Is Radioactive Decay Calculation Confusing?

[Sanosuke Sagara]

I have my doubt , solution and the question in the attachment that followed.
Thanks for anybody that spend some time on this question.

 

[learningphysics]

I have my doubt , solution and the question in the attachment that followed.
Thanks for anybody that spend some time on this question.

I think you made some careless errors.

Your answer for a is right.

I'd like to write out the equations...
\frac{-dn}{dt} = 5000(\frac{1}{2})^{t/8} where t is in hours

Just plug in t=1 into the right side and you get your answer for a. 4585

If we integrate both sides we get

n(t) = -\frac{5000}{ln(1/2)}(\frac{1}{2})^{t/8} or

n(t) = \frac{5000}{ln(2)} (\frac{1}{2})^{t/8}

Now since the time in the equations are hours... there's no need to convert to seconds.

n(0) = 5000/ln(2) = 7213
n(1) = (5000/ln(2)) (1/2)^(1/8) = 6615

So for b) n(0) - n(1) = 598

c) She comes back 24 hours later. So in the equation we'll need to plug in t=25 (24 hours after t=1) to see how much of the radioactive stuff is left.
n(25) = (5000/ln(2)) (1/2)^(25/8) = 827

To get the same treatment she needs to get 598 atoms just like part b). So the total needs to come down from 827 to 827-598=229 after the second treatment is finished

We solve n(t) = 229, I get t=40 hours.

So the second treatment starts at t=25 hours, and finished t=40 hours to get 15 hours of treatment.

Check the work yourself. I might have made some careless errors.

 

[Sanosuke Sagara]

Thanks for your explanation and comment to me for my unneccasary confusing way of calculation.I think I have understand what the question want and I will try to find out the answer.Thanks again,learningphysics.