# Substance decrease by factor of 100? (radioactivity)

1. Jul 19, 2010

### physics(L)10

1. The problem statement, all variables and given/known data
Radioactive iodine (131-I) has a half life of 8 days. How long does it take for iodine to be reduced by factor of 100?

2. Relevant equations
N(t)=N(o)e-kt
k=ln100/t1/2

3. The attempt at a solution

N(t)=N(o)e-kt

N(t)/N(o)=e-ln(100/8)t

Now I'm stuck. I would think to to take the ln of both sides, but there isn't any values given for N(t) or N(o). Also, the value of k I'm pretty sure is right...Any confirmation on this?

Thanks alot for your help :)

2. Jul 19, 2010

### Redbelly98

Staff Emeritus
Not true, you should look up this relation and use the correct one.

The phrase "to be reduced by factor of 100" suggests a relation between N(t) and N(o). It has nothing to do with the value of k.

3. Jul 19, 2010

### collinsmark

No, but you know the ratio of N(t)/N0 right? "...reduced by factor of 100.."

Sorry, something is not quite right for your k value. You know the half life is 8 days. In other words,

N(8 days)/N0 = 1/2 = e-k(8 days). Solve for k.

4. Jul 19, 2010

### physics(L)10

Ok I think I got it:

N(o)/100 = N(o)e-kt

1/100 = e-kt

ln(1/100) = e-kt

ln(1/100) = -[(ln2)/8](t)

53.1 days = t

5. Jul 19, 2010

### collinsmark

I think you forgot to express taking the natural log of the right side of the equation above in red (but it seems that you did later in the next step). (Any time you do something to one side of an equation, you must do the same thing to the other side, at the same time.) Anyway, your final answer is about the same as what I got (slightly different at the 4th significant digit).

6. Jul 20, 2010

### physics(L)10

Yes, I just forgot to put it in. Thanks for your help :)