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How would I solve the following problem?

A year ago there were 4 gram of a radioactive substance.Now there are 3 grams. How much was there 10 years ago?

I did the following

$4e^{k(1)}=3$ With four grams being my initial grams and this happened 1 year ago.

for my $k$ I got $k=\ln\frac{3}{4}$

Then I set up the equation

$$4e^{\ln\frac{3}{4}(t)}$$

But what should I use for my time or t I tired -9 because 1 year passed and you want to know how many grams you had 10 years ago.

And i go 53.271 grams but that seems wrong.

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4 Answers 4

up vote 2 down vote accepted

You are quite close, possibly off a bit only by a copying error or low percision: $$\left(\frac 43 \right)^{10}\cdot 3 = \left(\frac 43 \right)^{9}\cdot 4=\frac{1048576}{19683}\approx 53.27318$$

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The equation for radioactive decay is

$$A(t) = A(0)e^{kt}$$

where $A(t)$ is the amount after $t$ years, $A(0)$ is the initial amount, $k$ is the rate constant, and $t$ is the time in years.

which gives us $3=4e^{k}$ or $ln(0.75)=k=-0.2876$

Can you proceed now?

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You are doing well. $-9$ is correct because in your expression $t=0$ represents $1$ year ago, when there were $4$ grams. I get the same answer-why do you think it is wrong?

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I seemed like It decreased too much from 53 grams to 3 grams that why I thought it was wrong... –  Fernando Martinez Sep 4 '13 at 18:01
    
Nope, exponential decrease is quite fast. =) –  Patrick Da Silva Sep 4 '13 at 18:04
    
As $(\frac 34)^2=\frac 9{16} \approx \frac 12$ we have (almost) 5 half-lives. In 5 half-lives the quantity decreases by a factor $32$, so it would take $96$ grams at the start. This is not far off. –  Ross Millikan Sep 4 '13 at 18:04

If you're worried about the initial state because of using negative times, you could use the 10 years ago as the initial state:

$$y(t)=ce^{kt}$$

$$ce^{9k}=4$$ $$ce^{10k}=3$$

And find the value of the $c$, it will give you the same result: $c=53.27...$.

There's nothing wrong by the way with what you were doing.

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