- #1

nmacholl

- 26

- 0

## Homework Statement

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

## Homework Equations

Here are the equations given to us on the table.

A = A

_{o}[tex]^{-\lambda t}[/tex]

[tex]\lambda[/tex]=(ln2)/T[tex]_{1/2}[/tex]

A

_{o}=10,000 cts/sec

## 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) = -[tex]\lambda[/tex]*t + ln(A

_{o})

So I made a separate table using the natural log of the Activity data and graphed it which comes out linear with an R

^{2}=1, the best fit is: f(x)=-0.09x+9.2

I used the slope to determine T

_{1/2}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 = A

_{o}[tex]^{-\lambda t}[/tex] to ln(A) = -[tex]\lambda[/tex]*t + ln(A

_{o}) algebraically.

In essence my question is: Is the ln(A) = -[tex]\lambda[/tex]*t + ln(A

_{o}) correct for this particular case of atomic decay and how do I get to this equation from the equation given.

Thanks alot!