Isotope Half-Life and Sample Size Calculation

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Homework Help Overview

The problem involves the isotope 239Pu, which has a specified half-life, and requires calculating the initial size of a sample after a certain period and its size after a shorter duration. The context is centered around radioactive decay and the mathematical relationships governing it.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the algebraic manipulation of the decay equation and express confusion regarding negative values arising from the calculations. There are questions about the interpretation of variables and the correct application of the half-life concept.

Discussion Status

The discussion is ongoing, with participants seeking clarification on algebraic steps and the proper use of the decay formula. Some guidance has been offered regarding the structure of the equation and the importance of correctly handling the variables involved.

Contextual Notes

There appears to be uncertainty regarding the definitions of certain variables and the implications of the half-life on the calculations. Participants are encouraged to clarify their algebraic steps to resolve misunderstandings.

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Homework Statement



The isotope 239Pu has a half-life of 24,100 years. After 10,000 years, a sample of the isotope is reduced 1.6 grams. What was the intial size of the same (in grams)? How large was the sample after the first 1,000 years. Round your answer to four decimal places.

Homework Equations



A=Aoert

The Attempt at a Solution



Intially, for the value of A, I subbed in "Ao-1.6" and then did the following:

Ao-1.6=Aoer10,000

Divided each side by Ao, and then take the Ln of both sides.

That's where I get stuck. My calculator is indicating that it's not possible to take the Ln of a negative number.
 
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How did you get a negative number after your algebraic manipulation?
 
SteamKing said:
How did you get a negative number after your algebraic manipulation?

The final isotope is the initial mass - 1.6.

When you divide by Ao, the negative stays there, right?
 
AryRezvani said:
The final isotope is the initial mass - 1.6.

When you divide by Ao, the negative stays there, right?

You'd better show your algebra, exactly. You seem to be making an elementary mistake.
 
It's not clear (to me) what you're solving for. What is r? Since you're given a half-life, shouldn't there be a power of two involved? And a negative power of two at that.
 
You might find it easier to solve for the exponential part, THEN just gather the A0 together on one side (and yes, don't forget that the equation should be A0-1.6 = A0e-rt/ln2
 

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