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 t_{1/2} 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(M_{n}/ M_{0}) = (age/t_{1/2}) ´
log0.5

Thus, "age" = log(0.11)/log0.5 ´ t_{1/2}
= 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 ^{238}_{92}U decays all the way to
the stable nuclide ^{206}_{82}Pb after quite a few alpha
and beta decays. Knowing that alpha decay emits ^{4}_{2}He,
and beta decay emits ^{0}_{–1}e, you can find out the number
of alpha and beta decays for ^{238}_{92}U. What are they?
(Answer here!)