# Finding Half-Lives of a word problem

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• Mar 1st 2010, 07:16 PM
krzyrice
Finding Half-Lives of a substance
Find the half-lives of the substances:

1) Einsteinium-253, which delays at a rate of 3.406% per day.

2) A radioactive substance that decays at a continuous rate of 11% at minute.

I don't know where to start. For the first one is the formula something like this? Q = Q0(1.03406)^t, solve for t. But whenever I solve it the answer comes out to be negative.

Also the second problem I don't know how to set up the equation. Does it involve e? Q = Q0(e^0.11)^t or Q = Q0(e^0.89)^t and solve for t?

Please help me with these problems. Thank you so much.
• Mar 1st 2010, 08:36 PM
krzyrice
The answer to 2) is 6.301 minutes according to the back of the book. I still haven't figured out how to solve this.

Please any help would be greatly appreciated.
• Mar 1st 2010, 09:04 PM
pickslides
Quote:

Originally Posted by krzyrice
Find the half-lives of the substances:

1) Einsteinium-253, which delays at a rate of 3.406% per day.

The element is decaying therefore,

Lets say:

$Q = Q_0 \times (1-0.03406)^t$

We require the half life so

$\frac{Q_0 }{2}= Q_0 \times (1-0.03406)^t$

$\frac{1}{2}= \times 0.96594^t$

Now solve for $t$
• Mar 1st 2010, 09:16 PM
krzyrice
Thank you for your help pickslides. I got 20 days for that one. However when I apply the same concept to the second problem I get 5.948. This isn't correct because the answer in the back of the book says 6.301 minutes. How would I solve the second problem?
• Mar 1st 2010, 09:37 PM
pickslides
Quote:

Originally Posted by krzyrice
Thank you for your help pickslides. I got 20 days for that one. However when I apply the same concept to the second problem I get 5.948. This isn't correct because the answer in the back of the book says 6.301 minutes. How would I solve the second problem?

I am prepared to offer the following solution.

You have $Q = Q_0\times 0.89^t$ where you need $Q= Q_0 e^{-kt}$

Using this information

$Q_0\times 0.89^t = Q_0 e^{-kt}$

$0.89^t = e^{-kt}$

and choosing $t=1$

$0.89 = e^{-k} \implies k = -\ln(0.89) \approx 0.117$

So now you have

$Q= Q_0 e^{-0.117t}$

Now for a half life

$\frac{Q_0}{2}= Q_0 e^{-0.117t}$

$\frac{1}{2}= e^{-0.117t}$
• Mar 1st 2010, 10:22 PM
krzyrice
Please correct me if I'm wrong, but using the formula you gave me I got the answer 5.9243.

This is how I solved it,

0.5 = e^-0.117t
0.5 = 0.889585193t
log 0.5 = t log 0.889585193
t = 5.920586461

I have tried both log and ln and it came out the same way. Can you verify your work that it comes out to be 6.301?
• Mar 1st 2010, 10:29 PM
pickslides
Quote:

Originally Posted by krzyrice

0.5 = e^-0.117t
0.5 = 0.889585193t

That step is not correct

$\frac{1}{2}= e^{-0.117t}$

$\ln(0.5)= -0.117t$
• Mar 1st 2010, 10:31 PM
krzyrice
I have also tried this way:

0.5 = e^-0.117t
ln 0.5 = -0.117t
-0.69314718 = -0.117t
-0.69314718/-0.117 = t
t = 5.924334872
• Mar 1st 2010, 10:35 PM
pickslides
I have no more suggestions at this point.

If you are happy with the expanation shown you may be able to assume the answer provided in your book is incorrect.

Good luck.
• Mar 1st 2010, 11:54 PM
krzyrice
Thank you pickslides for guiding me through this problem. I think the answer you given me is right because if you plug in 6.301 into the formula it wouldn't come out to be 0.5.