# Question #116c1

##### 1 Answer

#### Explanation:

Your tool of choice here will be the equation

#color(blue)(ul(color(black)(A_t = A_0 * (1/2)^n)))#

Here

#A_t# is the amount of the radioactive substance that remainsundecayedafter a time period#t# #A_0# is theinitial amountof said substance#n# is thenumber of half-livesthat pass in the time period#t#

In your case, you know that the relic contains

#A_t = 27.3/100 * A_0#

The equation becomes

#27.3/100 * color(red)(cancel(color(black)(A_0))) = color(red)(cancel(color(black)(A_0))) * (1/2)^n#

#27.3/100 = (1/2)^n#

This can be rewritten as

#ln(27.3/100) = ln[(1/2)^n]#

#ln(27.3/100) = n * ln(1/2)#

which gets you

#n = ln(27.3/100)/ln(1/2) = 1.873#

Before moving on, take a second to check if this value makes sense.

As you know, the **half-life** of a radioactive nuclide tells you the time needed for **half** of an initial mass of said nuclide to decay.

In other words, the mass of the nuclide will be **halved** with *every passing half-life*. This implies that you have

#A_0 * 1/2 = A_0/2 = 50/100 * A_0 = 50% A_0 -># afterone half-life#A_0/2 * 1/2 = A_0/4 = 25/100 * A_0 = 25% A_0 -># aftertwo half-lives

In your case, you are left with *very close* to **half-lives**, i.e. the number of half-lives that passed is

So the fact that **half-lives** passed confirms that you are on the right track.

Now, you can say that

#n = "the total time that passed"/"the half-life" = t/t_"1/2"#

This means that you will have

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

which will get you

#t = 1.873 * "3000 years" = color(darkgreen)(ul(color(black)("5620 years")))#

I'll leave the answer rounded to three **sig figs**, but keep in mind that you only have one significant figure for the half-life of the radioactive substance.