# Question #d4494

##### 1 Answer

Here's what I got.

#### Explanation:

The thing to remember about an isotope's **nuclear half-life**, **half** of an initial sample of that isotope to undergo radioactive decay.

In other words, the half-life tells you how much must pass in order for your sample to be **halved**.

If you take **remains undecayed** after a time **initial amount** of that isotope, you can say that you will have

#A_t = A_0 * 1/2 = A_0/2 = A_0/2^color(red)(1) -># after#color(red)(1)# half-life#A_t = A_0/2 * 1/2 = A_0/4 = A_0/2^color(red)(2) -># after#color(red)(2)# half-lives#A_t = A_0/4 * 1/2 = A_0/8 = A_0/2^color(red)(3) -># after#color(red)(3)# half-lives

#vdots#

and so on. So with **every passing half-life**, you get to divide the *initial amount* by **half-lives** pass in a given period of tiem **times**.

#A_t =A_0/(underbrace(2 * 2 * ... * 2)_(color(black)(color(red)(n)color(white)(.)"times"))) = A_0/2^color(red)(n)#

This is equivalent to

#A_t = A_0 * (1/2)^color(red)(n)#

with

#color(red)(n) = t/t_"1/2"#

In your case, you start with

#25 color(red)(cancel(color(black)("g"))) = 200color(red)(cancel(color(black)("g"))) * (1/2)^color(red)(n)#

Divide both sides by

#25/200 = 1/2^color(red)(n)#

This is equivalent to

#1/4 = 1/2^color(red)(n)#

#1/2^2 = 1/2^color(red)(n) implies color(red)(n) = 2#

So you can say that in order for your sample to be reduced from **two half-lives** must pass.

This means that you have

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

#color(darkgreen)(ul(color(black)(t = 2 * "8.4 days" = "16.8 days")))#

I'll leave the answer rounded to three **sig figs**, but a more accurate answer would be

#t = "20 days"#

because you have only one significant figure for the initial mass of the sample.