Question

You have just isolated a new radioactive element. If you can determine its half-life, you will win the Nobel Prize in physic. You purify a sample of 2 grams.One of you colleges steals half of it, and 3 days later you find that 0.1 grams of the radioactive material is still left. What is the half life?

Heres what I did but I don't think its right I am very confused on how to set this up to solve. Please if this is not right help me set this up properly. Thank you.

R(t)=Ae^-kt
0.1=2e^-k3
0.1/2=e^-k3
ln 0.1/2=-k3
ln0.1/2/3=-k
-k=0.998577425/-1
k=0.998577425

Answer 1

That's a lot of work. Just think about it a bit. You start with 1 gram. 3 days later the amount left is 0.1 gram

If the half life is k days, then

(1/2)^3k = 1/10
3k = log(.1)/log(.5) = 3.32
So, k = 1.11
R(t) = (1/2)^(1.11t)
But, since we like things base e for our calculators, that is

e^((-ln2)*1.11)t) = e^-.77t

check: e^(-.77*3) = 0.1

You started off wrong, because you forgot that half of it was stolen. Your work is right, but you should have A=1, not A=2.