• July 10th 2006, 08:28 AM
kwtolley
The amount of a radioactive tracer remaining after t days is given by A=Ao e^-0.058t, where Ao is the starting amount at the beginning of the time period. How many days will it take for one half of the original amount to decay? My answer is 12 days is that right. Thanks for looking at my answer.
• July 10th 2006, 08:34 AM
CaptainBlack
Quote:

Originally Posted by kwtolley
The amount of a radioactive tracer remaining after t days is given by A=Ao e^-0.058t, where Ao is the starting amount at the beginning of the time period. How many days will it take for one half of the original amount to decay? My answer is 12 days is that right. Thanks for looking at my answer.

RonL
• July 10th 2006, 09:08 AM
kwtolley
missing table
Looks like I'm missing a table I need to use with this question. I need to find half - life Bismuth isotope Ao. Then maybe I can get a better answer. Thanks for your help.
• July 10th 2006, 10:14 AM
topsquark
Quote:

Originally Posted by kwtolley
Looks like I'm missing a table I need to use with this question. I need to find half - life Bismuth isotope Ao. Then maybe I can get a better answer. Thanks for your help.

You don't need a value for A0. Here's a hint: What is the definition of the term "half life?"

-Dan
• July 10th 2006, 12:06 PM
Soroban
Hello, kwtolley!

You probably did it correctly . . . it's hard to see your work from here.

Quote:

The amount of a radioactive tracer remaining after t days is given by: $A = A_o e^{-0.058t}$
where $A_o$ is the starting amount at the beginning of the time period.
How many days will it take for one half of the original amount to decay?

The question is: When is: $A = \frac{1}{2}A_o$ ?

The equation is: . $A_oe^{-0.058t} \:=\:\frac{1}{2}A_o\quad\Rightarrow\quad e^{-0.058t}\:=\:\frac{1}{2}$

Take logs: . $\ln\left(e^{-0.058t}\right) \:= \:\ln(0.5)\quad\Rightarrow\quad (-0.058t)\ln e\:=\:\ln(0.5)$

Therefore: . $t \:= \:\frac{\ln(0.5)}{-0.058} \;=\;11.95081346\:\approx\;12$

• July 11th 2006, 08:02 AM
kwtolley
thank you
Thanks for checking my problem.