of both sides of the equation and the significance of the ratio of N(t) to N(o).

Discussion Status

Some participants have provided guidance on the formulation of the decay equation and the calculation of the decay constant. There are multiple interpretations being explored regarding the setup of the problem and the application of logarithmic properties. The conversation reflects a collaborative effort to clarify the mathematical relationships involved.

Contextual Notes

Participants note the absence of specific values for N(t) and N(o) and emphasize the importance of understanding the ratio in the context of the problem. There is an acknowledgment of potential errors in the initial assumptions about the decay constant.

physics(L)10
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Homework Statement


Radioactive iodine (131-I) has a half life of 8 days. How long does it take for iodine to be reduced by factor of 100?


Homework Equations


N(t)=N(o)e-kt
k=ln100/t1/2


The Attempt at a Solution



N(t)=N(o)e-kt

N(t)/N(o)=e-ln(100/8)t

Now I'm stuck. I would think to to take the ln of both sides, but there isn't any values given for N(t) or N(o). Also, the value of k I'm pretty sure is right...Any confirmation on this?

Thanks a lot for your help :)
 
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physics(L)10 said:
k=ln100/t1/2
Not true, you should look up this relation and use the correct one.

Now I'm stuck. I would think to to take the ln of both sides, but there isn't any values given for N(t) or N(o).
The phrase "to be reduced by factor of 100" suggests a relation between N(t) and N(o). It has nothing to do with the value of k.
 
physics(L)10 said:
Now I'm stuck. I would think to to take the ln of both sides, but there isn't any values given for N(t) or N(o).

No, but you know the ratio of N(t)/N0 right? "...reduced by factor of 100.." :wink:

Also, the value of k I'm pretty sure is right...Any confirmation on this?

Sorry, something is not quite right for your k value. You know the half life is 8 days. In other words,

N(8 days)/N0 = 1/2 = e-k(8 days). Solve for k.
 
Ok I think I got it:

N(o)/100 = N(o)e-kt

1/100 = e-kt

ln(1/100) = e-kt

ln(1/100) = -[(ln2)/8](t)

53.1 days = t
 
physics(L)10 said:
Ok I think I got it:

N(o)/100 = N(o)e-kt

1/100 = e-kt

ln(1/100) = e-kt

ln(1/100) = -[(ln2)/8](t)

53.1 days = t

I think you forgot to express taking the natural log of the right side of the equation above in red (but it seems that you did later in the next step). (Any time you do something to one side of an equation, you must do the same thing to the other side, at the same time.) Anyway, your final answer is about the same as what I got (slightly different at the 4th significant digit). :smile:
 
Yes, I just forgot to put it in. Thanks for your help :)
 

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