# The half-life of a certain substance is 3.6 days. How long will it take for 20 g of the substance to decay to 7 g?

##### 1 Answer

#### Explanation:

You know that the **half-life** of a radioactive nuclide, **half** of the atoms present in a sample to undergo radioactive decay.

In this case, it takes **days** for *any* sample of this substance to decay to **half** of its initial mass.

More specifically, you will have

#"20 g" * 1/2 = "10 g " -># after,one half-life#1 xx "3.6 days"#

#"10 g" * 1/2 = "5 g "-># after,two half-lives#2xx "3.6 days"#

#"5 g" * 1/2 = "2.5 g " -># after,three half-lives#3 xx "3.6 days"#

#vdots#

and so on. In your case, you're interested in finding out how much time must pass for the initial **more than one** half-life and **less than two** half-lives, so

#overbrace(" 3.6 days ")^(color(purple)("one half-life")) < t < overbrace(" 7.2 days ")^(color(red)("two half-lives"))#

Since the numbers don't allow for a quick calculation, your tool of choice will be the equation

#color(blue)(|bar(ul(color(white)(a/a)A_"t" = A_0 * 1/2^ncolor(white)(a/a)|)))#

Here

**remains undecayed** after a period of time

**initial mass** of the sample

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

Rearrange the above equation to solve for

#2^n = A_0/A_"t"#

This will be equivalent to

#ln(2^n) = ln(A_0/A_"t")#

#n * ln(2) = ln(A_0/A_"t") implies n = ln(A_0/A_"t")/ln(2)#

Plug in your values to find

#n = ln( (20 color(red)(cancel(color(black)("g"))))/(7color(red)(cancel(color(black)("g"))))) * 1/ln(2) = 1.5146#

Since

#t = 1.5146 * "3.6 days" = color(green)(|bar(ul(color(white)(a/a)color(black)("5.5 days")color(white)(a/a)|)))#

I'll leave the answer rounded to two **sig figs**, but keep in mind that you only have one sig figs for the initial mass of the sample and for the mass that remains undecayed.