1. ## Another Growth/Decay Question

Please let me know if I am on the right track or way off tangent!

Plutonium 241: A(t)=Aoe^-.053t

Q: The Half Life of Plutonium 241 is approximately 13 years.

A: How much of a sample weighing 4g will remain after 100 years?

B: How much time is necessary for a sample weighing 4g to decay to .1g?

So for question A, this is what I thought:

1/2=4e^-.053 (100)

On the left hand side, I used 1/2 to represent the half life and on the RS, I plugged in 4 for Ao which is my initial amount? For T, I used 100 years for time. This is where I think I goofed up. I don't think I set up the equation correctly.

TIA

2. Originally Posted by mvho
Please let me know if I am on the right track or way off tangent!

Plutonium 241: A(t)=Aoe^-.053t

Q: The Half Life of Plutonium 241 is approximately 13 years.

A: How much of a sample weighing 4g will remain after 100 years?

B: How much time is necessary for a sample weighing 4g to decay to .1g?

So for question A, this is what I thought:

1/2=4e^-.053 (100)

On the left hand side, I used 1/2 to represent the half life and on the RS, I plugged in 4 for Ao which is my initial amount? For T, I used 100 years for time. This is where I think I goofed up. I don't think I set up the equation correctly.

TIA
(a) $A(100) = 4e^{-0.053 \cdot 100}$

$A(100) = 0.019$ grams

(b) $0.1 = 4e^{-0.053t}$

solve for t using logarithms

3. Thanks a lot ( How is it that everyone is able to type in the exponents?). To clarify, I would only set up the left hand side as 1/2 if they were looking for the half life correct?

And would you mind, breaking down the Right hand side after taking the natural ln?

Would it be:

-.053*100 ln4e?

4. Originally Posted by mvho
Thanks a lot ( How is it that everyone is able to type in the exponents?). To clarify, I would only set up the left hand side as 1/2 if they were looking for the half life correct?

And would you mind, breaking down the Right hand side after taking the natural ln?

Would it be:

-.053*100 ln4e?
Yes, the LHS is only 0.5 when you want to measure the time it will take for a substance to decay to half it's original amount which is the definition of half-life.

A) Skeeter solved this one directly

B) Set
• $A(t) = 0.1$
• $A_0 =$

As $A(t) = A_0e^{-0.053t}$

Divide by $A_0$: $\frac{A}{A_0} = e^-{0.053t}$

Take the log of both sides: $ln(\frac{A}{A_0}) = -0.053t$

Divide through by -0.053: $t = -\frac{1}{0.053} \times ln(\frac{A}{A_0}) = -\frac{ln(A) - ln(A_0)}{0.053}$

Put in your values for A and A_0

5. Thanks everyone.