The half life of iodine 131 is approximately 8 days. What is the decay rate of iodine 131?

Question

I'm studying for my final and I'm confused on this question.

The half life of iodine 131 is approximately 8 days. What is the decay rate of iodine 131?

This is the formula my teacher said to use.
A(t)= A0e^(kt). (That's suppose to be A sub 0) I know you have to plug 8 into t, but what steps do I take to get the decay rate? The Answer key shows the decay rate is -8.7%

Answer 1

Chase,

 

The equation your teacher has is quite valid, but I find it to be a bit more confusing than the one I am going to present and will then ultimately migrate to the form your teacher has given you.

 

What does it mean that the half life is 8 days.  Simply it means that due to the decay half of the material will statistically decay every eight days.  if you start with one kilo of iodine 131, eight days later you will have 1/2 kilo, in another 8 days you will have 1/4 kilo and in 8 more days 1/8 kilo, etc. etc. etc.

 

How do we represent this mathematically?

 

(1/2) ^ (t/8) or one-half raised to the power of t (in days) divided by 8 (the half life).  So in 8 days we have 1/2^(1) or 1/2.  In 16 days we have 1/2^2 or 1/4, ......

 

To use your teachers equation, we next to scale the constant k to reflect not only the 8 days but also the different base of e =2.718.

 

The scale factor to convert is 0.693, because e^(0.693) = 2.

 

So I used (1/2)^(t/8)=2^(-t/8)= e^(-.693t/8) = e^(-.0866)

 

To make this more formal A(t)=A0e^(-.0866), where A0 is the initial quantity of Iodine 131.

 

The -.086 in the -8.7% decay factor in your book.

Answer 2

Hi Chase,

 

OK so lets set up the equation

 

A(8) = e^8k

 

but we know the half life is 8 days so

 

.5 = e^8k

 

now when can take the natural log of both sides of the equation and solve for k

 

LN(.5) = LN(e^8k)

 

-.69315 = 8k         (LN e to any power is the power

 

k = - .08664

 

= - 8.7%

 

Hope this helps