# Carbon-14 has a half-life of 5770 years. If a fossil is 23,080 years old and it has 3 Kg of Carbon-14, how much carbon-14 did it originally have?

##### 1 Answer

#### Answer:

#### Explanation:

The **nuclear half-life** is simply the time needed for **half** of the atoms present in a sample of a radioactive substance to undergo radioactive decay.

In your case, carbon-14 is said to have a half-life of **years**, which means that the mass of *any* sample of carbon-14 will be **halved** with the passing of

Now, if **half-life from now** the mass of the sample will be **halved**, it follows that **half-life ago** it was **twice** its current value.

An interesting technique to use here is to *backtrack* from current time, when the sample is down to **years ago**.

You can thus say that **years before this point**, the mass of the sample was **twice** the current mass

#2 xx "3 kg" = "6 kg"#

*How about* **half-lives** ago?

#2 xx "5770 years" = "11,540 years ago"# The mass of the sample was

#2 xx "6 kg" = "12 kg"#

*How about* **half-lives** ago?

#3 xx "5770 years" = "17,310 years ago"# The mass of the sample was

#2 xx "12 kg" = "24 kg"#

*Finally, how about* **half-lives** ago?

#4 xx "5770 years" = "23,080 years ago"# The mass of the sample was

#2 xx "24 kg" = "48 kg"#

Therefore, **years ago**, the sample contained

When the values don't allow for such simple calculations, you can use the equation

#color(blue)(|bar(ul(color(white)(a/a)"A"_t = "A"_0 * 1/2^ncolor(white)(a/a)|)))#

Here

**remains** after a period of time

**initial mass** of the sample

**number of half-lives** that pass in the period of time

In your case, you have

#n = ("23,080" color(red)(cancel(color(black)("years"))))/(5770color(red)(cancel(color(black)("years")))) = 4#

You will thus have

#"A"_0 = "A"_t * 2^4#

#"A"_0 = "3 kg" * 2^4 = color(green)(|bar(ul(color(white)(a/a)color(black)("48 kg")color(white)(a/a)|)))#