• Oct 21st 2008, 07:57 PM
GreenandGold
A sample of a radioactive substance decayed to 91.5% of its original amount after a year. What is the half life of the substance? Also have to find the amount of time for it to decay to 30% of the original amount.
• Oct 21st 2008, 08:05 PM
Chris L T521
Quote:

Originally Posted by GreenandGold
A sample of a radioactive substance decayed to 91.5% of its original amount after a year. What is the half life of the substance? Also have to find the amount of time for it to decay to 30% of the original amount.

Radioactivity can be defined using this model:

$\displaystyle N=N_0e^{-kt}$, where

$\displaystyle N$ is the current amount of the substance
$\displaystyle N_0$ is the initial amount of the substance
$\displaystyle k$ is the decay constant
$\displaystyle t$ is the length of time

Now, we are told that we have $\displaystyle .915N_0$ after 1 year.

Thus, $\displaystyle .915N_0=N_0e^{-k}$

Solving for k, we see that $\displaystyle k=-\ln(.915)\approx 0.0888$

Now, we can find the half life:

Half life is defined as $\displaystyle \lambda=\frac{\ln 2}{k}$, where

$\displaystyle \lambda$ is the half life
$\displaystyle k$ is the decay constant

Thus, $\displaystyle \lambda=\frac{\ln 2}{.0888}\approx \color{red}\boxed{7.803\text{ years}}$

Now, our decay model is $\displaystyle N=N_0e^{-0.0888t}$

Can you find $\displaystyle t$ when $\displaystyle N=.3N_0$?

--Chris
• Oct 21st 2008, 08:19 PM
GreenandGold
take the natural log of both sides?
• Oct 21st 2008, 08:20 PM
Chris L T521
Quote:

Originally Posted by GreenandGold
take the natural log of both sides?

--Chris
• Oct 21st 2008, 08:28 PM
GreenandGold
I got -13 but i did something wrong. is it 7.803(ln(.3)/ln(2))
• Oct 21st 2008, 08:33 PM
GreenandGold
ok i added a negative to the decay since its negative. so i got 1.35535
• Oct 21st 2008, 08:49 PM
Chris L T521
Quote:

Originally Posted by GreenandGold
ok i added a negative to the decay since its negative. so i got 1.35535

Hmmm...

$\displaystyle N=N_0e^{-0.0888t}\implies.3N_0=N_0e^{-0.0888t}$ $\displaystyle \implies \ln(.3)=-0.0888t\implies t=-\frac{\ln(.3)}{0.0888}\approx\color{red}\boxed{13. 558\text{ years}}$

--Chris
• Oct 21st 2008, 08:57 PM
GreenandGold
ok i see now!!! thanks...