More about nuclear chemistry

The fearsome-looking (according to the fossils, of course) saber-tooth tiger has been extinct for quite a long time. A piece of 18.0-mg bone of this creature was found to contain 2.0 mg C-14. How old is this piece of bone? (The t1/2 of C-14 is 5730 years.)

First, you need to read this explanation, and work on the problem there.

Now, it should be easy!

The amount of C-14 in % is 2.0/18.0 = 0.11 (= 11%)
Since log(Mn/ M0) = (age/t1/2) ´ log0.5
Thus, "age" = log(0.11)/log0.5 ´ t1/2 = 18,250 years

<< Read a "bogus news" about saber-tooth tiger here, http://www.acclaimedmedia.com/voafa/bnn/825d.htm and relax a minute from your hard study!>>

Here is a challenging one:
The radio-active 23892U decays all the way to the stable nuclide 20682Pb after quite a few alpha and beta decays. Knowing that alpha decay emits 42He, and beta decay emits 01e, you can find out the number of alpha and beta decays for 23892U. What are they? (Answer here!)