Substance decrease by factor of 100? (radioactivity)

Click For Summary
SUMMARY

The discussion focuses on calculating the time required for radioactive iodine (131-I) to decrease by a factor of 100, given its half-life of 8 days. The correct formula to use is N(t) = N(o)e^(-kt), where k is derived from the half-life using k = ln(2)/t1/2. The final calculation shows that it takes approximately 53.1 days for the substance to reduce by this factor, confirming the relationship between N(t) and N(o) as well as the proper application of logarithmic functions in the solution.

PREREQUISITES
  • Understanding of radioactive decay and half-life concepts
  • Familiarity with the exponential decay formula N(t) = N(o)e^(-kt)
  • Knowledge of natural logarithms and their properties
  • Ability to manipulate equations involving logarithmic functions
NEXT STEPS
  • Study the derivation of the decay constant k for different half-lives
  • Learn about the applications of radioactive decay in medical imaging and treatment
  • Explore more complex decay problems involving multiple isotopes
  • Investigate the implications of radioactive decay in environmental science
USEFUL FOR

Students studying nuclear physics, chemistry, or environmental science, as well as professionals working in fields related to radioactivity and decay processes.

physics(L)10
Messages
101
Reaction score
0

Homework Statement


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?


Homework Equations


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


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 a lot for your help :)
 
Physics news on Phys.org
physics(L)10 said:
k=ln100/t1/2
Not true, you should look up this relation and use the correct one.

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).
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.
 
physics(L)10 said:
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).

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

Also, the value of k I'm pretty sure is right...Any confirmation on this?

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.
 
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
 
physics(L)10 said:
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

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

Similar threads

Replies
2
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 21 ·
Replies
21
Views
1K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
Replies
14
Views
3K