# You have found a fossilized leg bone of some unknown mammal. Based on the size of the bone, you determine that it should have contained about 100 g of carbon-14 when the animal was alive. The bone now contains 12.5 g of carbon-14. How old is the bone?

##### 1 Answer

#### Answer:

#### Explanation:

Nuclear half-life is simply a measure of how much time must pass in order for a sample of a radioactive substance to decrease to **half** of its initial value.

Simply put, in one nuclear half-life, **half** of the atoms in the initial sample undergo *radioactive decay* and the other **half** do not.

Since the problem doesn't provide the nuclear half-life of carbon-14, you will have to do a quick search.

You'll find it listed as

#t_"1/2" = "5730 years"#

https://en.wikipedia.org/wiki/Carbon-14

So, what does that tell you?

An initial sample of carbon-14, **halved** with the passing of **every** half-life, which in your case is **left with**

#A_0 * 1/2 -># after the passing ofone half-life

#A_0/2 * 1/2 = A_0/4 -># after the passing oftwo half-lives

#A_0/4 * 1/2 = A_0/8 -># after the passing ofthree half-lives

#A_0/8 * 1/2 = A_0/16 -># after the passing offour half-lives

#vdots#

and so on.

You can thus say that **remains undecayed**, will be equal to

#color(blue)(A = A_0 * 1/2^n)" "# , where

**number of half-lives** that pass in a given period of time

So, you know that you start with

This means that you can say

#overbrace(12.5 color(red)(cancel(color(black)("g"))))^(color(orange)("remaining mass")) = overbrace(100.0 color(red)(cancel(color(black)("g"))))^(color(purple)("initial mass")) * 1/2^n#

Rearrange to get

#12.5/100.0 = 1/2^n#

#1/8 = 1/2^n implies 2^n = 8#

Since

#2^n = 2^3 implies n = 3#

So, **three half-lives** must pass in order for your sample of carbon-14 to be reduced from

#color(blue)(n = "period of time"/"half-life" = t/t_"1/2")#

you can say that

#t = n xx t_"1/2"#

In your case,

#t = 3 xx "5730 years" = color(green)("17,190 years")#

The answer *should* be rounded to three sig fig, but I'll leave it as-in, just for good measure.