• Nov 11th 2010, 03:26 PM
Quantity of radioactive material after 20 years decayed to 50 grams, after 40 years it decayed to 20 grams.What was the amount of radioactive material to start with.

In problems like this usually we were finding half life or time but never initial amount.

N=No e^kt

I know that I have to find N at t=0. My problem is solving for k from the above equation.

Thank you.
• Nov 11th 2010, 03:30 PM
pickslides
For $\displaystyle N=N_0 e^{kt}$

$\displaystyle (t,N) =(20,50),(40,20)$ sub them in to find $\displaystyle N_0,k$
• Nov 11th 2010, 03:43 PM
I understand that.
50=Noe^k(20)

How to actually solve this.
• Nov 11th 2010, 05:14 PM
RiseAgainstMe
Simple answer, you don't solve that equation by itself. You need to use all of the data given, ie. $\displaystyle 20 = N_0 e^{40k}$
• Nov 11th 2010, 05:31 PM
pickslides
$\displaystyle 50=N_0 e^{20k}$ ...(1)

$\displaystyle 20=N_0 e^{40k}$ ...(2)

Can you proceed?
• Nov 11th 2010, 06:08 PM
No not really.
I don't understand what to do with equations.
• Nov 11th 2010, 06:26 PM
RiseAgainstMe
You have two equations and two unknowns, you need to eliminate one of them so that you can find the other. If you were going to do this which unknown factor do you think would be the easiest to eliminate?
• Nov 11th 2010, 06:30 PM
pickslides
$\displaystyle 50=N_0 e^{20k}$ ...(1) $\displaystyle \implies \frac{50}{ e^{20k}}= N_0$

$\displaystyle 20=N_0 e^{40k}$ ...(2) $\displaystyle \implies \frac{20}{ e^{40k}}= N_0$

equating left hand sides $\displaystyle \implies \frac{20}{ e^{40k}}=\frac{50}{ e^{20k}}$

Then

$\displaystyle 20e^{20k} = 50e^{40k}$

$\displaystyle \frac{20}{50} = \frac{e^{40k}}{e^{20k}}$

$\displaystyle \frac{2}{5} = e^{40k-20k}$

$\displaystyle \frac{2}{5} = e^{20k}$

$\displaystyle \ln \frac{2}{5} = 20k$

$\displaystyle \frac{1}{20}\ln \frac{2}{5} = k$

Now can you find $\displaystyle N_0$ ?
• Nov 11th 2010, 06:46 PM
So what is left to do is take k and plug it into one of the No equations?
50/e^20(-.0458)=No
• Nov 11th 2010, 06:51 PM
pickslides
Yep.
• Nov 11th 2010, 07:57 PM
CaptainBlack
Quote:

Originally Posted by pickslides
$\displaystyle 50=N_0 e^{20k}$ ...(1)

$\displaystyle 20=N_0 e^{40k}$ ...(2)

Can you proceed?

Quote:

You take natural logs of both equations which reduces them to linear equations in $\displaystyle L=\ln(N_0)$ and $\displaystyle$$k$. These you solve in the usual manner and then $\displaystyle N_0=\exp(L)$