# Radioactive Decay and Linear Graphing

1. Jan 22, 2009

### nmacholl

1. The problem statement, all variables and given/known data
This particular exercise has no "problem statement" but I'll explain it in detail using all the information provided to me.

The class was presented with a table of data containing values for time (in days) and activity (in cts/sec) for a radioactive isotope Iodine-131. We are to create a Linear graph using the data and fine Iodine-131's half life.

Here's the data, Time in days first - then the activity in cts/sec.
05 | 6523
10 | 4191
15 | 2736
20 | 1722
25 | 1114
30 | 0722
35 | 0507
40 | 0315

2. Relevant equations
Here are the equations given to us on the table.
A = Ao$$^{-\lambda t}$$
$$\lambda$$=(ln2)/T$$_{1/2}$$

Ao=10,000 cts/sec

3. The attempt at a solution
I've fiddled with the equation and wasn't getting anywhere near a y=mx+b solution so I decided to look up first order exponential decay on the web which landed me this equation.

ln(A) = -$$\lambda$$*t + ln(Ao)

So I made a separate table using the natural log of the Activity data and graphed it which comes out linear with an R2=1, the best fit is: f(x)=-0.09x+9.2

I used the slope to determine T1/2 or half life of the sample. I calculated a value of 7.7 days - the accepted value is barely over 8 days. I believe that the equation I found for first order exponential decay is in fact correct for this case but I have no idea how to go from A = Ao$$^{-\lambda t}$$ to ln(A) = -$$\lambda$$*t + ln(Ao) algebraically.

In essence my question is: Is the ln(A) = -$$\lambda$$*t + ln(Ao) correct for this particular case of atomic decay and how do I get to this equation from the equation given.

Thanks alot!

2. Jan 23, 2009

### LowlyPion

I think the equations are equivalent.

Recall that Bx = (elnB)x = exlnB

When you take the ln of both sides of A = Ao-λt

keeping in mind that B above is your Ao

Then you get lnA = -λ*t + lnAo

3. Jan 23, 2009

### nmacholl

Thanks a lot but I have a little problem.
Okay so if:
A = Aox // x = -λ*t
A = ex*ln(Ao)

then

ln(A) = x*ln(Ao)
ln(A) = -λ*t*ln(Ao)

Why is my version -λ*t*ln(Ao) as opposed to the correct version -λ*t+ln(Ao)? Why is it "+ln(Ao)"

4. Jan 23, 2009

### LowlyPion

This step is incorrect. Exponents add, not multiply.

5. Jan 23, 2009

### nmacholl

I'm a little confused, which step is incorrect?

A power raised to a power is multiplication, which is how you get A = ex*ln(Ao) no?

Since ln() is the inverse of e it shouldn't change the exponent of e correct? ln(A) = x*ln(Ao)

6. Jan 23, 2009

### LowlyPion

I think you should familiarize yourself with ln arithmetic.

eln(x) = x

ln(ex) = x

ln(ea*b) = a + b

7. Jan 23, 2009

### nmacholl

If I substitute numbers in for a and b using my calculator I don't get a true statement.
a = 2
b = 5

ln(e^(a*b)) = a + b
ln(e^(10)) = 7
10 = 7

What am I doing wrong? :(

8. Jan 23, 2009

### LowlyPion

Don't despair. Maybe I'm not remembering my ln math.

Sorry if I confused you at all. I'll have to look it up.

9. Jan 23, 2009

### LowlyPion

OK. I don't know what I was thinking, because of course it's equal to the product.

I think however that your original equation is

A = Ao*e-λt

Taking the ln of both sides:

ln(A) =ln(Ao) -λt

That makes sense. My fault. I should have recognized it immediately.

10. Jan 23, 2009

### nmacholl

That certainly makes more sense than before! The worksheet must have a typo on the equation, I double checked and the handout and the -λt is the exponent of Ao and Euler's number is nowhere to be found.

Thanks alot.