# Question #d5cfe

##### 1 Answer

#### Answer:

#### Explanation:

Your go-to equation here looks like this

#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#

Now, you know that **undecayed** after a period fo time

In such cases, you can express

#A_t = 5/100 * A_0#

This means that the amount left after a period of time#t# is equal to#5% = 5/100# of the initial amount

Plug this into the above equation to get

#5/100 * color(red)(cancel(color(black)(A_0))) = color(red)(cancel(color(black)(A_0))) * (1/2)^n#

Now, you can rewrite this as

#ln(5/100) = ln[(1/2)^n]#

This is done by taking thenatural logof both sides of the equation

This will be equivalent to

#ln(5/100) = n * ln(1/2)#

which gets you

#n = ln(5/100)/ln(1/2) = 4.32#

So, you know that in order for **undecayed**, i.e. in order for *decay*, a period of time equal to **half-lives** must pass.

Since you have

#n = "total time that passes"/"half life" = t/t_"1/2"#

you can say that

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

In your case, this will get you

#color(darkgreen)(ul(color(black)(t = 4.32 * "3 billion years" = "13 billion years")))#

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