Isotope Half-Life and Sample Size Calculation

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The discussion focuses on calculating the initial size of a sample of isotope 239Pu, which has a half-life of 24,100 years, after it has reduced by 1.6 grams over 10,000 years. The user attempts to solve the equation A = Ao * e^(-rt) but struggles with algebraic manipulation, particularly taking the natural logarithm of a negative number. Participants emphasize the importance of correctly applying the half-life concept and suggest re-evaluating the algebra to avoid confusion. They highlight that the equation should incorporate the half-life factor and clarify the steps needed to isolate the initial mass. Accurate calculations are essential for determining the sample size after specified time intervals.
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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
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

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