Substance decrease by factor of 100? (radioactivity)

AI Thread Summary
The discussion revolves around calculating the time it takes for radioactive iodine (131-I) to decrease by a factor of 100, given its half-life of 8 days. The key equation used is N(t) = N(o)e^(-kt), where k is derived from the half-life. Participants clarify that the ratio N(t)/N(o) can be used to express the reduction, leading to the equation 1/100 = e^(-kt). After some calculations, the time required for the reduction is determined to be approximately 53.1 days. The conversation emphasizes the importance of correctly applying logarithmic functions in the calculations.
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 :)
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top