# Radioactive Decay and Linear Graphing

• nmacholl
In summary: I was getting a little lost. In summary, The class was given a table of data containing values for time (in days) and activity (in cts/sec) for a radioactive isotope Iodine-131. They were to create a Linear graph using the data and find its half life. Using the slope of the graph, the class determined T1/2 or the time it would take for half of the Iodine-131 to decay. They found that the accepted value of 7.7 days was barely over the correct value of 8 days and so they used first order exponential decay to find the ln(A) equation. They substituted numbers in for a and b and found that the equation was not true
nmacholl

## 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 = Ao$$^{-\lambda t}$$
$$\lambda$$=(ln2)/T$$_{1/2}$$

Ao=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) = -$$\lambda$$*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!

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

LowlyPion said:
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

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)"

nmacholl said:
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)"

This step is incorrect. Exponents add, not multiply.

LowlyPion said:
This step is incorrect. Exponents add, not multiply.

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)

nmacholl said:
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)

I think you should familiarize yourself with ln arithmetic.

eln(x) = x

ln(ex) = x

ln(ea*b) = a + b

LowlyPion said:
I think you should familiarize yourself with ln arithmetic.

eln(x) = x

ln(ex) = x

ln(ea*b) = a + b

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? :(

nmacholl said:
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? :(

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.

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.

LowlyPion said:
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.

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.

## 1. What is radioactive decay?

Radioactive decay is the process by which an unstable atomic nucleus emits radiation in the form of particles or waves, resulting in the transformation of the nucleus into a more stable form.

## 2. How is radioactive decay measured?

Radioactive decay is measured using a unit called the half-life, which is the amount of time it takes for half of the radioactive substance to decay. This can be measured using a Geiger counter or other radiation-detecting instruments.

## 3. What is a linear graph?

A linear graph is a type of graph in which the relationship between two variables is represented by a straight line. It is commonly used to show the correlation between two quantitative variables.

## 4. How is radioactive decay represented on a linear graph?

The rate of radioactive decay can be represented on a linear graph by plotting the amount of radioactive substance remaining over time. The resulting line will have a negative slope, as the amount of substance decreases over time.

## 5. What can we learn from the linear graph of radioactive decay?

The linear graph of radioactive decay can provide insight into the half-life of a substance, the rate of decay, and the remaining amount of substance over time. It can also help to predict the future behavior of the substance and determine its stability.

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