Radioactive decay

**ado061** · November 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.
Radioactive decay is given by

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.

**pickslides** · November 11th 2010, 03:30 PM

For $N=N_0 e^{kt}$

$(t,N) =(20,50),(40,20)$ sub them in to find $N_0,k$

**ado061** · November 11th 2010, 03:43 PM

I understand that.
50=Noe^k(20)

How to actually solve this.

**RiseAgainstMe** · November 11th 2010, 05:14 PM

Simple answer, you don't solve that equation by itself. You need to use all of the data given, ie. $20 = N_0 e^{40k}$

**pickslides** · November 11th 2010, 05:31 PM

$50=N_0 e^{20k}$ ...(1)

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

Can you proceed?

**ado061** · November 11th 2010, 06:08 PM

No not really.
I don't understand what to do with equations.

**RiseAgainstMe** · November 11th 2010, 06:26 PM

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?

**pickslides** · November 11th 2010, 06:30 PM

$50=N_0 e^{20k}$ ...(1) $\implies \frac{50}{ e^{20k}}= N_0$

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

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

Then

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

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

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

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

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

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

Now can you find $N_0$ ?

**ado061** · November 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

**pickslides** · November 11th 2010, 06:51 PM

Yep.

**CaptainBlack** · November 11th 2010, 07:57 PM

Originally Posted by pickslides

$50=N_0 e^{20k}$ ...(1)

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

Can you proceed?

Originally Posted by ado061

No not really.
I don't understand what to do with equations.

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

CB

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